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I have problem with command awk in bash I want to find pattern with variable, but it doesn't work. Can you tell me what's wrong with this line?

awk -F" "  "/$PWD/ {print $1,$3}"  file.txt
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double quote allow expansion of $1,$3 as well aw $PWD, try awk '/'$PWD'/ {print $1,$3}' –  Archemar Mar 17 at 14:11
    
it says synax error –  applenic Mar 17 at 14:13
    
what is $PWD value ? –  Archemar Mar 17 at 14:14
    
it's adress of directory –  applenic Mar 17 at 14:16
2  
This is extremely similar to unix.stackexchange.com/a/190307/88983 where you received an answer that covers the reason why $PWD causes a syntax error. –  dhag Mar 17 at 14:40

1 Answer 1

up vote 3 down vote accepted

Since you have been using double quotes, any $-variables will be expanded by bash before they reach awk. That's the first problem: { print $1, $3 } will expand to { print , }.

Next, the awk syntax you are trying to use is /<regular expression>/ { <action> }: the regular expression is not allowed to contain any unquoted slashes, but $PWD certainly contains some. That's the second issue.

You could try something like this instead:

"/${PWD//\//\\/}/ { print \$1, \$3 }"

This asks bash to replace any occurrences of / in $PWD with \/, and also escapes the $ signs of $1 and $3 so that bash leaves them intact.

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it still says syntax error –  applenic Mar 17 at 14:40
    
Then it would be interesting for us to know exactly where your awk thinks there is an error. I tried with GNU awk 4.0.1, successfully. –  dhag Mar 17 at 14:42
    
It's working now.. Thanks –  applenic Mar 17 at 15:01
    
The syntax error is because you forgot a slash: ${PWD//\//\/} should be ${PWD//\//\\/}. I fixed it. –  terdon♦ Mar 17 at 15:42
    
I tested the actual code that I posted, and the extra backslash should make no difference in this case (because it is within double quotes and does not precede a character with special meaning). Thanks for the edit anyway; doubling the backslash is probably less confusing. –  dhag Mar 17 at 15:48

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