[RESOLVED] Regex backreferences

[john1704]

Hello,

I'm having a problem with regex backreferences. Now, I'm very close to solving my problem, but there is just one thing getting in the way. Here it is:

I have a pattern:

$pattern = '%(http://www\.[a-z]+\.[a-z]+/)(.*?)%';

This is supposed to match the first part of the URL as backreference 1, and the second part as backreference 2.

Let's look at an example URL:

http://www.example.org/images/moreimages/myfavoriteimage.jpg

Backreference 1 is:

http://www.example.org/

Backreference 2 is the rest of it:

images/moreimages/myfavoriteimage.jpg

However, when I do this:

$url2 = preg_replace($pattern, '$1', "http://www.example.org/images/moreimages/myfavoriteimage.jpg");

It matches and returns both backreferences instead of one! Why is it doing that?

Thanks.

[NogDog]

The "?" after the "" in the final sub-pattern is making the repetition "ungreedy". Therefore, it is not matching any of that part of the string since the most ungreedy solution to "." is nothing. So, you are only actually replacing the part of the string matched by the first sub-pattern. If the string being matched consists only of the URL, just get rid of the "?". If, instead, it is part of a larger string, you could perhaps use:

$pattern = '%(http://www\.[a-z]+\.[a-z]+/)([^\s]*)%';

This would match everything after the first part until some sort of white-space is encounterd. (You could expand the character class to [^\s\'"] if the URL might be in quotes.)

[john1704]

Thanks NogDog.