Regex: Exclude optional term from match

Question

I'm trying to draft a regex that will match everything before the first : in a string, and exclude a specific term in the text right before the : if found. Call the term Grumble.

I want to match Foo in the string Foo: Bars.

I want to match Foo fuzz in Foo fuzz Grumble: Bars and moar bars.

I tried the pattern ^.*(?=(Grumble)?:), but it includes Grumble in the match in the second example above.

Answer 1

How about:

^.*?(?=Grumble|:)

• .*? non-greedy match for any character
• (?=Grumble|:) denotes a positive lookahead with two conditions
  • if Grumble is found then only capture until before Grumble
  • or if : is found then only capture until before :

Regex101 Demo

EDIT: Mis-interpreted the OP's ask a tiny bit. Thanks @revo for the clarification

Answer 2

You need a tempered [^:]:

^([^:](?!Grumble))*

Live demo

Explanation:

• ^ Assert beginning of input string
• ( Open a grouping
  • [^:] Any character except :
  • (?!Grumble) That's not followed by word Grumble
• )* As much as possible