preg matching all hrefs and srcs in a string

Question

I'm trying to extract all the hrefs and srcs in a string like this :

$content = "
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium
voluptatum deleniti Image: <img src = 'http://example.com/check-3.png' /> Link: <a href ='http://example.com/test.xls'>test.xls</a>";

Basically what I want to do is change example.com to a to a different domain name (say test.com) and then extract all the filenames from hrefs and srcs. I was able to do the domain name replacement with a simple str_replace but now I'm stuck trying to extract the hrefs and srcs.

Here's what I tried using :

$regex = "/src=[\"' ]?([^\"' >]+)[\"' ]?[^>]*>.*?href=[\"' ]?([^\"' >]+)[\"' ]?[^>]*>/i";

This seems to work if there is no space between src (or href) and the = (e.g. ) but if there is space (e.g. ) it does not work. I've tried adding the space character but that fails the preg match. I don't want to use a heavy library like simple HTML dom, besides i don't think it will work as its not a proper HTML document. It's a string coming out of ckeditor.

"If I had a coin each time anybody tried to parse HTML with regexes..." - I advice you to with DomDocument and XPath - see stackoverflow.com/questions/1933631/how-do-i-parse-partial-html. — moonwave99
– moonwave99, Commented Aug 29, 2012 at 16:50

Andrew Cheong · Accepted Answer · 2012-08-29 16:47:16Z

1

Why not just add quantifiers on the space?

$regex = "/src *= *[\"' ]?([^\"' >]+)[\"' ]?[^>]*>.*?href=[\"' ]?([^\"' >]+)[\"' ]?[^>]*>/i";
               ^  ^

answered Aug 29, 2012 at 16:47

Andrew Cheong

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2 Comments

Ashesh Over a year ago

Why is there a space after the = ? Shouldn't it be /src*=* meaning any number of spaces betfore and after = ?

gen_Eric Over a year ago

The * modifies the previous character. src *= * means: "'src 'followed by any amount of spaces. followed by '=' followed by any amount of spaces.". src*=* means: "'sr' followed by any number of 'c's followed by any number of '='s".

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