How do I read a variable from a file?
I want to insert in my script a value (string) that I would read from a text file.
For example, instead of:
echo "Enter your name"
read name
I want to read a string from another text file so the interpreter should read the string from the file and not the user input.
bash scripts
add a comment |
I want to insert in my script a value (string) that I would read from a text file.
For example, instead of:
echo "Enter your name"
read name
I want to read a string from another text file so the interpreter should read the string from the file and not the user input.
bash scripts
1
What does your text file look like? Is the entire file 1 variable or are theyKEY=VALUE
pairs? The solutions are quite different (if it's the latter, Takkat's answer applies, the former Radu's)
– kiri
Oct 28 '13 at 11:37
1
@CiroSantilli it is not crossposted if it was posted by a different user, there is no such thing as a "cross-site duplicate". The only thing that should be avoided is the same question asked by the same user on different sites.
– terdon♦
Mar 24 '14 at 17:52
add a comment |
I want to insert in my script a value (string) that I would read from a text file.
For example, instead of:
echo "Enter your name"
read name
I want to read a string from another text file so the interpreter should read the string from the file and not the user input.
bash scripts
I want to insert in my script a value (string) that I would read from a text file.
For example, instead of:
echo "Enter your name"
read name
I want to read a string from another text file so the interpreter should read the string from the file and not the user input.
bash scripts
bash scripts
edited Oct 28 '13 at 10:03
Braiam
51.6k20136220
51.6k20136220
asked Oct 28 '13 at 8:35
user208413user208413
276133
276133
1
What does your text file look like? Is the entire file 1 variable or are theyKEY=VALUE
pairs? The solutions are quite different (if it's the latter, Takkat's answer applies, the former Radu's)
– kiri
Oct 28 '13 at 11:37
1
@CiroSantilli it is not crossposted if it was posted by a different user, there is no such thing as a "cross-site duplicate". The only thing that should be avoided is the same question asked by the same user on different sites.
– terdon♦
Mar 24 '14 at 17:52
add a comment |
1
What does your text file look like? Is the entire file 1 variable or are theyKEY=VALUE
pairs? The solutions are quite different (if it's the latter, Takkat's answer applies, the former Radu's)
– kiri
Oct 28 '13 at 11:37
1
@CiroSantilli it is not crossposted if it was posted by a different user, there is no such thing as a "cross-site duplicate". The only thing that should be avoided is the same question asked by the same user on different sites.
– terdon♦
Mar 24 '14 at 17:52
1
1
What does your text file look like? Is the entire file 1 variable or are they
KEY=VALUE
pairs? The solutions are quite different (if it's the latter, Takkat's answer applies, the former Radu's)– kiri
Oct 28 '13 at 11:37
What does your text file look like? Is the entire file 1 variable or are they
KEY=VALUE
pairs? The solutions are quite different (if it's the latter, Takkat's answer applies, the former Radu's)– kiri
Oct 28 '13 at 11:37
1
1
@CiroSantilli it is not crossposted if it was posted by a different user, there is no such thing as a "cross-site duplicate". The only thing that should be avoided is the same question asked by the same user on different sites.
– terdon♦
Mar 24 '14 at 17:52
@CiroSantilli it is not crossposted if it was posted by a different user, there is no such thing as a "cross-site duplicate". The only thing that should be avoided is the same question asked by the same user on different sites.
– terdon♦
Mar 24 '14 at 17:52
add a comment |
9 Answers
9
active
oldest
votes
To read variables from a file we can use the source
or .
command.
Lets assume the file contains the following line
MYVARIABLE="Any string"
we can then import this variable using
#!/bin/bash
source <filename>
echo $MYVARIABLE
15
One reason why you might not want to do it this way is because whatever is in<filename>
will be run in your shell, such asrm -rf /
, which could be very dangerous.
– Mark Ribau
Jan 10 '15 at 2:57
3
Late to the party, but, no, it doesn't deserve to be marked up OR as the answer. This sources another bash source file which is NOT what the OP asked for. He asked how to read the contents of a FILE into a variable. NOT how to execute another script which sets a variable. The correct answer is the one below. name=$(cat "$file") .
– RichieHH
Nov 24 '15 at 12:54
1
OTOH, it is the answer to the question I asked google and google brought me here. Weird times, eh?
– Michael Terry
Jul 10 '17 at 18:04
1
Great also works for multiple variables! Point of @MarkRibau is understood but I guess that if you right the file in a different script and just have to pass it over its fine and controlled.
– madlymad
Jul 19 '17 at 11:25
1
You can usesed
to addlocal
keyword and make the script a bit safer and not polute your global scope. So inside a a function docat global_variables.sh | sed -e 's/^/local /' > local_variables.sh
and then use. ./local_variables.sh
. Whatever you import in the function will only be available in that function. Note that it assumesglobal_variables.sh
only contains one variable per line.
– Nux
Jun 14 '18 at 11:46
|
show 2 more comments
Considering that you want all the content of your text file to be kept in your variable, you can use:
#!/bin/bash
file="/path/to/filename" #the file where you keep your string name
name=$(cat "$file") #the output of 'cat $file' is assigned to the $name variable
echo $name #test
Or, in pure bash:
#!/bin/bash
file="/path/to/filename" #the file where you keep your string name
read -d $'x04' name < "$file" #the content of $file is redirected to stdin from where it is read out into the $name variable
echo $name #test
I dont want to display the variable, I want that this variable should be read by the script for further processing. I have tried awk which reads line by line.
