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Capturing grouping for regex function replace in Go
func main() {
str := "abc foo:bar def baz:qux ghi"
re := regexp.MustCompile("([a-z]+):([a-z]+)")
result := ReplaceAllStringSubmatchFunc(re, str, func(groups []string) string {
return groups[1] + "." + groups[2]
})
fmt.Printf("'%s'\n", result)
}
import "regexp"
func ReplaceAllStringSubmatchFunc(re *regexp.Regexp, str string, repl func([]string) string) string {
result := ""
lastIndex := 0
for _, v := range re.FindAllSubmatchIndex([]byte(str), -1) {
groups := []string{}
for i := 0; i < len(v); i += 2 {
groups = append(groups, str[v[i]:v[i+1]])
}
result += str[lastIndex:v[0]] + repl(groups)
lastIndex = v[1]
}
return result + str[lastIndex:]
}
@vruzin
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vruzin commented Sep 17, 2020

bad function ))
change

groups = append(groups, str[v[i]:v[i+1]])

on

if(v[i]==-1 || v[i+1] == -1) {
	groups = append(groups, "")
}else{
	groups = append(groups, str[v[i]:v[i+1]])
}

Otherwise, if the group is not found, there will be an error.
For example:
(?:aa(xxx)bbb)?(yyy)

@StevenACoffman
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StevenACoffman commented Apr 21, 2021

See related blog post for slightly more background: http://elliot.land/post/go-replace-string-with-regular-expression-callback

Here is a go playground link (with @vruzin change): https://play.golang.org/p/yBcmsTS8Ycm

The "groups" argument to the callback is a little weird and deserves some documentation:

  1. The 0th element is the original full match
  2. The following slice elements are the nth string found by a regex parenthesized capture group (including named capturing groups)

This matches other languages to make things convenient for porting:

However, if you are not porting, it's a little unexpected.

Golang also has ReplaceAllStringFunc() which is more limited, as it returns a copy of src in which all matches of the Regexp have been replaced by the return value of function repl applied to the matched substring. If you need to look at the surrounding text, or at the position, in order to determine what to replace it with you need something else.

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