CODEWITHSUNDEEP
regex
Naigel
Naigel

Reputation: 9644

regexp match anything before and after a word, if it exists

I have sentences like these

word1 word2 abc word3 word4
word1 word2 word3 word4

And I want to get the matches:

  1. everything before abc
  2. abc or nothing
  3. everything after abc

I need 3 different matches because $1 and $3 will be copied to the output as they are, while $2 should be modified. I expect $2 not always to be there, as well as $3).

My current solution is this one

(.+?(?=abc))(abc)?(.*)

This doesn't work if abc is not part of the sentence. How can I express "match everything until that word if it exists" with regex?

P.S. I am using JS (not sure if it's relevant)

Upvotes: 1

Views: 588

Answers (1)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626738

You may use

^(.*?)(?:(abc)(.*))?$

If the string can have line breaks, use

^([\w\W]*?)(?:(abc)([\w\W]*))?$

See the regex demo. In JS, [\w\W] can be replaced with [^] to match any chars including line break chars.

Details

  • ^ - start of string
  • (.*?) - Group 1: any zero or more chars other than line break chars as few as possible
  • (?:(abc)(.*))? - an optional greedy pattern matching one or zero times:
    • (abc) - Group 2: abc
    • (.*) - Group 3: any zero or more chars other than line break chars as many as possible
  • $ - end of string.

Since the first group pattern is non-greedy, it will keep expanding from the start of the string until the first occurrence of the optional non-capturing group pattern (if present) or end of string, and if there is abc, the second and third groups will be populated, else the whole string will land in Group 1.

Upvotes: 1

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