– user208413
Oct 28 '13 at 8:48
1
Ok, deleteecho $name
. I use it just for test. Next you can use$name
variable for further processing anywhere you want in your script.
– Radu Rădeanu
Oct 28 '13 at 8:51
I cant use it anywhere it my script because dhe script doesn't know the value of the variable. In our case the variable name.
– user208413
Oct 28 '13 at 9:28
6
@user208413 The value of$name
variable is the string from your file assigned usingcat $file
command. What is not clear or so difficult to understand?
– Radu Rădeanu
Oct 28 '13 at 9:36
1
It's redundant to use a seperate$file
variable if you're only using it once to load the$name
variable, just usecat /path/to/file
– kiri
Oct 28 '13 at 10:05
add a comment |
From within your script you can do this:
read name < file_containing _the_answer
You can even do this multiple times e.g. in a loop
while read LINE; do echo "$LINE"; done < file_containing_multiple_lines
or for a file with multiple items on each line . . . grep -v ^# file |while read a b c; do echo a=$a b=$b c=$c; done
– gaoithe
Sep 1 '15 at 14:29
add a comment |
One alternative way to do this would be to just redirect standard input to your file, where you have all the user input in the order it's expected by the program. For example, with the program (called script.sh
)
#!/bin/bash
echo "Enter your name:"
read name
echo "...and now your age:"
read age
# example of how to use the values now stored in variables $name and $age
echo "Hello $name. You're $age years old, right?"
and the input file (called input.in
)
Tomas
26
you could run this from the terminal in one of the following two ways:
$ cat input.in | ./script.sh
$ ./script.sh < input.in
and it would be equivalent to just running the script and entering the data manually - it would print the line "Hello Tomas. You're 26 years old, right?".
As Radu Rădeanu has already suggested, you could use cat
inside your script to read the contents of a file into a avariable - in that case, you need each file to contain only one line, with only the value you want for that specific variable. In the above example, you'd split the input file into one with the name (say, name.in
) and one with the age (say, age.in
), and change the read name
and read age
lines to name=$(cat name.in)
and age=$(cat age.in)
respectively.
I have the following script: vh = awk "NR==3{print;exit}" /var/www/test/test.txt with this I read line 3 from text file named test.txt. Once i get a string I want to use this string as an input for creating a directory
– user208413
Oct 28 '13 at 10:18
1
Change that tovh=$(awk "NR==3 {print;exit}" /var/www/test.txt)
. You will then have the string saved in variable$vh
, so you can use it like so:mkdir "/var/www/$vh"
- this will create a folder with the specified name inside /var/www.
– Tomas Aschan
Oct 28 '13 at 11:09
I have done this way but it still cant create the folder :(
– user208413
Oct 28 '13 at 11:34
@user208413: What error message(s) do you get? "It doesn't work" is, unfortunately, not a very helpful problem description...
– Tomas Aschan
Oct 29 '13 at 11:02
add a comment |
Short answer:
name=`cat "$file"`
add a comment |
I found working solution here:
https://af-design.com/2009/07/07/loading-data-into-bash-variables/
if [ -f $SETTINGS_FILE ];then
. $SETTINGS_FILE
fi
The slides are no longer available.
– kris
Mar 12 '18 at 15:34
add a comment |
If you want to use multiple strings, you could go with:
path="/foo/bar";
for p in `cat "$path"`;
do echo "$p" ....... (here you would pipe it where you need it)
OR
If you want the user to indicate the file
read -e "Specify path to variable(s): " path;
for p in 'cat "$path"';
do echo "$p" ............................
add a comment |
name=$(<"$file")
From man bash:1785
, this command substitution is equivalent to name=$(cat "$file")
but faster.
add a comment |
#! /bin/bash
# (GPL3+) Alberto Salvia Novella (es20490446e)
variableInFile () {
variable=${1}
file=${2}
source ${file}
eval value=${${variable}}
echo ${value}
}
variableInFile ${@}
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "89"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2faskubuntu.com%2fquestions%2f367136%2fhow-do-i-read-a-variable-from-a-file%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
To read variables from a file we can use the source
or .
command.
Lets assume the file contains the following line
MYVARIABLE="Any string"
we can then import this variable using
#!/bin/bash
source <filename>
echo $MYVARIABLE
15
One reason why you might not want to do it this way is because whatever is in<filename>
will be run in your shell, such asrm -rf /
, which could be very dangerous.
– Mark Ribau
Jan 10 '15 at 2:57
3
Late to the party, but, no, it doesn't deserve to be marked up OR as the answer. This sources another bash source file which is NOT what the OP asked for. He asked how to read the contents of a FILE into a variable. NOT how to execute another script which sets a variable. The correct answer is the one below. name=$(cat "$file") .
– RichieHH
Nov 24 '15 at 12:54
1
OTOH, it is the answer to the question I asked google and google brought me here. Weird times, eh?
