Regex prob with interpolated patterns

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Monks,
I have a problem with a regex that contains interpolated patterns. I would expect the following pattern match

$string = 'string with word foo in it';
$string =~ s/(foo)/\1bar/g;
[download]

to produce the output

string with word foobar in it

which it happily does. But, I would like the "find" and "replace" elements of the regex to be interpolated into the pattern as variables (so, say, they could be provided by user input). For example

$string = 'string with word foo in it';
$find = '(foo)';
$replace = '\1bar';
$string =~ s/$find/$replace/g;
[download]

However this is instead giving me the output

string with word \1bar in it.

Any ideas what I can do to remedy this - I do not want to build up a string and then use eval on it.

Many Thanks

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Re: Regex prob with interpolated patterns by halley (Prior) on Apr 06, 2004 at 17:29 UTC
Use the `\1` form inside a regex, to refer to previously collected groups. Use the `$1` form anywhere else, including the right half of a `s///` operator. In your case, you will have to eval the results, but that's pretty easy: express the right hand as an expression, and add the /e option twice to your `s///` operator. (This will require careful attention to tainting if you're security-conscious.) -- `[ e d @ h a l l e y . c c ]`	[reply] [d/l] [select]
Re: Regex prob with interpolated patterns by Roy Johnson (Monsignor) on Apr 06, 2004 at 17:32 UTC
Backslash-backreferences aren't cricket in the replacement portion of s///. You would want '$1bar', and then you'd need to use the /ee flag to get it to eval properly: once to turn $bar into $1foo and once to sub for $1. But $1foo isn't a proper expression, you need to put a . to concatenate them, or double-quotes around it: `$replace = '$1.bar'; # or '"$1bar"' s/$find/$replace/eeg;` [download] Update: It might be more secure to put the quotes in the s/// expression itself: `$replace = 'warn "AARGH!"'; s/$find/qq(qq($replace))/eeg;` [download] The PerlMonk `tr///` Advocate	[reply] [d/l] [select]
Re: Regex prob with interpolated patterns by Enlil (Parson) on Apr 06, 2004 at 17:33 UTC
Here are two previous threads you can look at: RegEx - Using Templates $1 in variable regex replacement string -enlil	[reply]
Re: Regex prob with interpolated patterns by Paladin (Vicar) on Apr 06, 2004 at 17:33 UTC
You shouldn't use `\1` on the RHS of `s///`, you should be using `$1`. `\1` is used in LHS of `s///` in the regex itself. Take a look at the `/e` modifier for `s///`. Your code will look something like this: `$string = 'string with word foo in it'; $find = '(foo)'; $replace = '"${1}bar"'; $string =~ s/$find/$replace/eeg;` [download] Note the double quotes in `$replace` and the `/ee` on the `s///`. The first `e` causes `$replace` to be evaluated, which returns `"${1}bar"` which is then evaluated by the second `e` which returns `foobar`.	[reply] [d/l] [select]