How to enforce greediness preceding a conditional match in a regular expression

Question

I am trying to create a regex that will parse a string that is of either one of the two forms:

name (tag) @ name (tag) 
name @ name
* name and tag can be multiple words separated by spaces

The regex I wrote for this is:

(?<a>.+?)(\s\(.+\))?\s@\s(?<b>.+?)(\s\(.+\))?

This fails when parsing a string of the first form which includes tags surrounded by parentheses. Specifically, it will only match one character in the 2nd name, essentially ignoring the 2nd tag group. It seems as though the greediness of the ? quantifier for the 2nd tag group is ignored, or of lower priority than the lazy quantifier of the group.

Furthermore, if I add an end of string anchor, $, to the end of the pattern, it will put everything on the left side of @, including the tags, into the a group and everything to the right side into the b group, which again seems unexpected.

Why is this happening? Is there a way to account for the two groups without branching?

If it's not inside of other text, you could simply include "end of string" in your regex: `(?.+?)(\s(.?))?\s@\s(?.+?)(\s(.?))?$`. — Christoph Lütjen, Oct 13 '20 at 18:11
Try `(?\w+(?:\s+\w+)*)(?:\s*$[^()]*$)?\s*@\s*(?\w+(?:\s+\w+)*)(?:\s*$[^()]*$)?` — Wiktor Stribiżew, Oct 13 '20 at 18:25
Or perhaps... `^(?[^(]+\b)(\s*\(.+)?\s*@\s+(?[^(]+\b)(\s*\(.+)?\s*$` ? (should work with multiple whitespaces too) — Christoph Lütjen, Oct 13 '20 at 18:43
@WiktorStribiżew after taking some ideas from the comments I ended up with this regex: (?[^(]+?)(\s$.+$)?\s@\s(?[^(]+?)(\s$.+$)?$ — cubesnyc, Oct 14 '20 at 17:49
So, do I understand it right you want to match an entire string with this pattern? — Wiktor Stribiżew, Oct 14 '20 at 18:35

score -1 · Answer 1 · answered Oct 13 '20 at 19:01

-1

I don’t think it’s a good idea to use .+ to capture a name because all characters including whitespace will be captured, it’s better to use \w+ (word).

Also if you want to capture brackets they should be escaped \( otherwise you’re just defining another capture group. Using .? inside the brackets means zero or 1 character, you should use + (1 or more)

This is why after the @ your regex only on captures 1 character, the ? should come after the closing group bracket.

I tested the following on regexr.com and it works:

(?<a>\w+)(\s\(\w+\))?\s@\s(?<b>\w+)(\s\(\w+\))?

answered Oct 13 '20 at 19:01

Mark van den Beemt

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As for "*I tested the following on regexr.com*", please bear in mind that regexr does not support .NET regex syntax, so it is not a proof the regex works in the target environment. – Wiktor Stribiżew Oct 13 '20 at 19:42
Indeed, thanks for the reminder @Wiktor Stribiżew. You can use https://www.regular-expressions.info/reference.html to see if a feature is supported on .NET and also you can use http://regexstorm.net/tester to test in a .NET environment. – Mark van den Beemt Oct 14 '20 at 08:35
I have not tested this pattern, but I do not think it will work if or contain spaces, modifying /w to .* to account for spaces, will then capture the parenthesized tags in the a and b groups. @MarkvandenBeemt – cubesnyc Oct 14 '20 at 17:51
I didn't understand from your first example you wanted to include spaces in the name. But the * is greedy so your first capture group would capture everything up to the \s@...Also do you want to trim off any leading spaces? If I understand correctly the name can contain anything except @ and brackets, right? If so then I think (?([^\s()@]+ *)+) ($.*$ )?@ (?([^\s()@]+ *)+) *($.*$)? would accomplish that and also trim of leading and trailing spaces in the name, leaving only the spaces in the middle of the names. (I used spaces instead of \s because \s also includes line breaks.) – Mark van den Beemt Oct 14 '20 at 22:25

How to enforce greediness preceding a conditional match in a regular expression

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