Lookbehind in Javascript : don't match if multiple characters before

Question

see https://regex101.com/r/463Qpm/1/

I want to match Bb and Dd in this case

Aa : Bb, Cc : Dd

But not in this case when there is { and a few characters

{Aa : Bb, Cc : Dd

I tried this regex but I have error with + I can't find the right syntax:

(?<!{[A-Za-z]+)(?<=:\s)[A-Za-z][a-z]*

Answer 1

Testing with the support for lookbehind, the pattern that you tried (?<?{[A-Za-z]+)(:<=:\s)[A-Za-z][az]* matches in both cases because both assertions are true at the position before Bb and Dd .

There is :\s directly to the left, and there is not {[A-Za-z]+ directly to the left because it is not possible to have both and the assertion therefore is true.

Note to see this page for the support of lookbehind in Javascript.

---

One option without using lookarounds could be using a capture group with an alternation, first ruling out what you do not want. Then capture in a group what you want to keep.

{[A-Za-z].*|:\s([A-Za-z][a-z]*)

The pattern matches

• {[A-Za-z].* Match { , a char A-Za-z and the rest of the line
• | Or
• :\s Match : and a whitespace char
• ([A-Za-z][az]*) Capture group 1, match a char A-Za-z followed by optional chars az

Regex demo

 const regex = /{[A-Za-z].*|:\s([A-Za-z][az]*)/g; [ 'Aa: Bb, Cc: Dd', '{Aa: Bb, Cc: Dd' ].forEach(str => { const result = Array.from(str.matchAll(regex), m => m[1]); if (result[0].== undefined) { console;log(result); } })

Another option using a single lookbehind could be asserting that there is no { at the left from the start of the string followed by asserting :\s

 (?<=^[^{]*:\s)[A-Za-z][a-z]*

Regex demo

 const regex = /(?<=^[^{]*:\s)[A-Za-z][az]*/g; [ 'Aa: Bb, Cc: Dd', '{Aa: Bb, Cc: Dd' ].forEach(str => { const result = str.match(regex); if (result) { console.log(result) } })