preg_replace background URL in CSS attribute

Question

I'm using file_get_contents() to get source code of a page but I fail in some websites that uses CSS like this:

background: url("/media/image")
background: url(/media/image)
background: url('/media/image')
// etc...

Now, I want to know how can I edit that CSS and add my website, so it will look like this:

background: url("http: //example.com/media/image")
background: url('http: //example.com/media/image')
background: url(http://example/media/image)
// etc...

. Here is my code:

$regex = "-(src\s*=\s*['\"])(((?!'|\"|http://|https://|//).)*)(['\"])-i";

I'm doing this on the src attribute of an HTML tag. I hope I'm helping

I'm not sure what you mean, your comments are getting confusing. I thought this is what you were trying to accomplish echo preg_replace('~url$("|\')?/?(.*?)(?:\1)?$~', 'url($1http://www.example.com/$2$1)', $css) . "\n";, no? Where $css is one of your three examples. — chris85
– chris85, Commented Jun 30, 2015 at 1:37
@chris85 -(src\s*=\s*['\"])(((?!'|\"|http://|https://|//).)*)(['\"])-i how can I add href attribute to src please? — the helper
– the helper, Commented Jun 30, 2015 at 1:47
I want this regex code -(src\s*=\s*['\"])(((?!'|\"|http://|https://|//).)*)(['\"])-i to be working with href attribute too not only src :/ — the helper
– the helper, Commented Jun 30, 2015 at 1:48
Perhaps you should clerify what exactly you want? What are you trying to accomplish? Are you trying to edit the source code of what's imported with file_get_contents()? To me it pretty much sounds like you're importing someone elses website and replace the stuff you can't import. Something smells a bit to much like phishing here.. — icecub
– icecub, Commented Jun 30, 2015 at 2:03

5ervant - techintel.github.io · Accepted Answer · 2015-07-01 04:55:52Z

2

This $regex will replace the background: url(["|']?$URL["|']?) with ["|']?$img["|']?:

$homepage = file_get_contents('http://www.example.com/');
$img = "http://placehold.it/350x150";

$regex = '/(background: url\((["|\']?))(.+)(["|\']?\))/';
$replacement = "$1$img$4";
$homepage = preg_replace($regex, $replacement, $homepage);

echo $homepage;

answered Jul 1, 2015 at 4:55

5ervant - techintel.github.io

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5 Comments

Gifford N. Over a year ago

This regex: $regex = '/(background-image: url$\')(.+)("|\'$)/'; with this replacement: $replacement = "$1$img$3"; worked better for me. Basically leaves background-image: url(' and ') in place and insterts the URL between, not touching the quotation marks.

5ervant - techintel.github.io Over a year ago

@GiffordN. Is that regex meets the OP requirements?

Gifford N. Over a year ago

Probably not for the op, that's why I left as comment instead of editing your post. Unfortunately your regex caused a CSS error because it retains the first ' but not the second ', just the ) so it broke the CSS. I was just addressing that issue for whoever may experience a similar problem using your code, since this is how I fixed the issue.

5ervant - techintel.github.io Over a year ago

@GiffordN. I think your url$\')(.+)("|\'$ should become url$\'(.+)\'$.

5ervant - techintel.github.io Over a year ago

@GiffordN. You're welcome, but you should give an upvote if you think the post have helped you..

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