Question

I want to perform a regex using grouping. I am only interested in the grouping, its all I want returned. Is this possible?

$haystack = '<a href="/foo.php">Go To Foo</a>';
$needle = '/href="(.*)">/';
preg_match($needle,$haystack,$matches);
print_r($matches);

//Outputs
//Array ( [0] => href="/foo.php">  [1] => /foo.php ) 

//I want:
//Array ( [0] => /foo.php ) 

Answer 1

Actually this is possible with lookarounds. Instead of:

href="(.*)">

You want

(?<=href=").*(?=">)

Now this will match (and therefore capture into group 0) any .* that is preceded by href=" and followed by ">. Note that I highly suspect that you really need .*? instead, i.e. reluctant instead of greedy.

---A--Z---A--Z----
   ^^^^^^^^^^^
      A.*Z

In any case, it looks like PHP’s preg is PCRE, so it should support lookarounds.

regular-expressions.info links

• Lookaround
• Flavor comparison
  • PHP’s preg functions implement the PCRE flavor.
  • PCRE:
    • (?=regex) (positive lookahead): YES
    • (?<=text) (positive lookbehind): fixed + alternation

---

Demonstration

<?php

$haystack = '<a href="/foo.php">Go To Foo</a><a href="/bar.php">Go To Bar</a>';
$needle = '/(?<=href=").*?(?=">)/';
preg_match_all($needle,$haystack,$matches);
print_r($matches);     

?>

Running this on ideone.com produces:

Array
(
    [0] => Array
        (
            [0] => /foo.php
            [1] => /bar.php
        )

)

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