How to extract value between parentheses where before there is a specific string with AWK

Question

Im able to capture all values between parentheses with awk expression

'NR>1{print $1}' RS='(' FS=')'

But im struggling to match one in specific, where i do not want to match by line number but by a string. Not sure if that is possible with awk.

Original file content is

if ($remote_addr ~ ^(1.2.3.4|5.6.7.8)$) {
    set $maintenance off;
}

if ($maintenance = on) {
    return 503;
}

where there are multiple parentheses in different orders.

I need to extract content 1.2.3.4|5.6.7.8 from the line

if ($remote_addr ~ ^

(or simply in the line that contains $remote_addr).

Answer 1

Since it's a simple substitution on a single line I'd just use sed for that:

$ sed -n 's/.*$remote_addr[^(]*(\([^)]*\).*/\1/p' file
1.2.3.4|5.6.7.8

If you really want to use awk though, then you can do this with any awk:

$ awk 'sub(/.*\$remote_addr[^(]*\(/,"") && sub(/).*/,"")' file
1.2.3.4|5.6.7.8

Answer 2

Considering you want to extract the content of the inner parentheses of the if ($remote_addr ~ ( ... ) ) statement, the following awk program should to:

awk 'index($0,"$remote_addr"){sub(/^.*\(/,"");sub(/\).+$/,"");print}' inputfile

This will match the line that contains the string $remote_addr. In that line, it will remove everything from the start-of-line up to the last (, and everything from the first ) to the end-of-line. It then prints the remaining value on the line.

Answer 3

You could also try this sed

sed -En '/\$remote_addr/ s/.*\(([0-9.]*\|[0-9.]*)\).*/\1/p' $file

It will match any line with $remote_addr then extract the match.

The code should be able to match the pattern without explicitly matching $remote_addr beforehand.

sed -En 's/.*\(([0-9.]*\|[0-9.]*)\).*/\1/p' $file

Output

1.2.3.4|5.6.7.8

Answer 4

awk '/\$remote_addr/{for(i=1;i<=NF;i++){if($i ~ /\([0-9].*|[0-9].*\)/){print $i}}}' filename|awk -F "(" '{gsub(/\).*/,"",$2);print $2}

output

1.2.3.4|5.6.7.8