Regex to get everything before the first slash except between quotes

Question

I am looking to create a Regex to extract everything before the first slash except if it is in single or double quotes. Currently, I have:

^(.*?)/

Now, I am at a lost. Based on the different texts below, I just want the bolded part below:

Text

abc,def,ghi,jkl,mno /123
/abc,def,ghi,jkl,mno 123
abc,/def,"/ghi",jkl,mno /123
abc,def,"/ghi",jkl,mno /123
abc,def,'/ghi',jkl,mno /123

Wiktor Stribiżew · Accepted Answer · 2018-08-02T20:59:03.003

1

You may use

^(?:[^/"']|"[^"]*"|'[^']*')+

See the regex demo

Details

^ - start of string
(?:[^/"']|"[^"]*"|'[^']*')+ - 1 or more occurrences of
- [^/"'] - any char other than /, " and '
- | - or
- "[^"]*" - a ", any 0+ chars other than ", and then "
- | - or
- '[^']*' - a ', any 0+ chars other than ', and then '

edited Aug 02 '18 at 20:59

answered Aug 02 '18 at 20:57

Wiktor Stribiżew

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So beautiful! ;) – gtgaxiola Aug 02 '18 at 20:58
Why use a non-capturing group? Is there a benefit? – Nick Aug 02 '18 at 21:50
@Nick See [Are non-capturing groups redundant?](https://stackoverflow.com/questions/31500422/are-non-capturing-groups-redundant/31500517#31500517) – Wiktor Stribiżew Aug 03 '18 at 06:36

score 0 · Answer 2 · answered Aug 02 '18 at 20:55

0

How about:

^(.*?)\/(?=([^\"\']*\"[^\"\']*\")*[^\"\']*$)

See Regex Demo

Using Group 1

answered Aug 02 '18 at 20:55

gtgaxiola

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Regex to get everything before the first slash except between quotes

2 Answers2