How to put a character as end of expression

Question

I have this regular expression

url\s*\((?:\'|"|)((\S*|\?)(?<!\'|\"))(?=\'\)|"\)|\))

It works well for what I want in these cases:

url("../../Common/common/fonts/lato/lato-black-webfont.svg?#lato_blackregular")

But when I have a situation like this it does not work the way I want it

lato_blackregular;src:url("../../Common/common/fonts/lato/lato-black-webfont.eot");src:url("../../Common/common/fonts/lato/lato-black-webfont.eot?#iefix")

How can I define ";" such as a line break, whenever you have a ";" it for the combination and start over?

Demonstration of what is happening: https://regex101.com/r/2YHl2C/1/

Use \S*? does not work because it stops combining correctly with background:url(/uploads/2019/03/0002-image(thumbnail_product).jpg)

It does not work for my case because it stops combining correctly with url (/ uploads / 2019/03/0002-image (thumbnail_product) .jpg) — Bruno Andrade, Apr 10 '19 at 16:30
No. He does not get /uploads/2019/03/0002-image(thumbnail_product).jpg — Bruno Andrade, Apr 10 '19 at 16:58
Can you please explain the requirements for the pattern? It is clear that you want to match from `url(`, then there may be an optional quote, then *some string* up to `)`. What can *some string* contain? Or, do you want to match up to `)` followed with `;` or whitepsace/end of string? Try `url\s*\(['"]?(.*?)['"]?\)(?![^;\s])`, see https://regex101.com/r/yzvxzM/1 — Wiktor Stribiżew, Apr 10 '19 at 17:01
Thank you very much, that's exactly what I want. Write as an answer and I mark as the solution of the question. — Bruno Andrade, Apr 10 '19 at 17:04

score 2 · Accepted Answer · answered Apr 10 '19 at 17:11

2

You may use

url\s*\(['"]?(.*?)['"]?\)(?![^;\s])

See the regex demo.

The pattern will match substrings starting with url(, then there may be an optional single or double quotation mark, then will capture into Group 1 zero or more chars up to the first ) followed with ; or whitepsace/end of string.

See the Regulex graph:

Details

url - an url string
\s* - 0+ whitespace chars
\( - a ( char
['"]? - 1 or 0 quotation marks
(.*?) - Group 1: any zero or more chars other than line break chars, as few as possible
['"]? - 1 or 0 quotation marks
\) - a ) char
(?![^;\s]) - there must be ;, whitespace or end of string immediately to the right of the current location.

answered Apr 10 '19 at 17:11

Wiktor Stribiżew

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Why not use a recursive way? – Jan Apr 10 '19 at 18:03
@Jan It would not yield expected results. – Wiktor Stribiżew Apr 10 '19 at 18:08
See my answer, why not? – Jan Apr 10 '19 at 18:11
@Jan It won't extract the right string from `background:url(/uploads/2019/03/0002-imagethumbnail_product).jpg)` – Wiktor Stribiżew Apr 10 '19 at 18:14
True, bit that's not a valid string. – Jan Apr 10 '19 at 18:18
@Jan `background:url('/uploads/2019/03/0002-imagethumbnail_product).jpg')` and `background:url("/uploads/2019/03/0002-imagethumbnail_product).jpg")` are – Wiktor Stribiżew Apr 10 '19 at 18:20

score 1 · Answer 2 · answered Apr 10 '19 at 17:58

1

While you already have your answer, what about the imo cleaner (recursive) way:

url
(\(
    (?P<url>(?:[^()]*|(?1))+)
\))

This only needs a tenth of the steps needed by @Wiktor's expression (~4000 vs 400), see your modified demo on regex101.com.

answered Apr 10 '19 at 17:58

Jan

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That sounds good too. The difference to @Wiktor is that his does not get the quotes. Which is the result I wanted. A group with only the url path – Bruno Andrade Apr 10 '19 at 21:36

How to put a character as end of expression

2 Answers2