Regex to find a string between something in javascript

Question

I have this two strings:

(username~contains~'ren'~and~status~contains~false)
(status~contains~false~and~username~contains~'ren')

i need one regex that find what is the status value after contains~

I try something like:

/(?<=status~contains~).*?(?=[)])|(?<=status~contains~).*?(?=~)/gi

pass in exemple 1: (username~contains~'ren'~and~status~contains~false)
fail in exemple 2: (status~contains~false~and~username~contains~'ren')

I need only the return true or false, but in the second exemple i get false~and~username~contains~'ren'

The fourth bird · Accepted Answer · 2019-10-29 20:17:37Z

1

The thing that you could change is to match either a ~ or )

(?<=status~contains~).*?(?=[)~])

Regex demo

Instead of using a lookbehind (which is not yet fully supported across browsers), you could also use a capturing group instead and make sure there are opening and closing parenthesis.

The value is in group 1.

\((?:[^~\r\n]+~)*status~contains~([^~\n]+)(?=[^()]*\))

\( Match an opening (
(?: Non capturing group
- [^~\r\n]+~ Match 1+ times any char except ~ or a newline
)* Close goup, repeat 0+ times
status~contains~ Match literally
( Capture group 1
- [^~\n]+ Match 1+ times any char except ~ or a newline
) Close group
(?=[^()]*\)) Positive lookahead, assert a closing )

Regex demo

const regex = /\((?:[^~\r\n]+~)*status~contains~([^~\n]+)(?=[^()]*\))/g;
const str = `(username~contains~'ren'~and~status~contains~false)
(status~contains~false~and~username~contains~'ren')`;
let m;

while ((m = regex.exec(str)) !== null) {
  // This is necessary to avoid infinite loops with zero-width matches
  if (m.index === regex.lastIndex) {
    regex.lastIndex++;
  }
  console.log(m[1]);
}

edited Oct 29, 2019 at 20:17

answered Oct 29, 2019 at 20:04

The fourth bird

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3 Comments

bla Over a year ago

the first example looks good, but the second without lookbehind don't pass, test it in reg site

The fourth bird Over a year ago

@Jhonny In the second example the value is in the first capturing group. This is an example with a different engine selected which will highlight the groups in green. regex101.com/r/YNLdOO/1 You can get those matches using the example Javascript and it does not use lookarounds.

bla Over a year ago

ooh, thanks for the explication. i will accept your answer because is the most explained

Jorge Mussato · Accepted Answer · 2019-10-29 20:12:03Z

1

Try this regex:

/(?<=status~contains)~([^~)]+)[~)]/

and get group 1 from result.

answered Oct 29, 2019 at 20:12

Jorge Mussato

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Comments

score 1 · Accepted Answer · 2019-10-29 20:22:02Z

1

True or false /^(?=.*status~contains~(true|false)).*/
Use the multi-line option and group 1 is the value.

https://regex101.com/r/N41U56/1

edited Oct 29, 2019 at 20:22

answered Oct 29, 2019 at 20:16

user12097764

Comments

Booboo · Accepted Answer · 2019-10-29 20:29:32Z

1

/(?<=status~contains~)[^~)]+(?=[~)])/g

This uses a lookbehind of: status~contains~ and a lookahead of either ~ or ). Therefore we can match 1 or more characters that are neither ~ nor ) between the lookbehind and lookahead. This is simpler and more efficient than using .*?.

let re = /(?<=status~contains~)[^~)]+(?=[~)])/g;
let text = "(username~contains~'ren'~and~status~contains~false)" +
           "(status~contains~false~and~username~contains~'ren')";
let m;
while (m = re.exec(text))
    console.log(m[0]);

See Regex Demo

edited Oct 29, 2019 at 20:29

answered Oct 29, 2019 at 20:23

Booboo

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