RegEx match numbers which have n repeating digits at the end
RegEx match numbers which have n repeating digits at the end
I want to match numbers which have exact n repeating digits at the end in Javascript. However, my regex matches n or more than n digits at the end, and I can't seem to fix that.
i.e n=3, match these:
12333
222
1233334333
12333
222
1233334333
Not match these:
11
12344
122233
123333
11
12344
122233
123333
My regexs(don't work):(d)1{2}$
[^1](d)1{2}$
(d){3}(?!1)$
(d)1{2}$
[^1](d)1{2}$
(d){3}(?!1)$
^(?:(d)(?!1{3}$))*?(d)2{2}$
3 Answers
3
Try this - match the digit right before the repeating digits start, use negative lookahead for said digit, then match 3 repeating digits:
const strs = [
'12333',
'222',
'1233334333',
'11',
'12344',
'123333'];
const re = /(^|(d)(?!2))(d)3{2}$/;
strs.forEach(str => {
if (re.test(str)) console.log('pass ' + str);
});
Way cooler than mine.
– kabanus
Jun 30 at 10:04
Yep, its awesome. I literaly never used the n thing in this part of regex, only for replace. I think this is like the en.wikipedia.org/wiki/Candle_problem : If you have been told you can use them this way, it looks so straightforward, but if not you probably never figure it.
– n3ko
Jun 30 at 13:59
Make sure there is no digit after or before:
/[^1]1{3}$|[^2]2{3}$|[^3]3{3}$|.../gm
where the ... mean repeat for each digit. You can generate this expression with a loop quite easily, and use can use a negative look behind for the first non series digit if preferred.
...
fiddle.
REPEATING digits, and only match digits at the end of the number
– Manually Overridden
Jun 30 at 9:54
@ManuallyOverridden Sorry, I missed that - fixed now.
– kabanus
Jun 30 at 10:02
I think this gives false negative for 111 222 etc.
– n3ko
Jun 30 at 10:16
@n3ko For first line, yes. You can add an optional
? to fix that.– kabanus
Jun 30 at 10:19
?
Whatever. There IS a regexp which answers the question, and without the
1,2,3... nonsens.– n3ko
Jun 30 at 13:50
1,2,3...
While this expression is almost resolve you need (it gives false positive for 4 or more repetitions) and maybe this can be fixed too ... i suggest to solve this with no or not only with regex. Maybe a [0-9]{n} and an exact repetition check with a backward loop on the string.
let re= /1{3}$|2{3}$|3{3}$|4{3}$|5{3}$|6{3}$|7{3}$|8{3}$|9{3}$|0{3}$/;
var div = document.getElementById('out');
div.innerHTML += "12333".match(re)+" -- <br/>"
div.innerHTML += "222".match(re) +" -- <br/>"
div.innerHTML += "1233334333".match(re) +" -- <br/>"
// Not match these:
div.innerHTML += "11".match(re) +" -- <br/>"
div.innerHTML += "12344".match(re) +" -- <br/>"
div.innerHTML += "122233".match(re)+" -- <br/>"
div.innerHTML += "123333".match(re)+" -- <br/>"
<div id=out></div>
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Another variant:
^(?:(d)(?!1{3}$))*?(d)2{2}$– bobble bubble
Jun 30 at 14:51