golang regexp break after first match

Asked Sep 13 '16 at 14:28

Active Sep 13 '16 at 14:28

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I have the following go code:

s := `prefix<b>Group ID:</b><br/>
G123<br/>temporary<br/>suffix`

re := regexp.MustCompile(`(?s)(.*)<b>Group ID:</b><br/>(.*)<br/>(.*)`)

m := re.FindAllStringSubmatch(s, -1)
fmt.Println(m[0][2])

playground link: https://play.golang.org/p/hz2kg7Pe2Z

The above program prints \nG123<br/>temporary because the last/second <br/> after the pattern is matched, as per the regex. Is there a way to get the first <br/> to be matched so that the output will be, just: G123. In addition to the first br match, the newline/whitespace in the beginning should also be stripped. Any suggestions ?

asked Sep 13 '16 at 14:28

Sankar

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Use `.*?` instead of the `.*` instead of the first two `.*` – Wiktor Stribiżew Sep 13 '16 at 14:30
Thanks Wiktor. I was about to ping you as you have already answered some of my regex questions but could not find such an option in stackoverflow. – Sankar Sep 13 '16 at 14:38
I think my first comment is a bit corrupt, I meant to say: *Use `.*?` (lazy quantifier) instead of the first two `.*`*. – Wiktor Stribiżew Sep 13 '16 at 14:41
No worries. I got it working based on the answer in the duplicate link. Thanks for your help. – Sankar Sep 13 '16 at 15:02

golang regexp break after first match

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