Parsing regex with escaped pipe delimiter

Question

Im trying to parse

|123|create|item|1497359166334|Sport|Some League|\|Team\| vs \|Team\||1497359216693|

With regex (https://regex101.com/r/KLzIOa/1/)

I currently have

[^|]++

Which is parsing everything correctly except \|Team\| vs \|Team\|

I would expect this to be parsed as |Team| vs |Team|

If i change the regex to

[^\\|]++

It parses the Teams separately instead of together with the escaped pipe

Basically i want to parse the fields between the pipes however, if there are any escaped pipes i would like to capture them. So with my example i would expect

["123", "create", "item", "1497359166334", "Sport", "Some League", "|Team| vs |Team|", "1497359216693"]

CertainPerformance · Accepted Answer · 2020-01-18T10:20:11.463

3

You can alternate between:

\\. - A literal backslash followed by anything, or
[^|\\]+ - Anything but a pipe or backslash

(?:\\.|[^|\\]+)+

https://regex101.com/r/KLzIOa/2

Note that there's no need for the possessive quantifier, because no backtracking will occur.

If you also want to replace \|s with |s, then do that afterwards: match \\\| and replace with |.

edited Jan 18 '20 at 10:20

answered Jan 18 '20 at 10:14

CertainPerformance

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In this situation `\|Team\| vs \|Team\|` is captured. Would that mean i would have to remove the `\` myself after manually? – Jack Wilkinson Jan 18 '20 at 10:17
A regular expression can generally *match* substrings, or *replace* part of a match, but not do both at the same time. If you want to replace the `\|`s with `|`s, you'll have to do that afterwards. – CertainPerformance Jan 18 '20 at 10:18
what if I want to capture empty values as well? I mean when pipe is placed after another one – Egor Baranov Jul 04 '23 at 17:42
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@EgorBaranov Alternate the whole pattern with another that matches between two pipes. `(?<=\|)(?=\|)` – CertainPerformance Jul 04 '23 at 19:19

Niet the Dark Absol · Answer 2 · 2020-01-18T10:18:44.873

1

To handle escaping, you should match a backslash and the character after it as a single "item".

(?:\\.|[^|])++

This conveniently also works for escaping the backslashes themselves!

To then remove the backslashes from the results, use a simple replacement:

Replace: \\(.)
With: $1

edited Jan 18 '20 at 10:18

answered Jan 18 '20 at 10:14

Niet the Dark Absol

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In this situation `\|Team\| vs \|Team\|` is captured. Would that mean i would have to remove the `` myself after manually? – Jack Wilkinson Jan 18 '20 at 10:17
Yes, you would indeed need to do that. Replace `\\(.)` with `$1`. – Niet the Dark Absol Jan 18 '20 at 10:18

Toto · Answer 3 · 2020-01-18T10:28:04.547

1

Use:

(?:\\\||[^|])+

Demo & explanation

edited Jan 18 '20 at 10:28

answered Jan 18 '20 at 10:15

Toto

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@Tomalak: True, I've just copied OP's regex. – Toto Jan 18 '20 at 10:28

Parsing regex with escaped pipe delimiter

3 Answers3