– Michael Terry
Jul 10 '17 at 18:04
1
Great also works for multiple variables! Point of @MarkRibau is understood but I guess that if you right the file in a different script and just have to pass it over its fine and controlled.
– madlymad
Jul 19 '17 at 11:25
1
You can usesed
to addlocal
keyword and make the script a bit safer and not polute your global scope. So inside a a function docat global_variables.sh | sed -e 's/^/local /' > local_variables.sh
and then use. ./local_variables.sh
. Whatever you import in the function will only be available in that function. Note that it assumesglobal_variables.sh
only contains one variable per line.
– Nux
Jun 14 '18 at 11:46
|
show 2 more comments
To read variables from a file we can use the source
or .
command.
Lets assume the file contains the following line
MYVARIABLE="Any string"
we can then import this variable using
#!/bin/bash
source <filename>
echo $MYVARIABLE
15
One reason why you might not want to do it this way is because whatever is in<filename>
will be run in your shell, such asrm -rf /
, which could be very dangerous.
– Mark Ribau
Jan 10 '15 at 2:57
3
Late to the party, but, no, it doesn't deserve to be marked up OR as the answer. This sources another bash source file which is NOT what the OP asked for. He asked how to read the contents of a FILE into a variable. NOT how to execute another script which sets a variable. The correct answer is the one below. name=$(cat "$file") .
– RichieHH
Nov 24 '15 at 12:54
1
OTOH, it is the answer to the question I asked google and google brought me here. Weird times, eh?
– Michael Terry
Jul 10 '17 at 18:04
1
Great also works for multiple variables! Point of @MarkRibau is understood but I guess that if you right the file in a different script and just have to pass it over its fine and controlled.
– madlymad
Jul 19 '17 at 11:25
1
You can usesed
to addlocal
keyword and make the script a bit safer and not polute your global scope. So inside a a function docat global_variables.sh | sed -e 's/^/local /' > local_variables.sh
and then use. ./local_variables.sh
. Whatever you import in the function will only be available in that function. Note that it assumesglobal_variables.sh
only contains one variable per line.
– Nux
Jun 14 '18 at 11:46
|
show 2 more comments
To read variables from a file we can use the source
or .
command.
Lets assume the file contains the following line
MYVARIABLE="Any string"
we can then import this variable using
#!/bin/bash
source <filename>
echo $MYVARIABLE
To read variables from a file we can use the source
or .
command.
Lets assume the file contains the following line
MYVARIABLE="Any string"
we can then import this variable using
#!/bin/bash
source <filename>
echo $MYVARIABLE
edited Oct 11 '18 at 2:29
mook765
3,95421330
3,95421330
answered Oct 28 '13 at 9:44
TakkatTakkat
106k35249375
106k35249375
15
One reason why you might not want to do it this way is because whatever is in<filename>
will be run in your shell, such asrm -rf /
, which could be very dangerous.
– Mark Ribau
Jan 10 '15 at 2:57
3
Late to the party, but, no, it doesn't deserve to be marked up OR as the answer. This sources another bash source file which is NOT what the OP asked for. He asked how to read the contents of a FILE into a variable. NOT how to execute another script which sets a variable. The correct answer is the one below. name=$(cat "$file") .
– RichieHH
Nov 24 '15 at 12:54
1
OTOH, it is the answer to the question I asked google and google brought me here. Weird times, eh?
– Michael Terry
Jul 10 '17 at 18:04
1
Great also works for multiple variables! Point of @MarkRibau is understood but I guess that if you right the file in a different script and just have to pass it over its fine and controlled.
– madlymad
Jul 19 '17 at 11:25
1
You can usesed
to addlocal
keyword and make the script a bit safer and not polute your global scope. So inside a a function docat global_variables.sh | sed -e 's/^/local /' > local_variables.sh
and then use. ./local_variables.sh
. Whatever you import in the function will only be available in that function. Note that it assumesglobal_variables.sh
only contains one variable per line.
– Nux
Jun 14 '18 at 11:46
|
show 2 more comments
15
One reason why you might not want to do it this way is because whatever is in<filename>
will be run in your shell, such asrm -rf /
, which could be very dangerous.
– Mark Ribau
Jan 10 '15 at 2:57
3
Late to the party, but, no, it doesn't deserve to be marked up OR as the answer. This sources another bash source file which is NOT what the OP asked for. He asked how to read the contents of a FILE into a variable. NOT how to execute another script which sets a variable. The correct answer is the one below. name=$(cat "$file") .
– RichieHH
Nov 24 '15 at 12:54
1
OTOH, it is the answer to the question I asked google and google brought me here. Weird times, eh?
– Michael Terry
Jul 10 '17 at 18:04
1
Great also works for multiple variables! Point of @MarkRibau is understood but I guess that if you right the file in a different script and just have to pass it over its fine and controlled.
– madlymad
Jul 19 '17 at 11:25
1
You can usesed
to addlocal
keyword and make the script a bit safer and not polute your global scope. So inside a a function docat global_variables.sh | sed -e 's/^/local /' > local_variables.sh
and then use. ./local_variables.sh
. Whatever you import in the function will only be available in that function. Note that it assumesglobal_variables.sh
only contains one variable per line.
