How to tell a RegEx to be greedy on an 'Or' Expression

Question

Text:

[A]I\'m an example text [] But I want to be included [[]]
[A]I\'m another text without a second part []

Regex:

\\[A\\][\\         


        

Answer 1

You may use

\[A][\s\S]*?(?=\[A]|$)

See the regex demo.

Details

• \[A] - a [A] substring
• [\s\S]*? - any 0+ chars as few as possible
• (?=\[A]|$) - a location that is immediately followed with [A] or end of string.

In C#, you actually may even use a split operation:

Regex.Split(s, @"(?!^)(?=\[A])")

See this .NET regex demo. The (?!^)(?=\[A]) regex matches a location in a string that is not at the start and that is immediately followed with [A].

If instead of A there can be any letter, replaces A with [A-Z] or [A-Z]+.

Answer 2

I have changed your regex (actually simpler) to do what you want:

\[A\].*\[?\[\]\]?

It starts by matching the '[A]', then matches any number of any characters (greedy) and finally one or two '[]'.

Edit:

This will prefer double Square brackets:

\[A\].*(?:\[\[\]\]|\[\])