regex (regular expression): parse the optional end of the string if it exists

Question

Please help me to find the end of the string and parse it.

I have the string like this "Some words" and "Some another words||MyTag". Please help me with regex to check any of this strings. But in the second case please extract "MyTag". So please make two groups: "Some words" or "Some another words" and MyTag or empty string.

I tried "^([\W\w]+)(?:||(.*))?" but without any success.

The fourth bird · Accepted Answer · 2022-01-21 14:14:35Z

1

You can create 2 capture groups:

^(.+?)(?:\|\|(.+))?$

^ Start of string
(.+?) Capture group 1 Match any character except newlines as least as possible
(?: Non capture group
- \|\|(.+) Match || and capture 1+ chars in group 2
)? Close non capture group and make it optional
$ End of string

If the second part is not present, then there will be no group 2 value.

Regex demo

You might also consider to split on ||

Another option without a non greedy dot is to not allow matching |, only when it is not directly followed by | using a negative lookahead.

^([^\r\n|]*(?:\|(?!\|)[^\r\n|]*)*)(?:\|\|(.+))?$

Regex demo

edited Jan 21, 2022 at 14:14

answered Jan 21, 2022 at 14:07

The fourth bird

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