– Nux
Jun 14 '18 at 11:46
15
15
One reason why you might not want to do it this way is because whatever is in
<filename>
will be run in your shell, such as rm -rf /
, which could be very dangerous.– Mark Ribau
Jan 10 '15 at 2:57
One reason why you might not want to do it this way is because whatever is in
<filename>
will be run in your shell, such as rm -rf /
, which could be very dangerous.– Mark Ribau
Jan 10 '15 at 2:57
3
3
Late to the party, but, no, it doesn't deserve to be marked up OR as the answer. This sources another bash source file which is NOT what the OP asked for. He asked how to read the contents of a FILE into a variable. NOT how to execute another script which sets a variable. The correct answer is the one below. name=$(cat "$file") .
– RichieHH
Nov 24 '15 at 12:54
Late to the party, but, no, it doesn't deserve to be marked up OR as the answer. This sources another bash source file which is NOT what the OP asked for. He asked how to read the contents of a FILE into a variable. NOT how to execute another script which sets a variable. The correct answer is the one below. name=$(cat "$file") .
– RichieHH
Nov 24 '15 at 12:54
1
1
OTOH, it is the answer to the question I asked google and google brought me here. Weird times, eh?
– Michael Terry
Jul 10 '17 at 18:04
OTOH, it is the answer to the question I asked google and google brought me here. Weird times, eh?
– Michael Terry
Jul 10 '17 at 18:04
1
1
Great also works for multiple variables! Point of @MarkRibau is understood but I guess that if you right the file in a different script and just have to pass it over its fine and controlled.
– madlymad
Jul 19 '17 at 11:25
Great also works for multiple variables! Point of @MarkRibau is understood but I guess that if you right the file in a different script and just have to pass it over its fine and controlled.
– madlymad
Jul 19 '17 at 11:25
1
1
You can use
sed
to add local
keyword and make the script a bit safer and not polute your global scope. So inside a a function do cat global_variables.sh | sed -e 's/^/local /' > local_variables.sh
and then use . ./local_variables.sh
. Whatever you import in the function will only be available in that function. Note that it assumes global_variables.sh
only contains one variable per line.– Nux
Jun 14 '18 at 11:46
You can use
sed
to add local
keyword and make the script a bit safer and not polute your global scope. So inside a a function do cat global_variables.sh | sed -e 's/^/local /' > local_variables.sh
and then use . ./local_variables.sh
. Whatever you import in the function will only be available in that function. Note that it assumes global_variables.sh
only contains one variable per line.– Nux
Jun 14 '18 at 11:46
|
show 2 more comments
Considering that you want all the content of your text file to be kept in your variable, you can use:
#!/bin/bash
file="/path/to/filename" #the file where you keep your string name
name=$(cat "$file") #the output of 'cat $file' is assigned to the $name variable
echo $name #test
Or, in pure bash:
#!/bin/bash
file="/path/to/filename" #the file where you keep your string name
read -d $'x04' name < "$file" #the content of $file is redirected to stdin from where it is read out into the $name variable
echo $name #test
I dont want to display the variable, I want that this variable should be read by the script for further processing. I have tried awk which reads line by line.
– user208413
Oct 28 '13 at 8:48
1
Ok, deleteecho $name
. I use it just for test. Next you can use$name
variable for further processing anywhere you want in your script.
– Radu Rădeanu
Oct 28 '13 at 8:51
I cant use it anywhere it my script because dhe script doesn't know the value of the variable. In our case the variable name.
– user208413
Oct 28 '13 at 9:28
6
@user208413 The value of$name
variable is the string from your file assigned usingcat $file
command. What is not clear or so difficult to understand?
– Radu Rădeanu
Oct 28 '13 at 9:36
1
It's redundant to use a seperate$file
variable if you're only using it once to load the$name
variable, just usecat /path/to/file
– kiri
Oct 28 '13 at 10:05
add a comment |
Considering that you want all the content of your text file to be kept in your variable, you can use:
#!/bin/bash
file="/path/to/filename" #the file where you keep your string name
name=$(cat "$file") #the output of 'cat $file' is assigned to the $name variable
echo $name #test
Or, in pure bash:
#!/bin/bash
file="/path/to/filename" #the file where you keep your string name
read -d $'x04' name < "$file" #the content of $file is redirected to stdin from where it is read out into the $name variable
echo $name #test
I dont want to display the variable, I want that this variable should be read by the script for further processing. I have tried awk which reads line by line.
– user208413
Oct 28 '13 at 8:48
1
Ok, deleteecho $name
. I use it just for test. Next you can use$name
variable for further processing anywhere you want in your script.
– Radu Rădeanu
Oct 28 '13 at 8:51
I cant use it anywhere it my script because dhe script doesn't know the value of the variable. In our case the variable name.
– user208413
Oct 28 '13 at 9:28
6
@user208413 The value of$name
variable is the string from your file assigned usingcat $file
command. What is not clear or so difficult to understand?
