preg_match: javascript url string

Question

I am trying to construct a REGEX pattern, to convert relative URLs to absolute URLs in javascript scripts.

Example: I would like to replace ALL instances of the following (taken from js script):

url('fonts/fontawesome-webfont.eot?v=4.2.0');

And return the following:

url('http://example.com/fonts/fontawesome-webfont.eot?v=4.2.0');

Example pattern that works for HTML tags (link):

$pattern = "#(<\s*a\s+[^>]*href\s*=\s*[\"'])(?!http)([^\"'>]+)([\"'>]+)#"

Preferred use of this pattern:

$result = preg_replace($pattern,'$1http://example.com/$2$3', $result);

The closest I have managed to guess (unsuccessfully) is:

$pattern = "#(url\s*=\s*[\"'])(?!http)([\"'])#";

You don't have an =s in your search string, or not in the form url =.. Try (url\(["'])(?!http). — chris85
– chris85, Commented Jun 16, 2016 at 19:21

Andreas Louv · Accepted Answer · 2016-06-16 19:36:35Z

2

Something like this might work for you:

/(?<=url\((['"]))(?!http)(?=.*?\1)/

With replacement:

http://example.com/

Regex Demo

The above regex will match the position just after url(' where the quote can also be a double quote.

(?<=...) # is a positive lookbehind
(?!...)  # is a negative lookahead
(?=...)  # is a positive lookahead
\1       # refers to capturing group 1, in this case either ' or "

Note that lookbehinds isn't supported in JavaScript but are in PHP.

edited Jun 16, 2016 at 19:36

answered Jun 16, 2016 at 19:25

Andreas Louv

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2 Comments

leenremm Over a year ago

Thanks, this is a great help!

Andreas Louv Over a year ago

@LeenRemmelzwaal I hope I have given enough information on how this regex works, you can consider looking up lookarounds on google :-)

acupofjose · Accepted Answer · 2016-06-16 19:30:16Z

0

My testing shows that this will work: /url\$['\"](?!http[s]?:\\/\\/)(.+)['|\"]\$/

So your replace should look like:

preg_replace("/url\$['\"](?!http[s]?:\\/\\/)(.+)['|\"]\$/",'http://example.com/$1', $result)

Not a regex master tho - so take this with a grain of salt

edited Jun 16, 2016 at 19:30

answered Jun 16, 2016 at 19:22

acupofjose

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1 Comment

chris85 Over a year ago

A character class is a list of optional characters. So the s doesn't need to be in one and the ['|\"] doesn't need the or the or actual makes the pipe optional as well. You also could capture the first quote type and use a back reference for the second to make sure you match. e.g. regex101.com/r/sY5sN2/1 (if using single quotes for the regex encapsulation, if double the ``s need to be doubled)

leenremm · Accepted Answer · 2016-06-16 19:53:55Z

0

The solution I've used (thanks to help from andlrc and chris85):

$result = preg_replace("#(?<=url\((['\"]))(?!https?)#",'http://example.com/', $result);

edited Jun 16, 2016 at 19:53

answered Jun 16, 2016 at 19:36

leenremm

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4 Comments

chris85 Over a year ago

The [s] is requiring an s did you mean to make the s optional?

leenremm Over a year ago

Yes please. How do you recommend I make the 's' optional?

Andreas Louv Over a year ago

Use s? the question mark will make s match zero or one time

chris85 Over a year ago

Like this https?. Also your second and third capture groups are empty, replacing with http://example.com/ will yield the same result. eval.in/590456 The ? makes the previous character, or group, optional.

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