Is there a way to match repeated characters that are not in a sequence?
Let's say I'm looking for at least two 6s in the string.
var string = "61236";
I can only find quantifiers that match either zero or one or characters in a sequence.
Thanks.
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Can they be in sequence as well?– user557597Commented Jan 16, 2014 at 16:29
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Yes. But does not have to.– Karl PokusCommented Jan 17, 2014 at 8:57
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4 Answers
Here is a regexp that matches a character that is followed somehwere in the string by the same digit:
/(.)(?=.*?\1)/
Usage:
var test = '0612364';
test.match(/(.)(?=.*?\1)/)[0] // 6
DEMO: https://regex101.com/r/xU1rG4/4
Here is one that matches it repeated at least 3 times (total of 4+ occurances)
/(.)(?=(.*?\1){3,})/
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1Maybe add a lazy qualifier so that the
.*doesn't match all the way to the right before looking for a backreference match:/(\d)(?=.*?\1)/Commented Jan 16, 2014 at 16:22 -
@MikeSamuel Added, thanks. I can't seem to add a maximum amount of occurances to the regex. Does anyone have any ideas? I tried [^\1] instead of ., but that is the character 1, tried {3,5}, but that just matches the \d with ..– TibosCommented Jan 16, 2014 at 16:28
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to match a maximum, you need to ensure that the
.*?does not match\1:(?:(?!=\1).)*?Commented Jan 16, 2014 at 16:33 -
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@MikeSamuel I tried that in the meantime, but it doesn't seem to work: regex101.com/r/iL2dQ1 Perhaps i made a stupid mistake?– TibosCommented Jan 16, 2014 at 16:41
Have a try with this regex:
/(.).*?\1/
Match 6 and anything up until another 6
!!"61236".match("6.*6")
// returns true
Match 6 and anything up until another the same as the first group (which is a 6)
!!"61236".match(/(6).*\1/)
// returns true
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1@KarlPokus it converts the value returned by match to a boolean. If a match is found, the resulting array gets converted to true, otherwise null gets converted to false.– TibosCommented Jan 16, 2014 at 16:44
Characters that are not in sequence and appear more than once:
/(.)(.+\1)+/
