Regex to accept 6 or 8 digits followed by 0, 1 or 2 letters

Question

I am writing a regular expression to accept first 6 or 8 digit then d, m, md, DM or no letter at all e.g.

I have written

/(^(\d{6}|\d{8}))(d|m|dm|md|)$/gi

is it correct or wrong?

You could probably do the suffix as (dm?|md?). Apart from that, looks a-ok. You just need to test it. — Stephen Quan, Nov 20 '17 at 07:25
May I know why this question is downvoted while OP has shown his efforts? — Mohammad Usman, Nov 20 '17 at 07:48
The *is it correct or wrong?* question is the same as "what does my pattern mean/what can my pattern match", and is thus a dupe. If you find any issues with the pattern, specify what is wrong, update the question. — Wiktor Stribiżew, Nov 20 '17 at 08:21

PJProudhon · Answer 1 · 2017-11-20T08:22:38.080

0

You should try with:

^\d{6}(?:\d{2})?(?:md?|dm?)?$

(?:\d{2})? stands for 2 optional more digits.
(?:md?|dm?)? stands for optional m or d, followed or not by its mate.

edited Nov 20 '17 at 08:22

answered Nov 20 '17 at 07:38

PJProudhon

4

Won't this make `121212mm` and `121212dd` valid ? – Hassan Imam Nov 20 '17 at 07:46
Absolutely right. I edited my answer to fix it. – PJProudhon Nov 20 '17 at 08:23

Adrien De Peretti · Answer 2 · 2017-11-20T07:29:59.950

-2

You should do something like \d{6,8}[a-zA-Z]{2}. Or [0-9]{2}[a-zA-Z]{2}

Have a nice day

edited Nov 20 '17 at 07:29

answered Nov 20 '17 at 07:27

Adrien De Peretti

This will allow any letters at the end - not just d and m. – sideroxylon Nov 20 '17 at 07:29
1

And it'll allow 7 digits. – SamWhan Nov 20 '17 at 07:30
Yes you right you can change the [a-zAZ] by (md?|dm?) tell me if its ok – Adrien De Peretti Nov 20 '17 at 07:34
You need `(md?|dm?| )/gi` to allow for no letters. – sideroxylon Nov 20 '17 at 07:35
Sorry i miss understood ^^ in this case your’s ok – Adrien De Peretti Nov 20 '17 at 07:41

2 Answers2