qx("perl -p -i -e \"s/(\@Filetyp\\s*?:)/\\1$ext/\" /path/test.c")

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qx() is interpolated the same as a double quoted string so after interpolation the string "perl -p -i -e "s/(@Filetyp\s*?:)/\1$ext/" /path/test.c" is passed to the shell to execute.

You should probably use single quotes around the substitution so that the shell doesn't mess with it, unless you are on Windows.

The \1 capture variable should be $1 in the replacement string. \1 is only allowed inside the regular expression.

To sum up, your code would be better written as (untested): 

my $ext = 'c';
qx(perl -p -i -e 's/\\Q@Filetyp\\E\\s*?:\\K/$ext/' /path/test.c)

[download]

Although you may be better off using the solution that haukex provied.

In reply to Re: perl one liner to replace matching string by jwkrahn
in thread perl one liner to replace matching string by vinoth.ree

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