To match any number greater than 15 using regexp

Question

I tried to match any number greater than 15 using the following regexp:

0*[1-9][6-9][0-9]*

But I can match only 2 digit number eg. I can unmatch successfully 12 or 13 (less than 15) , whereas I am unable to match 105, 124 etc..

Anyone help me out how to resolve this.

Exactly @cmbuckley why using regex when conversion to int + comparison will keep a code maintenable and readable — Cid, Oct 08 '18 at 11:24
You could try it with lookaheads `^(?!0*(\d|1[0-5])$)\d+$`. See live demo here https://regex101.com/r/iDJVLv/1 — revo, Oct 08 '18 at 11:39

JGNI · Accepted Answer · 2018-10-08T11:21:27.233

4

Any number greater than 15 is

Any number with 3 or more digits with possible leading 0's
Any number with 2 digits where the first digit in in the character class [2-9]
Any number with 2 digits where the first digit is 1 and the second digit in in the character class [6-9]

From these three rules we can build the regex, assumes that what we are matching only contains digits

/^0*(?:[1-9][0-9]{2,}|[2-9][0-9]|1[6-9])$/

If you can't use an extended regex then the following should work

/^0*[1-9][0-9][0-9][0-9]*|0*[2-9][0-9]|0*1[6-9]$/

edited Oct 08 '18 at 11:21

answered Oct 08 '18 at 11:05

JGNI

This assumes extended regular expressions, i.e.`-r` option with `sed` or `-E` option with `grep` – Hkoof Oct 08 '18 at 11:08
This should work fine `^[1-9]{3,}|1[6-9]|[2-9][0-9]$` – IftekharDani Oct 08 '18 at 11:13
This won't match 100 – Cid Oct 08 '18 at 11:14
Add `0*` at the begining to match any leading 0 – Cid Oct 08 '18 at 11:17
0100 won't match – Cid Oct 08 '18 at 11:17
Dam numbers are hard :-) – JGNI Oct 08 '18 at 11:18
2

Well, regex are maybe not suited for that purpose :) – Cid Oct 08 '18 at 11:18

1 Answers1