How do I use a regular expression to put the beginning of a string, preceding some user-specified character, into a group?

Question

This question was asked here, but I do not want a python specific solution. It should work in JavaScript, and any other mainstream language. It must be a regular expression.

Suppose that the delimiter was "!"

The table below shows the desired behavior for a few sample inputs

+------------------------+--------+
|         INPUT          | OUTPUT |
+------------------------+--------+
| "aaaa!bbbb"            | "aaaa" |
| "aaaa"                 | "aaaa" |
| "aaaa!bbbb!ccccc!dddd" | "aaaa" |
| "aaaa!"                | "aaaa" |
| "!aaaaa"               | ""     |
+------------------------+--------+

I tried the following to no avail:

"^[^!]*$(!)"

score 6 · Accepted Answer · answered Oct 25 '19 at 20:44

6

You don't need (!) at the end. Just use ^([^!]*).

answered Oct 25 '19 at 20:44

Barmar

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score 0 · Answer 2 · answered Oct 25 '19 at 20:44

0

You can use a capturing group for the part up to (but not including) ! and replace the match with the first group:

^([^!]+)($|!.*$)

Regex 101 demo

answered Oct 25 '19 at 20:44

Frank Schmitt

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You can just use `.*` in the second group, there's no need for the details. – Barmar Oct 25 '19 at 20:56
But the question never said anything about needing to capture the rest of the string. – Barmar Oct 25 '19 at 20:56
This won't work for the last example in the table. `+` should be `*` to allow for an empty string before `!`. – Barmar Oct 25 '19 at 20:57
@Barmar I don't' need to capture the test of the string, only the beginning up until the delimiter. – Toothpick Anemone Oct 25 '19 at 21:38
1

@SamuelMuldoon Isn't that what I said? – Barmar Oct 25 '19 at 21:42

How do I use a regular expression to put the beginning of a string, preceding some user-specified character, into a group?

2 Answers2