Regex to extract CSS URL function parameter

Question

I am trying to get the url part of the following line:

url(images/ui-bg_highlight-soft_75_cccccc_1x100.png)

So the required part is if images/ui-bg_highlight-soft_75_cccccc_1x100.png.

I currently have this:

url\((?<url>.*)\)

But it seems to suffocate in the following example:

url(images/ui-bg_flat_0_aaaaaa_40x100.png) 50% 50% repeat-x; opacity: .30;filter:Alpha(Opacity=30)

The result is images/ui-bg_flat_0_aaaaaa_40x100.png) 50% 50% repeat-x; opacity: .30;filter:Alpha(Opacity=30...

I would like to make sure that it supports as many options as possible (extra spaces, etc.).

Thank!
Kiron

Edit

Also, I need to ignore optional quotes or "...

Now it looks like this:

url\(['|"]?(?<url>[^)]+)\)

I can't make him stop / ignore the last quote ...

Answer 1

- *, .

url\((?<url>[^)]*)\)

url\((?<url>.*?)\)

.

: , ,

url\(['"]?(?<url>[^)]+?)['"]?\)

| .

Answer 2

.* [^\)]+

Answer 3

:

url\(['"]?(?<url>.*?)['"]?\)

:

url\s*\(['"\s]*(?<url>.*?)['"\s]*\)

, , CSS .

: | ['|"] OR, |. OR .

Answer 4

url\((. * \. png)

url\\((.[^)]*)\\)