preg_replace a query string parameter

Question

I'm using file_get_contents on a fairly large file. In it, I need to find every instance of:

http://www.example.com/?foo=1&bar=3

and change it to:

http://www.example.com?foo=1&bar=4

My problem is not understand how preg_replace will replace only a partial match on my regex and not the entire string. For example, pseudocode looks like the following:

 $content = file_get_contents($filename);
 $pattern = '/http:\/\/www\.example\.com/\?foo=1\&bar=(\d+)';
 preg_replace($pattern, "4", $content);
 file_put_contents($filename, $content);

I'm almost certain preg_replace($pattern, "4", $content); is wrong in this case. What's the right way to just replace the '3' with the '4'here?

Because bar=4 is not static. it might be bar=10, bar=12341324 etc. — randombits
– randombits, Commented Feb 10, 2016 at 0:06
you have missed closing regexp slash, and preg_replace will change whole pattern to '4', in your case. if you need to change just a one character, you can use @Enissay solution, but be ready, that his solution will change any digits with any length to '4'. If you need replacing exactly that string, you need additional limiting condition. — Wizard
– Wizard, Commented Feb 10, 2016 at 0:26

Enissay · Accepted Answer · 2016-02-10 00:11:42Z

3

Use the \K: resets the starting point of the reported match. Any previously consumed characters are no longer included in the final match

$pattern = '/http:\/\/www\.example\.com\/\?foo=1\&bar=\K\d+/';
preg_replace($pattern, "4", $content);

DEMO

You can also use a lookbehind:

$pattern = '/(?<=http:\/\/www\.example\.com\/\?foo=1\&bar=)\d+/';
preg_replace($pattern, "4", $content);

DEMO

edited Feb 10, 2016 at 0:11

answered Feb 10, 2016 at 0:05

Enissay

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Comments

Ashish GamezAddicted · Accepted Answer · 2016-02-10 00:15:48Z

0

$content = file_get_contents($filename);
$pattern = '/http:\/\/www\.example\.com\/\?foo=1\&bar=(\d+)/i';
$content = preg_replace($pattern, "http://www.example.com/?foo=1&bar=4", $content);
file_put_contents($filename, $content);

answered Feb 10, 2016 at 0:15

Ashish GamezAddicted

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