Regex match valid Phone Number

Question

I'm quite new to regex, and not sure what I'm doing wrong exactly.

I'm looking for a regex that match the following number format:

Matching requirements:

Must start with either 0 or 3
Must be between 7 to 11 digits
Must not allow ascending digits. e.g. 0123456789, 01234567
Must not allow repeated digits. e.g. 011111111, 3333333333, 0000000000

This is what I came up with:

^(?=(^[0,3]{1}))(?!.*(\d)\1{3,})(?!^(?:0(?=1|$))?(?:1(?=2|$))?(?:2(?=3|$))?(?:3(?=4|$))?(?:4(?=5|$))?(?:5(?=6|$))?(?:6(?=7|$))?(?:7(?=8|$))?(?:8(?=9|$))?9?$).{7,11}$

The above regex fails the No. (4) condition. Not sure why though.

Any help would be appreciated.

Thanks

Don't use regex to validate telephone numbers? https://stackoverflow.com/a/4338544/989920 — evolutionxbox, Feb 25 '20 at 12:08
Its a requirement to use regex unfortunately. The application will be used in a specific location. @evolutionxbox — JOJO, Feb 25 '20 at 12:11
Do you mean like this? `^(?!.*(\d)\1)(?!(?:0(?=1|$))?(?:1(?=2|$))?(?:2(?=3|$))?(?:3(?=4|$))?(?:4(?=5|$))?(?:5(?=6|$))?(?:6(?=7|$))?(?:7(?=8|$))?(?:8(?=9|$))?9?$)[03]\d{6,10}$` https://regex101.com/r/4EWkQJ/1 — The fourth bird, Feb 25 '20 at 12:27
@Thefourthbird Unfortunately, it doesnt match the following case: `022682673` — JOJO, Feb 25 '20 at 12:50
@JOJO You could add a quantifier `{3}` to repeat the backreference to assert not 4 times the same char https://regex101.com/r/MWPzfO/1 — The fourth bird, Feb 25 '20 at 12:52

score 0 · Answer 1 · answered Feb 25 '20 at 13:00

A solution for a JS flavor of PCRE would be

/^[03](?!123456(7(8(9|$)|$)|$))(?!(?<d>.)\k<d>+$)[0-9]{6,10}$/

Explanations

^[03] starts at the beginning of the string, then reads either 0 or 3
(?!123456(7(8(9|$)|$)|$)) makes sure that, after this first char, there is no sequence (if a sequence can be read, then the negative lookahead fails
(?!(?<d>.)\k<d>+$) is another negative lookahead : it ensures that the first char read (flagged d) is not repeated again and again until end of string
[0-9]{6,10}$/ finally reads 6 to 10 digits (first one already read)

A few tests:

"0123456789: No match"
"01234567: No match"
"01234568: No match"
"011111111: No match"
"33333333: No match"
"333333233 is valid"
"042157891023 is valid"
"019856: No match"
"0123451245 is valid"

Because a sequence would have been 0**1**23456789. Without the 1 there is a jump between 0 and 2 hence it's not a sequence, is it? — David Amar, Feb 26 '20 at 17:07

The fourth bird · Accepted Answer · 2020-02-25T16:56:31.923

0

A few notes about the pattern that you tried

You can omit the {1} and the comma in [0,3]
In the lookahead (?!.*(\d)\1{3,}) the (\d) is the second capturing group because this (?=(^[0,3]{1})) contains the first capturing group so it should be \2 instead of \1
In the lookahead, you can omit the comma in {3,}
In the match itself you use .{7,11} where the dot would match any character except a newline. You could use \d instead to match only digits

You pattern might look like

^(?=(^[03]))(?!.*(\d)\2{3})(?!^(?:0(?=1|$))?(?:1(?=2|$))?(?:2(?=3|$))?(?:3(?=4|$))?(?:4(?=5|$))?(?:5(?=6|$))?(?:6(?=7|$))?(?:7(?=8|$))?(?:8(?=9|$))?9?$)\d{7,11}$

Regex demo

Or leaving out the first lookahead and move that to the match, changing the quantifier to \d{6,10} and repeating capture group \1 instead of \2

^(?!.*(\d)\1{3})(?!(?:0(?=1|$))?(?:1(?=2|$))?(?:2(?=3|$))?(?:3(?=4|$))?(?:4(?=5|$))?(?:5(?=6|$))?(?:6(?=7|$))?(?:7(?=8|$))?(?:8(?=9|$))?9?$)[03]\d{6,10}$

Regex demo

Edit

Based on the comments, the string not having 4 ascending digits:

^(?!.*(\d)\1{3})[03](?!\d*(?:0123|1234|2345|3456|4567|5678|6789))\d{6,10}$

Regex demo

edited Feb 25 '20 at 16:56

answered Feb 25 '20 at 13:01

The fourth bird

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I appreciate your help. However why is it accepting 023456789? – JOJO Feb 25 '20 at 16:04
Sorry my mistake. I meant is shouldnt accept ascending order – JOJO Feb 25 '20 at 16:13
For example: it should reject if: `012345678`, `07123456`, `07271234`, `016385678` – JOJO Feb 25 '20 at 16:15
@JOJO This will work then right? https://regex101.com/r/yhmwll/1 – The fourth bird Feb 25 '20 at 16:16
It won't because it only accept number starting with `02`. Remember it has to start either with 0 or 3, following by anything. But reject ascending order from any point of the number – JOJO Feb 25 '20 at 16:20
But that also matches `07271234` where `1234` is something you don't want to allow right? – The fourth bird Feb 25 '20 at 16:23
Yes, if you can improve on what I have there to reject that. Then it will be done – JOJO Feb 25 '20 at 16:27
Because it contains `45` which should not be allowed right? – The fourth bird Feb 25 '20 at 16:34
Yes, but the increase is only 45, which is only 2 numbers. It should reject if it was 073645675. i.e 4 numbers (4567) – JOJO Feb 25 '20 at 16:38
You could extend the alternation in that case https://regex101.com/r/lbCf5J/1 – The fourth bird Feb 25 '20 at 16:39
So do I have to manually extend it? Can we not make the regex look out for this ascending pattern starting from anywhere until the asending reach more than 4 numbers, in which case it should reject? – JOJO Feb 25 '20 at 16:42
I am not sure you can dot that. If there are longer ranges you could use a chacter class, but `[0-3]` is 5 chars and `0123` is 4 – The fourth bird Feb 25 '20 at 16:49
Yh I see. Im accepting your answer. Really apprciate your help :) – JOJO Feb 25 '20 at 16:52
You are welcome. will update the answer with the latest pattern and clean up some of the comments. – The fourth bird Feb 25 '20 at 16:53

Regex match valid Phone Number

2 Answers2