CODEWITHSUNDEEP
phppreg-replace
Mertafor
Mertafor

Reputation: 414

preg_replace get only matched

I'm having trouble with preg_replace and I'm not sure that I use correct function.

I'm usign function below in order to change youtube links into youtube embed videos. But I couldn't findout how I only get matched part and remove the rest?

I mean for instance :

http://www.vimeo.com/3124234&feature=1&v=1

when I use this function it's change to matched part an embed code. But I couldnt achieve to remove "&feature=1" part.

Should I use preg_replace for that or any other function can do what I'm trying?

Cheers

function convert_videos($string) {
    $rules = array(
'#http://(www\.)?vimeo\.com/(\w+)?#i' => '<object width="450" height="320" data="http://vimeo.com/moogaloop.swf?clip_id=$2&amp;server=vimeo.com&amp;show_title=1&amp;show_byline=1&amp;show_portrait=0&amp;color=&amp;fullscreen=1"></object>'
    );

    foreach ($rules as $link => $player)
        $string = preg_replace($link, $player, $string);


    return $string;
}

Upvotes: 0

Views: 520

Answers (3)

rabusmar
rabusmar

Reputation: 4143

Your pattern doesn't match the entire string, and preg_replace will only replace what your pattern matches. You can use this pattern instead:

'#^http://(www\.)?vimeo\.com/(\w+)(&.*)?$#i'

Upvotes: 0

prdatur
prdatur

Reputation: 1009

edit: typo + backslash

You can use backreference for your (.*) values in your case

#http://(www\.)?vimeo\.com/(\w+)?#i

you can use \\1 for getting www. and \\2 for getting the values which you catch by (\w+)

or was this not what you want?

Upvotes: 0

greut
greut

Reputation: 4373

& isn't part of \w that's it. I'd go with \S (not a space) instead.

#http://(www\.)?vimeo\.com/(\S+)?#i

Upvotes: 1

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