– Radu Rădeanu
Oct 28 '13 at 9:36
1
It's redundant to use a seperate$file
variable if you're only using it once to load the$name
variable, just usecat /path/to/file
– kiri
Oct 28 '13 at 10:05
add a comment |
Considering that you want all the content of your text file to be kept in your variable, you can use:
#!/bin/bash
file="/path/to/filename" #the file where you keep your string name
name=$(cat "$file") #the output of 'cat $file' is assigned to the $name variable
echo $name #test
Or, in pure bash:
#!/bin/bash
file="/path/to/filename" #the file where you keep your string name
read -d $'x04' name < "$file" #the content of $file is redirected to stdin from where it is read out into the $name variable
echo $name #test
Considering that you want all the content of your text file to be kept in your variable, you can use:
#!/bin/bash
file="/path/to/filename" #the file where you keep your string name
name=$(cat "$file") #the output of 'cat $file' is assigned to the $name variable
echo $name #test
Or, in pure bash:
#!/bin/bash
file="/path/to/filename" #the file where you keep your string name
read -d $'x04' name < "$file" #the content of $file is redirected to stdin from where it is read out into the $name variable
echo $name #test
edited Feb 8 '15 at 11:49
muru
1
1
answered Oct 28 '13 at 8:44
Radu RădeanuRadu Rădeanu
116k34247323
116k34247323
I dont want to display the variable, I want that this variable should be read by the script for further processing. I have tried awk which reads line by line.
– user208413
Oct 28 '13 at 8:48
1
Ok, deleteecho $name
. I use it just for test. Next you can use$name
variable for further processing anywhere you want in your script.
– Radu Rădeanu
Oct 28 '13 at 8:51
I cant use it anywhere it my script because dhe script doesn't know the value of the variable. In our case the variable name.
– user208413
Oct 28 '13 at 9:28
6
@user208413 The value of$name
variable is the string from your file assigned usingcat $file
command. What is not clear or so difficult to understand?
– Radu Rădeanu
Oct 28 '13 at 9:36
1
It's redundant to use a seperate$file
variable if you're only using it once to load the$name
variable, just usecat /path/to/file
– kiri
Oct 28 '13 at 10:05
add a comment |
I dont want to display the variable, I want that this variable should be read by the script for further processing. I have tried awk which reads line by line.
– user208413
Oct 28 '13 at 8:48
1
Ok, deleteecho $name
. I use it just for test. Next you can use$name
variable for further processing anywhere you want in your script.
– Radu Rădeanu
Oct 28 '13 at 8:51
I cant use it anywhere it my script because dhe script doesn't know the value of the variable. In our case the variable name.
– user208413
Oct 28 '13 at 9:28
6
@user208413 The value of$name
variable is the string from your file assigned usingcat $file
command. What is not clear or so difficult to understand?
– Radu Rădeanu
Oct 28 '13 at 9:36
1
It's redundant to use a seperate$file
variable if you're only using it once to load the$name
variable, just usecat /path/to/file
– kiri
Oct 28 '13 at 10:05
I dont want to display the variable, I want that this variable should be read by the script for further processing. I have tried awk which reads line by line.
– user208413
Oct 28 '13 at 8:48
I dont want to display the variable, I want that this variable should be read by the script for further processing. I have tried awk which reads line by line.
– user208413
Oct 28 '13 at 8:48
1
1
Ok, delete
echo $name
. I use it just for test. Next you can use $name
variable for further processing anywhere you want in your script.– Radu Rădeanu
Oct 28 '13 at 8:51
Ok, delete
echo $name
. I use it just for test. Next you can use $name
variable for further processing anywhere you want in your script.– Radu Rădeanu
Oct 28 '13 at 8:51
I cant use it anywhere it my script because dhe script doesn't know the value of the variable. In our case the variable name.
– user208413
Oct 28 '13 at 9:28
I cant use it anywhere it my script because dhe script doesn't know the value of the variable. In our case the variable name.
