Replace spaces with sed and regexp grouping not working

Question

Somewhy replace empty chars (space, tab, etc.) is not working with sed and regexp grouping:

sed -i "s|dir(\s*)=(\s*)'${site_path_source}'|dir$1=$2'${site_path_target}'|g"

The whole line does replace a string in a file:

sudo find "${config_path}" -type f -exec sed -i "s|dir(\s*)=(\s*)'${site_path_source}'|dir$1=$2'${site_path_target}'|g" {} \;

Any idea how to fix it?

Thanks!

EDIT1: Now using -r to enable extended regular expressions, but still doesn't replace:

sudo find "${config_path}" -type f -exec sed -ir "s|dir(\s*)=(\s*)'${site_path_source}'|dir$1=$2'${site_path_target}'|g" {} \;

But still doesn't do the replace.

EDIT2 (WORKING SOLUTION): When using \( \) for grouping, it WORKS:

sudo find "${config_path}" -type f -exec sed -i "s|dir\(\s*\)=\(\s*\)'${site_path_source}'|dir\1=\2'${site_path_target}'|g" {} \;

But I don't understand, why the EDIT1 solution with -r extended regular expressions is not working.

EDIT3 (WORKING SOLUTION2): Using -r to enable extended regular expressions, it WORKS NOW:

sudo find "${config_path}" -type f -exec sed -r -i "s|dir(\s*)=(\s*)'${site_path_source}'|dir$1=$2'${site_path_target}'|g" {} \;

Explanation:

• Using "sed -ir SUFFIX" is wrong, because -i needs a suffix.
• Using "sed -r -i SUFFIX" is GOOD, because -r and -i are used separately, and -i is followed by a suffix.

Answer 1

The command

sed -i "s|dir(\s*)=(\s*)'${site_path_source}'|dir$1=$2'${site_path_target}'|g"

looks for lines containing d i r ( zero or more whitespace ) = ( zero or more whitespace ) ' etc. If you are trying to use grouping you must either use \( \) or say sed -r to enable extended regular expressions.

Note that $1 and $2 are the first two positional parameters; if you want the first and second groups in the matched string you must use \1 and \2.