Regex in IF condition in awk





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I have awk script file as below. I need to add another condition in the if statement to check if the string contains atleast one alphabet. How can I add the extra condition to the present if statement?



Required regex condition: [[ "$1" =~ [A-Za-z] ]]



BEGIN { FS = ";"; counter=0}



{ 

 if ((length($1) != 10 && length($1) != 12))

   { 

    counter++

    print counter, $1;

    if ($counter -gt 2){

        print "Invalid input file";

        exit;

    }    

   }

}


I am getting error if I use the same condition which I have posted. How to add the condition?






awk regular-expression







share|improve this question




















  • 1





    See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage

    – steeldriver
    12 hours ago











  • How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just [A-Za-z]. What are you adding to your awk?

    – terdon♦
    12 hours ago


















2















I have awk script file as below. I need to add another condition in the if statement to check if the string contains atleast one alphabet. How can I add the extra condition to the present if statement?



Required regex condition: [[ "$1" =~ [A-Za-z] ]]



BEGIN { FS = ";"; counter=0}



{ 

 if ((length($1) != 10 && length($1) != 12))

   { 

    counter++

    print counter, $1;

    if ($counter -gt 2){

        print "Invalid input file";

        exit;

    }    

   }

}


I am getting error if I use the same condition which I have posted. How to add the condition?






awk regular-expression







share|improve this question




















  • 1





    See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage

    – steeldriver
    12 hours ago











  • How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just [A-Za-z]. What are you adding to your awk?

    – terdon♦
    12 hours ago














2












2








2








I have awk script file as below. I need to add another condition in the if statement to check if the string contains atleast one alphabet. How can I add the extra condition to the present if statement?



Required regex condition: [[ "$1" =~ [A-Za-z] ]]



BEGIN { FS = ";"; counter=0}



{ 

 if ((length($1) != 10 && length($1) != 12))

   { 

    counter++

    print counter, $1;

    if ($counter -gt 2){

        print "Invalid input file";

        exit;

    }    

   }

}


I am getting error if I use the same condition which I have posted. How to add the condition?






awk regular-expression







share|improve this question
















I have awk script file as below. I need to add another condition in the if statement to check if the string contains atleast one alphabet. How can I add the extra condition to the present if statement?



Required regex condition: [[ "$1" =~ [A-Za-z] ]]



BEGIN { FS = ";"; counter=0}



{ 

 if ((length($1) != 10 && length($1) != 12))

   { 

    counter++

    print counter, $1;

    if ($counter -gt 2){

        print "Invalid input file";

        exit;

    }    

   }

}


I am getting error if I use the same condition which I have posted. How to add the condition?






awk regular-expression




awk regular-expression






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 11 hours ago









muru

37.9k590166




37.9k590166










asked 12 hours ago









LaxmanLaxman

254




254








  • 1





    See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage

    – steeldriver
    12 hours ago











  • How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just [A-Za-z]. What are you adding to your awk?

    – terdon♦
    12 hours ago














  • 1





    See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage

    – steeldriver
    12 hours ago











  • How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just [A-Za-z]. What are you adding to your awk?

    – terdon♦
    12 hours ago








1




1





See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage

– steeldriver
12 hours ago





See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage

– steeldriver
12 hours ago













How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just [A-Za-z]. What are you adding to your awk?

– terdon♦
12 hours ago





How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just [A-Za-z]. What are you adding to your awk?

– terdon♦
12 hours ago










1 Answer
1






active

oldest

votes


















7














You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.



BEGIN { FS = ";"; counter=0}



{ 

 if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)

   { 

    counter++

    print counter, $1;

    if (counter > 2){

        print "Invalid input file";

        exit;

    }    

   }

}


Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.






share|improve this answer


























  • Yes. I was using =~ format.

    – Laxman
    12 hours ago











  • Why couldnt you move the condition to be the pattern? length($1) != 10 && ... && $1 ~ /[A-Za-z]/ { counter++ ... } seems more « awkish » to me, avoiding some seriously nested braces

    – D. Ben Knoble
    7 hours ago






  • 1





    @D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!

    – terdon♦
    7 hours ago












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7














You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.



BEGIN { FS = ";"; counter=0}



{ 

 if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)

   { 

    counter++

    print counter, $1;

    if (counter > 2){

        print "Invalid input file";

        exit;

    }    

   }

}


Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.






share|improve this answer


























  • Yes. I was using =~ format.

    – Laxman
    12 hours ago











  • Why couldnt you move the condition to be the pattern? length($1) != 10 && ... && $1 ~ /[A-Za-z]/ { counter++ ... } seems more « awkish » to me, avoiding some seriously nested braces

    – D. Ben Knoble
    7 hours ago






  • 1





    @D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!

    – terdon♦
    7 hours ago
















7














You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.



BEGIN { FS = ";"; counter=0}



{ 

 if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)

   { 

    counter++

    print counter, $1;

    if (counter > 2){

        print "Invalid input file";

        exit;

    }    

   }

}


Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.






share|improve this answer


























  • Yes. I was using =~ format.

    – Laxman
    12 hours ago











  • Why couldnt you move the condition to be the pattern? length($1) != 10 && ... && $1 ~ /[A-Za-z]/ { counter++ ... } seems more « awkish » to me, avoiding some seriously nested braces

    – D. Ben Knoble
    7 hours ago






  • 1





    @D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!

    – terdon♦
    7 hours ago














7












7








7







You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.



BEGIN { FS = ";"; counter=0}



{ 

 if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)

   { 

    counter++

    print counter, $1;

    if (counter > 2){

        print "Invalid input file";

        exit;

    }    

   }

}


Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.






share|improve this answer















You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.



BEGIN { FS = ";"; counter=0}



{ 

 if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)

   { 

    counter++

    print counter, $1;

    if (counter > 2){

        print "Invalid input file";

        exit;

    }    

   }

}


Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.







share|improve this answer














share|improve this answer



share|improve this answer








edited 7 hours ago

























answered 12 hours ago









terdon♦terdon

134k33270450




134k33270450













  • Yes. I was using =~ format.

    – Laxman
    12 hours ago











  • Why couldnt you move the condition to be the pattern? length($1) != 10 && ... && $1 ~ /[A-Za-z]/ { counter++ ... } seems more « awkish » to me, avoiding some seriously nested braces

    – D. Ben Knoble
    7 hours ago






  • 1





    @D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!

    – terdon♦
    7 hours ago



















  • Yes. I was using =~ format.

    – Laxman
    12 hours ago











  • Why couldnt you move the condition to be the pattern? length($1) != 10 && ... && $1 ~ /[A-Za-z]/ { counter++ ... } seems more « awkish » to me, avoiding some seriously nested braces

    – D. Ben Knoble
    7 hours ago






  • 1





    @D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!

    – terdon♦
    7 hours ago

















Yes. I was using =~ format.

– Laxman
12 hours ago





Yes. I was using =~ format.

– Laxman
12 hours ago













Why couldnt you move the condition to be the pattern? length($1) != 10 && ... && $1 ~ /[A-Za-z]/ { counter++ ... } seems more « awkish » to me, avoiding some seriously nested braces

– D. Ben Knoble
7 hours ago





Why couldnt you move the condition to be the pattern? length($1) != 10 && ... && $1 ~ /[A-Za-z]/ { counter++ ... } seems more « awkish » to me, avoiding some seriously nested braces

– D. Ben Knoble
7 hours ago




1




1





@D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!

– terdon♦
7 hours ago





@D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!

– terdon♦
7 hours ago


















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