CODEWITHSUNDEEP
regexgoregex-greedy
Garlic Xu
Garlic Xu

Reputation: 173

Fail the regex match at the end of the string in Go

I'm trying to test a string against this pattern: "At least one square bracket pair, wrapping 2 digits, followed by at lease one character". For example, [11][22][33]dd should match while [11][22][33] shouldn't.

I've tried this regex: (\[\d{2}])+.+. However, when it is tested against with [11][22][33], which should have failed, it still passes that test. The first + quantifier only matches two groups [11] and [22], while the rest part [33] is matched by .+.

I thought the "greedy" behaviour of the + quantifier would exhaust all the matching segments of the group it modifies; however it seems that the regex engine would place the "exhaust all matching possibilities" principle above the "greedy quantifier" rule, not the way I'd expected.

How should I achieve my goal?

(This question is actually language-agnostic, though tagged with "golang" which is the language I'm currently using.)

Upvotes: 0

Views: 1509

Answers (2)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626870

You may use

re := regexp.MustCompile(`(?:\[\d{2}])+(.*)`)
match := re.FindStringSubmatch(s)
if len(match) > 1 {
    return match[1] != ""
}
return false

The (?:\[\d{2}])+(.*) pattern matches 1+ occurrences of [, 2 digits, ] and then captures any 0 or more chars other than line break chars into Group 1. Then, if the match was found (if len(match) > 1), true should be returned if the Group 1 value is not empty (match[1] != ""), otherwise, false is returned.

See Go demo:

package main

import (
    "fmt"
    "regexp"
)

func main() {
    strs := []string{
        "[11][22][33]",
        "___[11][22][33]",
        "[11][22][33]____",
        "[11][22]____[33]",
    }
    for _, str := range strs {
        fmt.Printf("%q - %t\n", str, match(str))
    }
}

var re = regexp.MustCompile(`(?:\[\d{2}])+(.*)`)

func match(s string) bool {
    match := re.FindStringSubmatch(s)
    if len(match) > 1 {
        return match[1] != ""
    }
    return false
}

Output:

"[11][22][33]" - false
"___[11][22][33]" - false
"[11][22][33]____" - true
"[11][22]____[33]" - true

Upvotes: 3

Klienblat Moshe
Klienblat Moshe

Reputation: 347

the regular expression you need to use is i presume goes like that:

(\[[0-9]{2}\]){1,}[a-z]{1,}

Upvotes: 0

Related Questions