– user208413
Oct 28 '13 at 9:28
6
6
@user208413 The value of
$name
variable is the string from your file assigned using cat $file
command. What is not clear or so difficult to understand?– Radu Rădeanu
Oct 28 '13 at 9:36
@user208413 The value of
$name
variable is the string from your file assigned using cat $file
command. What is not clear or so difficult to understand?– Radu Rădeanu
Oct 28 '13 at 9:36
1
1
It's redundant to use a seperate
$file
variable if you're only using it once to load the $name
variable, just use cat /path/to/file
– kiri
Oct 28 '13 at 10:05
It's redundant to use a seperate
$file
variable if you're only using it once to load the $name
variable, just use cat /path/to/file
– kiri
Oct 28 '13 at 10:05
add a comment |
From within your script you can do this:
read name < file_containing _the_answer
You can even do this multiple times e.g. in a loop
while read LINE; do echo "$LINE"; done < file_containing_multiple_lines
or for a file with multiple items on each line . . . grep -v ^# file |while read a b c; do echo a=$a b=$b c=$c; done
– gaoithe
Sep 1 '15 at 14:29
add a comment |
From within your script you can do this:
read name < file_containing _the_answer
You can even do this multiple times e.g. in a loop
while read LINE; do echo "$LINE"; done < file_containing_multiple_lines
or for a file with multiple items on each line . . . grep -v ^# file |while read a b c; do echo a=$a b=$b c=$c; done
– gaoithe
Sep 1 '15 at 14:29
add a comment |
From within your script you can do this:
read name < file_containing _the_answer
You can even do this multiple times e.g. in a loop
while read LINE; do echo "$LINE"; done < file_containing_multiple_lines
From within your script you can do this:
read name < file_containing _the_answer
You can even do this multiple times e.g. in a loop
while read LINE; do echo "$LINE"; done < file_containing_multiple_lines
answered Oct 28 '13 at 22:35
thomthom
4,72731624
4,72731624
or for a file with multiple items on each line . . . grep -v ^# file |while read a b c; do echo a=$a b=$b c=$c; done
– gaoithe
Sep 1 '15 at 14:29
add a comment |
or for a file with multiple items on each line . . . grep -v ^# file |while read a b c; do echo a=$a b=$b c=$c; done
– gaoithe
Sep 1 '15 at 14:29
or for a file with multiple items on each line . . . grep -v ^# file |while read a b c; do echo a=$a b=$b c=$c; done
– gaoithe
Sep 1 '15 at 14:29
or for a file with multiple items on each line . . . grep -v ^# file |while read a b c; do echo a=$a b=$b c=$c; done
– gaoithe
Sep 1 '15 at 14:29
add a comment |
One alternative way to do this would be to just redirect standard input to your file, where you have all the user input in the order it's expected by the program. For example, with the program (called script.sh
)
#!/bin/bash
echo "Enter your name:"
read name
echo "...and now your age:"
read age
# example of how to use the values now stored in variables $name and $age
echo "Hello $name. You're $age years old, right?"
and the input file (called input.in
)
Tomas
26
you could run this from the terminal in one of the following two ways:
$ cat input.in | ./script.sh
$ ./script.sh < input.in
and it would be equivalent to just running the script and entering the data manually - it would print the line "Hello Tomas. You're 26 years old, right?".
As Radu Rădeanu has already suggested, you could use cat
inside your script to read the contents of a file into a avariable - in that case, you need each file to contain only one line, with only the value you want for that specific variable. In the above example, you'd split the input file into one with the name (say, name.in
) and one with the age (say, age.in
), and change the read name
and read age
lines to name=$(cat name.in)
and age=$(cat age.in)
respectively.
I have the following script: vh = awk "NR==3{print;exit}" /var/www/test/test.txt with this I read line 3 from text file named test.txt. Once i get a string I want to use this string as an input for creating a directory
– user208413
Oct 28 '13 at 10:18
1
Change that tovh=$(awk "NR==3 {print;exit}" /var/www/test.txt)
. You will then have the string saved in variable$vh
, so you can use it like so:mkdir "/var/www/$vh"
- this will create a folder with the specified name inside /var/www.
– Tomas Aschan
Oct 28 '13 at 11:09
I have done this way but it still cant create the folder :(
– user208413
Oct 28 '13 at 11:34
@user208413: What error message(s) do you get? "It doesn't work" is, unfortunately, not a very helpful problem description...
– Tomas Aschan
Oct 29 '13 at 11:02
add a comment |
One alternative way to do this would be to just redirect standard input to your file, where you have all the user input in the order it's expected by the program. For example, with the program (called script.sh
)
#!/bin/bash
echo "Enter your name:"
read name
echo "...and now your age:"
read age
# example of how to use the values now stored in variables $name and $age
echo "Hello $name. You're $age years old, right?"
and the input file (called input.in
)
Tomas
26
you could run this from the terminal in one of the following two ways:
$ cat input.in | ./script.sh
$ ./script.sh < input.in
and it would be equivalent to just running the script and entering the data manually - it would print the line "Hello Tomas. You're 26 years old, right?".
As Radu Rădeanu has already suggested, you could use cat
inside your script to read the contents of a file into a avariable - in that case, you need each file to contain only one line, with only the value you want for that specific variable. In the above example, you'd split the input file into one with the name (say, name.in
) and one with the age (say, age.in
), and change the read name
and read age
lines to name=$(cat name.in)
and age=$(cat age.in)
respectively.
I have the following script: vh = awk "NR==3{print;exit}" /var/www/test/test.txt with this I read line 3 from text file named test.txt. Once i get a string I want to use this string as an input for creating a directory
– user208413
Oct 28 '13 at 10:18
1
Change that tovh=$(awk "NR==3 {print;exit}" /var/www/test.txt)
. You will then have the string saved in variable$vh
, so you can use it like so:mkdir "/var/www/$vh"
- this will create a folder with the specified name inside /var/www.
– Tomas Aschan
Oct 28 '13 at 11:09
I have done this way but it still cant create the folder :(
– user208413
Oct 28 '13 at 11:34
@user208413: What error message(s) do you get? "It doesn't work" is, unfortunately, not a very helpful problem description...
– Tomas Aschan
Oct 29 '13 at 11:02
add a comment |
One alternative way to do this would be to just redirect standard input to your file, where you have all the user input in the order it's expected by the program. For example, with the program (called script.sh
)
#!/bin/bash
echo "Enter your name:"
read name
echo "...and now your age:"
read age
# example of how to use the values now stored in variables $name and $age
echo "Hello $name. You're $age years old, right?"
and the input file (called input.in
)
Tomas
26
you could run this from the terminal in one of the following two ways:
$ cat input.in | ./script.sh
$ ./script.sh < input.in
and it would be equivalent to just running the script and entering the data manually - it would print the line "Hello Tomas. You're 26 years old, right?".
