replace the second match using preg_replace

electroniceric

Hi.
I'm having a bit of trouble getting preg_replace to replace only subpattern match and not the whole string match.

I'm writing a script that reads an HTML file and replaces the contents of HTML tags like href="somefile.html"
with href="myscript.php?somefile.html"

I match the tag with:

$html_tag_pattern = "/<[\/!]?[^<>]?(?:src|href|background)=\"?([^{\s\"])\"?[^<>]?>/i";}

in my understanding of PCRE, \0 is the whole tag, and \1 is the part inside the href="", <a href="\1">

I want to replace only \1, not \0 with a string I've put together.

Can I do this with a single preg_replace? How?

As a futher complication, as I do the above I need to ensure that I add quotes to any elements that are specified without quotes, i.e.: background=img.gif

Thanks in advance for any advice.

Eric

Weedpacket

Use extra () to match the bits before and after; then you can use \2 to represent to the contents of the href (instead of \1), \1 to represent the bit before it, and \3 to represent the bit after.

electroniceric

If I understand right, you're saying basically use () to capture all the matches in the pattern and then re-include them?

If that's it, seems like a good idea. Thanks!

Weedpacket

Yes, that is what I meant.

PCRE regexps do allow for "assertions" - which are statements to the effect that certain patterns must or must not appear, without actually "matching" them, but there is a limitation that "look-behind" assertions, describing patterns that must appear before a match, can only be of a fixed length. Basically, no *, + or ?, in other words. In that case, you do have to match the pattern, (and capture it so's you can put it back when you replace the match).