Can Anyone Complete this regex?

Question

KELECTRIC.BNK.20160526

compare above expression and take 8 digits after 2nd full stop and total of 22 characters,

below expression is working fine till 2nd full stop but not taking number after full stop

^(KELECTRIC)\.(BNK)+\.$

`^(KELECTRIC)\.(BNK)\.([0-9]{8})$` – Kuba Wyrostek Jun 10 '16 at 06:49 — Kuba Wyrostek, Jun 10 '16 at 06:49
thanks working fine :) – Ahsan Aziz Abbasi Jun 10 '16 at 07:51 — Ahsan Aziz Abbasi, Jun 10 '16 at 07:51
little modified :-- ^KELECTRIC\.BNK\.\K([0-9]{8})$ – Shekhar Khairnar Jun 10 '16 at 08:58 — Shekhar Khairnar, Jun 10 '16 at 08:58

score 2 · Answer 1 · edited Jun 10 '16 at 07:05

2

You can accept number of 8 digits after 2nd full stop like this

^(KELECTRIC)\.(BNK)\.(\d{8})$

edited Jun 10 '16 at 07:05

Kuba Wyrostek

answered Jun 10 '16 at 06:50

Muhammad Hassan

2

Perhaps it would be better with (\d{8}) so you can get the number from one group instead of 8? – Jørn Jensen Jun 10 '16 at 06:59
I think both are same. I have checked and both are working fine. – Muhammad Hassan Jun 10 '16 at 07:03

score 1 · Answer 2 · answered Jun 10 '16 at 06:53

1

Try ^(KELECTRIC)\.(BNK)\.[0-9]{8}$ or ^([A-Z]{1,})\.([A-Z]{1,})\.[0-9]{8}$

answered Jun 10 '16 at 06:53

Marcin Wadoń

score 0 · Answer 3 · edited Jun 10 '16 at 08:20

0

^(KELECTRIC)\.(BNK)\.([0-9]{8})$

this one answered by Kuba Wyrostek, and some other guys

thanks

edited Jun 10 '16 at 08:20

Kuba Wyrostek

answered Jun 10 '16 at 07:52

Ahsan Aziz Abbasi

It will accept every character at position where only `.` can come. You should use `\.` instead. `KELECTRIC-BNK-20160526` will also be a valid string for this. – Muhammad Hassan Jun 10 '16 at 07:57
1

This is equivalent to answer by @MHassan, so it is better to accept his answer instead of writing your own. – Kuba Wyrostek Jun 10 '16 at 08:23

3 Answers3