As Radu Rădeanu has already suggested, you could use cat
inside your script to read the contents of a file into a avariable - in that case, you need each file to contain only one line, with only the value you want for that specific variable. In the above example, you'd split the input file into one with the name (say, name.in
) and one with the age (say, age.in
), and change the read name
and read age
lines to name=$(cat name.in)
and age=$(cat age.in)
respectively.
One alternative way to do this would be to just redirect standard input to your file, where you have all the user input in the order it's expected by the program. For example, with the program (called script.sh
)
#!/bin/bash
echo "Enter your name:"
read name
echo "...and now your age:"
read age
# example of how to use the values now stored in variables $name and $age
echo "Hello $name. You're $age years old, right?"
and the input file (called input.in
)
Tomas
26
you could run this from the terminal in one of the following two ways:
$ cat input.in | ./script.sh
$ ./script.sh < input.in
and it would be equivalent to just running the script and entering the data manually - it would print the line "Hello Tomas. You're 26 years old, right?".
As Radu Rădeanu has already suggested, you could use cat
inside your script to read the contents of a file into a avariable - in that case, you need each file to contain only one line, with only the value you want for that specific variable. In the above example, you'd split the input file into one with the name (say, name.in
) and one with the age (say, age.in
), and change the read name
and read age
lines to name=$(cat name.in)
and age=$(cat age.in)
respectively.
edited Apr 13 '17 at 12:25
Community♦
1
1
answered Oct 28 '13 at 9:57
Tomas AschanTomas Aschan
1,39772754
1,39772754
I have the following script: vh = awk "NR==3{print;exit}" /var/www/test/test.txt with this I read line 3 from text file named test.txt. Once i get a string I want to use this string as an input for creating a directory
– user208413
Oct 28 '13 at 10:18
1
Change that tovh=$(awk "NR==3 {print;exit}" /var/www/test.txt)
. You will then have the string saved in variable$vh
, so you can use it like so:mkdir "/var/www/$vh"
- this will create a folder with the specified name inside /var/www.
– Tomas Aschan
Oct 28 '13 at 11:09
I have done this way but it still cant create the folder :(
– user208413
Oct 28 '13 at 11:34
@user208413: What error message(s) do you get? "It doesn't work" is, unfortunately, not a very helpful problem description...
– Tomas Aschan
Oct 29 '13 at 11:02
add a comment |
I have the following script: vh = awk "NR==3{print;exit}" /var/www/test/test.txt with this I read line 3 from text file named test.txt. Once i get a string I want to use this string as an input for creating a directory
– user208413
Oct 28 '13 at 10:18
1
Change that tovh=$(awk "NR==3 {print;exit}" /var/www/test.txt)
. You will then have the string saved in variable$vh
, so you can use it like so:mkdir "/var/www/$vh"
- this will create a folder with the specified name inside /var/www.
– Tomas Aschan
Oct 28 '13 at 11:09
I have done this way but it still cant create the folder :(
– user208413
Oct 28 '13 at 11:34
@user208413: What error message(s) do you get? "It doesn't work" is, unfortunately, not a very helpful problem description...
– Tomas Aschan
Oct 29 '13 at 11:02
I have the following script: vh = awk "NR==3{print;exit}" /var/www/test/test.txt with this I read line 3 from text file named test.txt. Once i get a string I want to use this string as an input for creating a directory
– user208413
Oct 28 '13 at 10:18
I have the following script: vh = awk "NR==3{print;exit}" /var/www/test/test.txt with this I read line 3 from text file named test.txt. Once i get a string I want to use this string as an input for creating a directory
– user208413
Oct 28 '13 at 10:18
1
1
Change that to
vh=$(awk "NR==3 {print;exit}" /var/www/test.txt)
. You will then have the string saved in variable $vh
, so you can use it like so: mkdir "/var/www/$vh"
- this will create a folder with the specified name inside /var/www.– Tomas Aschan
Oct 28 '13 at 11:09
Change that to
vh=$(awk "NR==3 {print;exit}" /var/www/test.txt)
. You will then have the string saved in variable $vh
, so you can use it like so: mkdir "/var/www/$vh"
- this will create a folder with the specified name inside /var/www.– Tomas Aschan
Oct 28 '13 at 11:09
I have done this way but it still cant create the folder :(
– user208413
Oct 28 '13 at 11:34
I have done this way but it still cant create the folder :(
– user208413
Oct 28 '13 at 11:34
@user208413: What error message(s) do you get? "It doesn't work" is, unfortunately, not a very helpful problem description...
– Tomas Aschan
Oct 29 '13 at 11:02
@user208413: What error message(s) do you get? "It doesn't work" is, unfortunately, not a very helpful problem description...
