How to match all strings with a specific REGEX pattern that do not start with a defined character without using a negative lookbehind

Question

I currently have a usecase on which I want to match all http:// and https:// strings in a text but only when they do not start with a " or ' using JavaScript. If they start with another character, e.g., a whitespace, I still only want to match the http:// or https:// without the preceding character.

My current regex uses a negative lookbehind but I just realized that this is not supported in Safari:

/(?<!["'])(https?:\/\/)/gm

So what would be an alternative for using a negative lookbehind to match the following strings in a text:

http:// -> should match http://
https:// -> should match https://
xhttps:// -> should match https:// whereby x can be any character except " and '
"https:// -> should NOT match at all

[Javascript: negative lookbehind equivalent?](https://stackoverflow.com/questions/641407) — adiga, May 22 '21 at 06:04
@anubhava This would match only the first `https?://` in a text. My goal would be to match all `https?://` even when they are in the middle of a text. — Ferdinand Frank, May 22 '21 at 06:19
@anubhava Thanks, looks good! It just also matches the preceding character. So for `xhttps://` I only want to match `https://` but it matches the full `xhttps://`. — Ferdinand Frank, May 22 '21 at 06:33

score 1 · Accepted Answer · answered May 22 '21 at 20:08

No need of lookbebind here, use character class and groups:

const vars = ['http://', 'https://', 'xhttps://', '"https://']
const re = /(?:[^'"]|^)(https?:\/\/)/
vars.forEach(x => 
   console.log(x, '- >', (x.match(re) || ['',''])[1])
)

Regex:

(?:[^'"]|^)(https?:\/\/)

EXPLANATION

--------------------------------------------------------------------------------
  (?:                      group, but do not capture:
--------------------------------------------------------------------------------
    [^'"]                    any character except: ''', '"'
--------------------------------------------------------------------------------
   |                        OR
--------------------------------------------------------------------------------
    ^                        the beginning of the string
--------------------------------------------------------------------------------
  )                        end of grouping
--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    http                     'http'
--------------------------------------------------------------------------------
    s?                       's' (optional (matching the most amount
                             possible))
--------------------------------------------------------------------------------
    :                        ':'
--------------------------------------------------------------------------------
    \/                       '/'
--------------------------------------------------------------------------------
    \/                       '/'
--------------------------------------------------------------------------------
  )                        end of \1

mahdi paydarpuya · Answer 2 · 2021-05-22T09:41:47.420

0

You should use this regex: /^[a-z]+:\/\//gm

Example:

const pattern = /^[a-z]+:\/\//gm
const string = `http://
https://
xhttp://
"http://
'https://`
console.log(string.match(pattern));
// Output: [ 'http://', 'https://', 'xhttp://' ]

edited May 22 '21 at 09:41

answered May 22 '21 at 09:35

mahdi paydarpuya

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How to match all strings with a specific REGEX pattern that do not start with a defined character without using a negative lookbehind

2 Answers2