How to match regex the regex only at the beginning of each line?

Question

I am using the regex to match the word with colon present only at the beginning of each line:

/[^\r\n:-][^\r\n:]*:\s*/g

Following is the text I want to match with. The problem is, it also matches the 2nd Blah: on 3rd row. Which I don't want.

Blah Blah: dsfdffsdfsdfsds

Blah: dsffsffsdfsdfds

Blah: dsffsfds Blah:fsfsfssdfs

You should look at non-greedy quantifiers and the usage of `^` to match the beginning of your string. — K.Dᴀᴠɪs, Dec 03 '18 at 05:39
@K.Dᴀᴠɪs Not much experience with regex, how can I do that ? — Hari Das, Dec 03 '18 at 05:40
@K.Dᴀᴠɪs A negative character set with a greedy quantifier followed by a character not permitted by the set is definitely more efficient than lazy repetition - fewer steps are required. — CertainPerformance, Dec 03 '18 at 05:41
/^[^\r\n:-][^\r\n:]*:\s*/gm worked for me. Somebody post it as answer I will accept it. — Hari Das, Dec 03 '18 at 05:45

score -1 · Answer 1 · answered Dec 03 '18 at 05:42

-1

you can use this regex ^[:] to do it.

answered Dec 03 '18 at 05:42

kumaranc

1 Answers1