How to pass IF condition via shell varibale in awk?

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07-22-2016

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How to pass IF condition via shell varibale in awk?

Hello,

I have one query regarding passing IF condition shell variable inside awk. Here is the case-

File content of keydefn.exp

Code:

201~2~LM Limit 02-current value~Limit 02    ~Limit02~Current~Value  ~N~Y~S~0~9999999
201~3~LM Limit 03-current value~Limit 03~Limit03~Current~Value~N~Y~S~-9999999~9999999

Case 1:

Code:

awk 'BEGIN{IGNORECASE=1;FS="~"}; { if ($2=="2") { print $0}}' keydefn.exp  ==> This is working fine!

Output:

Code:

201~2~LM Limit 02-current value~Limit 02    ~Limit02~Current~Value  ~N~Y~S~0~9999999

Case 2:
But when we move the IF condition in some variable like con='$2 == "2"' then it's not working

Code:

awk -v cond="$con" 'BEGIN{IGNORECASE=1;FS="~"}; { if (cond) { print $0}}' keydefn.exp  ==> This is not working

Output:

Code:

201~2~LM Limit 02-current value~Limit 02    ~Limit02~Current~Value  ~N~Y~S~0~9999999
201~3~LM Limit 03-current value~Limit 03~Limit03~Current~Value~N~Y~S~-9999999~9999999

Please suggest the reason!

Last edited by Scrutinizer; 07-22-2016 at 05:30 AM.. Reason: code tags

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I think it can not be done with -v option this way. Wat awk sees is a 9-character string $2 == "2" and it tests if it not null or an empty string. This is the case, so it will always evaluate to true. There is no eval mechanism like in the shell so that this somehow gets re-interpreted.

To see that it does not get evaluated: you get the same result with a condition that would normally lead to an error message (divide by zero):

Code:

$ cond='$27==1/0';  awk -v cond="$cond" 'BEGIN{IGNORECASE=1;FS="~"}; { if (cond) { print $0}}' infile 
201~2~LM Limit 02-current value~Limit 02    ~Limit02~Current~Value  ~N~Y~S~0~9999999
201~3~LM Limit 03-current value~Limit 03~Limit03~Current~Value~N~Y~S~-9999999~9999999

(The $-sign at the beginning of the lines is not part of the code and represents the command prompt)

Code:

$ awk 'BEGIN{IGNORECASE=1;FS="~"}; { if ($27==1/0) { print $0}}' infile
awk: division by zero
 input record number 1,

--
What you could do is pass a regex string, like so:

Code:

$ regex='^2$'; awk -v regex="$regex" 'BEGIN{IGNORECASE=1;FS="~"}; { if ($2~regex) { print $0}}' infile
201~2~LM Limit 02-current value~Limit 02    ~Limit02~Current~Value  ~N~Y~S~0~9999999

or a value:

Code:

$ val=2; awk -v val="$val" 'BEGIN{IGNORECASE=1;FS="~"}; { if ($2==val) { print $0}}' infile
201~2~LM Limit 02-current value~Limit 02    ~Limit02~Current~Value  ~N~Y~S~0~9999999

--
Or, do not use the -v option and let the shell put things in place before awk interprets:

Code:

$ cond='$2 == "2"'; awk 'BEGIN{IGNORECASE=1;FS="~"}; { if ('"$cond"') { print $0}}' infile
201~2~LM Limit 02-current value~Limit 02    ~Limit02~Current~Value  ~N~Y~S~0~9999999

Last edited by Scrutinizer; 07-22-2016 at 06:45 AM..

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What you could do - but this is far from comfortable nor is it bullet proof - is to analyze the variable's contents and react accordingly, like

Code:

awk -F~ -v cond="$con" '
BEGIN   {gsub (/[$"]/, _, cond) 
         split (cond, T, " ")
        }

T[2] == "==" && $T[1] == T[3] ||
T[2] == ">=" && $T[1] >= T[3] ||
T[2] == "<=" && $T[1] <= T[3] ||
T[2] == "!=" && $T[1] != T[3] ||
T[2] == ">"  && $T[1] >  T[3] ||
T[2] == "<"  && $T[1] <  T[3]           {print $0}
' file

With con='$2 == "2"' it yields

Code:

201~2~LM Limit 02-current value~Limit 02    ~Limit02~Current~Value  ~N~Y~S~0~9999999

, and with con='$2 >= "3"' it yields

Code:

201~3~LM Limit 03-current value~Limit 03~Limit03~Current~Value~N~Y~S~-9999999~9999999

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