Regex for finding value inside url in background-url

Question

I have $('#tab1').css("background-image", "url(../CARImages/Icons/tab1.png)"); I need to extract ../CARImages/Icons/tab1.png from this.
I may have background-image:url(../Images/Themes/tile_bg.png); syntax in css. I need to extract ../Images/Themes/tile_bg.png from this.
Syntax can be background:url("../Images/bg_topnav.gif"); I need to extract ../Images/bg_topnav.gif from this.

4.. Syntax can be background:url('../Images/bg_topnav.gif'). I need to extract ../Images/bg_topnav.gif from this.

Can I have same Regex for the above syntax to extract image reference.

Output should not contain inverted comas, which are present in the original syntax.

You have at least 2 questions that you've duplicated each -- please don't do this. If your question isn't getting attention, edit it with additional details and it will be bumped back up in the queue. Posting it a second time isn't acceptable. — jprofitt
– jprofitt, Commented Feb 28, 2014 at 5:26

aelor · Accepted Answer · 2014-02-13 12:26:05Z

2

use this:

/url\([\"\']?(.*?)[\"\']?\)/

demo : http://rubular.com/r/EXrGNXnmnn

for eg :

var str = '$(\'#tab1\').css("background-image", "url(../CARImages/Icons/tab1.png)");';
var res = /url\([\"\']?(.*?)[\"\']?\)/.exec(str);
console.log(res[1]);

regex with look behind

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

edited Feb 13, 2014 at 12:26

answered Feb 13, 2014 at 11:27

aelor

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7 Comments

Kabir Over a year ago

I dont want url to get prefixed to the image reference. I just need the value inside parenthesis. In this, I am getting url("../Images/bg_topnav.gif"). I just want ../Images/bg_topnav.gif.

aelor Over a year ago

you will get that only if you look at the matched groups carefully

Kabir Over a year ago

I had this Regex (?<=url().*?(?=)), But I used to get inverted comas also with this regex. If possible, can you please edit this regex, so as to not get inverted comas, but only Image reference.

Kabir Over a year ago

This wont work in this case, Please provide Regex to get only Image reference inside url:("").

Andre Figueiredo Over a year ago

You can use this regex to find value of background-image attribute, then after you extract url(), url(''), etc.

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