Regular expression to match defined length number with leading zeros

Question

I would like to use a regular expression to match a number with a defined length and leading zeros. For example with number length of 6:

"000123" //Match
"002535" //Match
"2654" //No match
"000021" //Match

Which regular expression can I use for this?

`\d{6}` allows for only numbers with 6 digits, regardless of what the digits are. — RevanProdigalKnight, Jul 23 '14 at 21:22

Bohemian · Answer 1 · 2014-07-24T00:01:27.200

8

Surely this is the simplest:

^0\d{5}$

Although not stated in the question as a requirement, to exclude an all-zero input, use a negative look ahead for that case:

^(?!0+$)0\d{5}$

edited Jul 24 '14 at 00:01

answered Jul 23 '14 at 21:28

Bohemian

+1, duh... Much better than mine (deleting it). But needs anchors. – zx81 Jul 23 '14 at 21:30
@zx81 I've used your idea to post a working regex wich rejects `000000` ;) – AlexR Jul 23 '14 at 21:32
@Bohemian very clever ;) I usually refrain from lookahead and lookbehind in java because their implementations are / were seriously bugged. (+1) – AlexR Jul 24 '14 at 08:26

score 3 · Answer 2 · edited May 23 '17 at 10:33

If 000000 is a valid match, use the answer by @Bohemian♦.

If 000000 is not allowed, used this variation on his answer (and please accept his answer not mine):

^(?=0+[1-9])0\d{5}$

See the matches in the demo.

Explanation

The ^ anchor asserts that we are at the beginning of the string
The lookahead (?=0+[1-9]) asserts that what follows is any number of zeroes followed by one char that is a 1 or a 9
0 matches one zeroes
\d{5} matches five digits
The $ anchor asserts that we are at the end of the string

Reference

Your first RegEx matches `000000` as well, since `\d` also matches `0` — AlexR, Jul 23 '14 at 21:28

AlexR · Answer 3 · 2014-07-23T21:34:30.853

1

Actually if you allow any combination of exactly six (6) digits from 0 to 9, the regex would be

^\d{6}$

If you only allow five (5) arbitrary digits and at least one leading zero, use

^0\d{5}$

instead.

If you want to disallow 000000 you need to do lookahead:

^(?=\d{6}$)\d*[1-9]\d*

And finally, if you want to disallow 000000 and force a leading 0, use

^(?=\d{6}$)0\d*[1-9]\d*

edited Jul 23 '14 at 21:34

answered Jul 23 '14 at 21:29

AlexR

There is a much easier way to disallow `000000`. Simply use it as the first pattern of an or, then capture the other pattern. `^000000$|^(0\d{5})$` – Adam Smith Jul 24 '14 at 00:06

3 Answers3