I want to validate a 5 digit number which should not be 00000. All numbers except 00000 are allowed.

examples : 01201 , 00001, 21436 , 45645 are valid numbers and 1, 12, 123, 1234, 00000 are invalid numbers.

I tried with ^([0-9]{5}[^(0){5}])$ but it was of no use.

Any help on this will be much appreciated.

Thanks.

I want to validate a 5 digit number which should not be 00000. All numbers except 00000 are allowed.

examples : 01201 , 00001, 21436 , 45645 are valid numbers and 1, 12, 123, 1234, 00000 are invalid numbers.

I tried with ^([0-9]{5}[^(0){5}])$ but it was of no use.

Any help on this will be much appreciated.

Thanks.

• javascript
• regex

• 4 what about string.length=5 && string != "00000"? [Edit: OK, testing for numberness requires regex, see @mplungjan's answer] – Jasper Commented Apr 7, 2014 at 12:09

4 Answers 4

Using negative positive lookahead:

/^(?!0{5})\d{5}$/

/^(?!0{5})\d{5}$/.test('01201')
// => true
/^(?!0{5})\d{5}$/.test('00001')
// => true
/^(?!0{5})\d{5}$/.test('21436')
// => true
/^(?!0{5})\d{5}$/.test('1')
// => false
/^(?!0{5})\d{5}$/.test('12')
// => false
/^(?!0{5})\d{5}$/.test('00000')
// => false

No need for negative lookahead

Live Demo

function isvalid5(str) {
  return str != "00000" && /^\d{5}$/.test(str);
}

You can use a negative look ahead:

^(?!0{5})\d{5}$

The negative loo ahead (?!...) will fail the whole regex if what's inside it matches, \d is a shortcut for [0-9].

for regex engines that do not support lookahead:

^(0000[1-9]|000[1-9][0-9]|00[1-9][0-9]{2}|0[1-9][0-9]{3}|[1-9][0-9]{4})$