Comments (2)
To accomplish this you'll want to use FindNextMatch
since it'll re-use the converted []rune
for you instead of re-generating a new []rune
each time.
func regexp2FindAllString(re *regexp2.Regexp, s string) []string {
var matches []string
m, _ := re.FindStringMatch(s)
for m != nil {
matches = append(matches, m.String())
m, _ = re.FindNextMatch(m)
}
return matches
}
I'm going to close this issue since this looks like a pretty simple function to implement. However, I will update the README.md to give an example usage of FindNextMatch
.
from regexp2.
OK, it turns out I was wrong and both Index
and Length
are in runes. They are consistent. So my only feedback is points 2 & 3 above 😃
Here is the final working regexp2FindAllString
implementation for anyone else who may be looking for one:
func regexp2FindAllString(re *regexp2.Regexp, s string) []string {
var (
matches []string
r = []rune(s)
)
for {
match, _ := re.FindStringMatch(string(r))
if match == nil {
break
} else {
matches = append(matches, match.String())
r = r[match.Index+match.Length:]
}
}
return matches
}
(I would send this as a PR, but I realize it is incomplete without the n
parameter and proper tests)
from regexp2.
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from regexp2.