PHP preg_match_all remove url parameters on the fly

Question

I want to fetch only the 'cleaner' version of the url without any parameters. IOW... if there is a question mark inside the url remove it and everything afterwards.

Here is my current line :

preg_match_all('/<a(.*?)href=("|\'|)(.*?)("|\'| )(.*?)>/s',$content,$ahref);

And just to be more clear here... I'm expecting that this url (for example):

/go/page/mobile_download_apps.html?&who=r,6GDewh28SCW3/fUSqmWqR_E9ljkcH1DheIMqgbiHjlX3OBDbskcuCZ22iDvk0zeZR7BEthcEaXGFWaQ4Burmd4eKuhMpqojjDE6BrCiUtLClkT32CejpMIdnqVOUmWBD

Would be :

/go/page/mobile_download_apps.html

Wouldn't this do the trick ? /(<a href=")(.*)(\?.*)/s (missing some info to give a more detailed answer...) — Benoît
– Benoît, Commented Jan 26, 2015 at 0:30
@Benoît Yes it would. But only in this example, not for other cases, e.g. when ? was absent. (And OP will just keep reasking those regex questions without trying to understand what they do.) — mario
– mario, Commented Jan 26, 2015 at 0:32
The best way to go is to get the url using some DOM parser, then use regex to remove that trailing part... ie: getting everything till the first ? => regex101.com/r/mD3sB1/1 — Enissay
– Enissay, Commented Jan 26, 2015 at 0:36
This is very easy to remove everything after the ? but I was asking if it's possible to do it on the fly directly via regex — davidmarko
– davidmarko, Commented Jan 26, 2015 at 0:39

Casimir et Hippolyte · Accepted Answer · 2015-01-26 01:04:42Z

5

With DOMDocument, strpos, substr:

$dom = new DOMDocument;
$dom->loadHTML($content);

$linkNodeList = $dom->getElementsByTagName('a');

foreach($linkNodeList as $linkNode) {
    $href = $linkNode->getAttribute('href');

    if ( false !== ($offset = strpos($href, '?')) )
        $linkNode->setAttribute('href', substr($href, 0, $offset));
}

$newContent = $dom->saveHTML();

or with explode:

$linkNode->setAttribute('href', explode('?', $href)[0]);

answered Jan 26, 2015 at 1:04

Casimir et Hippolyte

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Andie2302 · Accepted Answer · 2015-01-26 01:10:08Z

0

Do you mean this behavior:

<a\s+href\s*=\s*"\K[^"?]+


$result = preg_replace('/<a\s+href\s*=\s*"\K[^"?]+/im', '', $text);

edited Jan 26, 2015 at 1:10

answered Jan 26, 2015 at 1:00

Andie2302

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AMDcze · Accepted Answer · 2015-01-26 01:14:20Z

0

As mentioned in the comments, you shouldn't get the tag with regex, you should use a parser. Nevertheless, here you go:

<a[^>]+href=("|')([^"'?]*)[^"']*\1[^>]*>

Demo: https://regex101.com/r/tV5pP8/3

edited Jan 26, 2015 at 1:14

answered Jan 26, 2015 at 1:01

AMDcze

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1 Comment

mario Over a year ago

Backreferences [^\1] don't work in character classes.

davidmarko · Accepted Answer · 2015-01-26 01:08:52Z

-1

Opps... Lack of concentration from my side :)

Solved it by myself... (It was super easy)

Here is the final line :

preg_match_all('/<a(.*?)href=("|\'|)(.*?)(\?|"|\'| )(.*?)>/s',$content,$ahref);

answered Jan 26, 2015 at 1:08

davidmarko

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