Regex | Regular Expressions | Getting anything from start of character to second instance of different character

Question

I'm having a bit of trouble getting a regular expression to work, I want it to get everything starting from a character and ending on the second instance of another character.

Ex. "matchVersion":"6.22.165.3566" should result in = 6.22

I have tried an answer that rvalvik posted here , regex match till a character from a second occurance of a different character , but it only solves part of the problem as shown below.

Example

(?<=:")[^:]+?(?=")

I would like the expresion to do someting like this instead.

Example

It have been trying to get everything after :" and before second occurance of .

Thanks for any help.

Try [`(?<=:")[^".]+\.[^".]+`](https://regex101.com/r/yDxJbf/1) — Wiktor Stribiżew, Nov 12 '16 at 18:07

score 1 · Accepted Answer · answered Nov 12 '16 at 18:12

It have been trying to get everything after :" and before second occurance of .

You may use negated character class:

(?<=:")[^".]+\.[^".]+

See the regex demo

(?<=:") - asserts that there is a :" before the current position
[^".]+ - 1+ chars other than . and "
\. - a dot
[^".]+ - 1+ chars other than . and "

Regex | Regular Expressions | Getting anything from start of character to second instance of different character

1 Answers1