Need a regex that makes sure the number starts with 01 02 or 08 and is 10 or 11 digits long

Question

Am having hard time crediting a regex to check if a number starts with 01, 02 or 08 and is ten or eleven digits long.

For example I want numbers formatted like this to pass:

01614125745
02074125475
0845895412
08004569321

and numbers other forms to fail, number like:

0798224141
441544122444
0925456754

and so on. Basically, any number that does not start with 01 02 or 08.

Solution 1

The following pattern follows the criteria:

/^(?=\d{10,11}$)(01|02|08)\d+/   # This pattern can easily be extended

/^(?=\d{10,11}$)0[128]\d{8,9}/   # Same effect

Explanation:

^                Begin of string
(?=\d{10,11}$)   Followed by 10 or 11 digits, followed by the end of the string
(01|02|08)       01, 02 or 08
\d+              The remaining digits

Solution 2

Rob W's answer is correct. But I'd use a simpler expression like this: (No need for lookahead)

/^0[128][0-9]{8,9}$/

Comments

[Roger Lindsjö]

An explanation of the parts in the regex would make the answer evenbetter. Especially when using non capturing groups and greedy quantifiers other than ? and * as those are exotic feature to most developers.

[Wiseguy]

I would've just done /^(01|02|08)\d{8,9}$/, but this works too and could be more flexible later.

[Rob W]

@Wiseguy I saw: "10 or 11 digits long, starting with ...". For this reason, I decided to suplly a pattern which can easily be re-used and extended (what if the following criteria is added: or starts with 112) ;)

[user1053865]

Thanks that worked great am i right in say if i have/^(?=\d{10,11}$)(01|02|08|03|07)\d+/ it will mean it allows 03 and 07 as well ? if so that's great massive help thank you

[Rob W]

@user1053865 Yes, that works as well. If the prefixes are of a fixed size, you can also use the following pattern (although, for an unexperienced user, its effect is not as obvious as the previous pattern): ^0[12378]\d{8,9}$ (= START 0 one of 12378 8 or 9 digits END).

[ridgerunner]

Why would someone downvote this answer? It is simple, fast and accurate. Go figure...

[Davyd Geyl]

Great answer. Helped me to see other ways to construct regex for this task.