– Tomas Aschan
Oct 29 '13 at 11:02
add a comment |
Short answer:
name=`cat "$file"`
add a comment |
Short answer:
name=`cat "$file"`
add a comment |
Short answer:
name=`cat "$file"`
Short answer:
name=`cat "$file"`
answered Sep 17 '15 at 14:53
nedimnedim
1234
1234
add a comment |
add a comment |
I found working solution here:
https://af-design.com/2009/07/07/loading-data-into-bash-variables/
if [ -f $SETTINGS_FILE ];then
. $SETTINGS_FILE
fi
The slides are no longer available.
– kris
Mar 12 '18 at 15:34
add a comment |
I found working solution here:
https://af-design.com/2009/07/07/loading-data-into-bash-variables/
if [ -f $SETTINGS_FILE ];then
. $SETTINGS_FILE
fi
The slides are no longer available.
– kris
Mar 12 '18 at 15:34
add a comment |
I found working solution here:
https://af-design.com/2009/07/07/loading-data-into-bash-variables/
if [ -f $SETTINGS_FILE ];then
. $SETTINGS_FILE
fi
I found working solution here:
https://af-design.com/2009/07/07/loading-data-into-bash-variables/
if [ -f $SETTINGS_FILE ];then
. $SETTINGS_FILE
fi
answered Jan 24 '16 at 15:33
LukaszLukasz
1212
1212
The slides are no longer available.
– kris
Mar 12 '18 at 15:34
add a comment |
The slides are no longer available.
– kris
Mar 12 '18 at 15:34
The slides are no longer available.
– kris
Mar 12 '18 at 15:34
The slides are no longer available.
– kris
Mar 12 '18 at 15:34
add a comment |
If you want to use multiple strings, you could go with:
path="/foo/bar";
for p in `cat "$path"`;
do echo "$p" ....... (here you would pipe it where you need it)
OR
If you want the user to indicate the file
read -e "Specify path to variable(s): " path;
for p in 'cat "$path"';
do echo "$p" ............................
add a comment |
If you want to use multiple strings, you could go with:
path="/foo/bar";
for p in `cat "$path"`;
do echo "$p" ....... (here you would pipe it where you need it)
OR
If you want the user to indicate the file
read -e "Specify path to variable(s): " path;
for p in 'cat "$path"';
do echo "$p" ............................
add a comment |
If you want to use multiple strings, you could go with:
path="/foo/bar";
for p in `cat "$path"`;
do echo "$p" ....... (here you would pipe it where you need it)
OR
If you want the user to indicate the file
read -e "Specify path to variable(s): " path;
for p in 'cat "$path"';
do echo "$p" ............................
If you want to use multiple strings, you could go with:
path="/foo/bar";
for p in `cat "$path"`;
do echo "$p" ....... (here you would pipe it where you need it)
OR
If you want the user to indicate the file
read -e "Specify path to variable(s): " path;
for p in 'cat "$path"';
do echo "$p" ............................
answered Jun 10 '15 at 16:47
GeorgeGeorge
1
1
add a comment |
add a comment |
name=$(<"$file")
From man bash:1785
, this command substitution is equivalent to name=$(cat "$file")
but faster.
add a comment |
name=$(<"$file")
From man bash:1785
, this command substitution is equivalent to name=$(cat "$file")
but faster.
add a comment |
name=$(<"$file")
From man bash:1785
, this command substitution is equivalent to name=$(cat "$file")
but faster.
name=$(<"$file")
From man bash:1785
, this command substitution is equivalent to name=$(cat "$file")
but faster.
edited Oct 11 '18 at 0:42
abu_bua
3,27981126
3,27981126
answered Oct 10 '18 at 22:11
hellorkhellork
1011
1011
add a comment |
add a comment |
#! /bin/bash
# (GPL3+) Alberto Salvia Novella (es20490446e)
variableInFile () {
variable=${1}
file=${2}
source ${file}
eval value=${${variable}}
echo ${value}
}
variableInFile ${@}
add a comment |
#! /bin/bash
# (GPL3+) Alberto Salvia Novella (es20490446e)
variableInFile () {
variable=${1}
file=${2}
source ${file}
eval value=${${variable}}
echo ${value}
}
variableInFile ${@}
add a comment |
#! /bin/bash
# (GPL3+) Alberto Salvia Novella (es20490446e)
variableInFile () {
variable=${1}
file=${2}
source ${file}
eval value=${${variable}}
echo ${value}
}
variableInFile ${@}
#! /bin/bash
# (GPL3+) Alberto Salvia Novella (es20490446e)
variableInFile () {
variable=${1}
file=${2}
source ${file}
eval value=${${variable}}
echo ${value}
}
variableInFile ${@}
answered 4 hours ago
Alberto Salvia NovellaAlberto Salvia Novella
19611
19611
add a comment |
add a comment |
Thanks for contributing an answer to Ask Ubuntu!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2faskubuntu.com%2fquestions%2f367136%2fhow-do-i-read-a-variable-from-a-file%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
What does your text file look like? Is the entire file 1 variable or are they
KEY=VALUE
pairs? The solutions are quite different (if it's the latter, Takkat's answer applies, the former Radu's)– kiri
Oct 28 '13 at 11:37
1
@CiroSantilli it is not crossposted if it was posted by a different user, there is no such thing as a "cross-site duplicate". The only thing that should be avoided is the same question asked by the same user on different sites.
– terdon♦
Mar 24 '14 at 17:52