I do know that '+' is a special character, but in any other place besides the first position, it works correctly. I have previously tried "\+44", and received the same error message.

But as you suggest, I changed the code to

However, the result was incorrect: Result: |\+44|\|

Thank you.

The problem is that your test is wrong. Try this:

$ perl -le '$out = "+44"; $in=quotemeta $out; $out =~ s/$in//; print "
+Result: |$in|$out|\n";'
Result: |\+44||
[download]

Note that '$in' is the pattern, not the original value of '$out'.

Alternatively, you may write:

$ perl -le '$in = "+44"; $out = $in; $out =~ s/\Q$in//; print "Result:
+ |$in|$out|\n";'
Result: |+44||
[download]

Is this what you looking for??

 perl -e '$in=quotemeta("+44");$out=$in;
$out=~s/\Q$in\E//;print "Result: |$in|$out|\n";'

Result: |\+44||
[download]


Or



perl -e '$in="+44";$out=$in;
$out=~s/\Q$in\E//;print "Result: |$in|$out|\n";'

Result: |+44||
[download]

Both JavaFan and suhailck solutions work correctly.

The quotemata or \Q is what I need in my code. The $in / $out were just to show a sample of the problem. Sorry, I didn't noticed the reversing of pattern at the time, since I got the same error message.

The actual code has only 1 variable, but without the backslash of '+' my solution didn't work in all case.

Thank you.

but in any other place besides the first position, it works correctly.

It does not work; it just does not die with a syntax error; there is a difference.

Hi,
Can you give the expected result which you want for

perl -e '$in="+44"; $out=$in; $out =~ s/$in//; print "Result: |$in|$ou
+t|\n";'
[download]

--
Thanks&Regards,
Gobi S.