Regex - Find numbers between 2000 and 3000

Question

I have a need to search all numbers with 4 digits between 2000 and 3000.

It can be that letters are before and after.

I thought I can use [2000-3000]{4}, but doesnt work, why?

thank you.

Answer 1

How about

^2\d{3}|3000$

Or as Amarghosh & Bart K. & jleedev pointed out, to match multiple instances

\b(?:2[0-9]{3}|3000)\b

If you need to match a3000 or 3000a but not 13000, you would need lookahead and lookbefore like

(?<![0-9])(?:2[0-9]{3}|3000)(?![0-9])

Answer 2

Regular expressions are rarely suitable for checking ranges since for ranges like 27 through 9076 inclusive, they become incredibly ugly. It can be done but you're really better off just doing a regex to check for numerics, something like:

^[0-9]+$

which should work on just about every regex engine, and then check the range manually.

In toto:

def isBetween2kAnd3k(s):
    if not s.match ("^[0-9]+$"):
        return false
    i = s.toInt()
    if i < 2000 or i > 3000:
        return false
    return true

What your particular regex [2000-3000]{4} is checking for is exactly four occurrences of any of the following character: 2,0,0,0-3,0,0,0 - in other words, exactly four digits drawn from 0-3.

With letters before an after, you will need to modify the regex and check the correct substring, something like:

def isBetween2kAnd3kWithLetters(s):
    if not s.match ("^[A-Za-z]*[0-9]{4}[A-Za-z]*$"):
        return false
    idx = s.locate ("[0-9]")
    i = s.substring(idx,4).toInt()
    if i < 2000 or i > 3000:
        return false
    return true

---

As an aside, a regex for checking the range 27 through 9076 inclusive would be something like this hideous monstrosity:

^2[7-9]|[3-9][9-9]|[1-9][0-9]{2}|[1-8][0-9]{3}|90[0-6][0-9]|907[0-6]$

I think that's substantially less readable than using ^[1-9][0-9]+$ then checking if it's between 27 and 9076 with an if statement?