Please help with regular expression

[paulsmith5]

Hi,

I'm developing a web app and I want to validate a users input on a
form. I need a regular expression to validate a string which must be
from 1 to 5 characters in length - each character must be a number from
0 to 9 however the string can't be comprised entirely of 0's (i.e.
invalid entries would be 0, 00, 000, 0000, and 00000).

Hope someone can help.

Thanks,

Paul

 

[loyal_caper]

I'm no regex expert, but this seems to work:

(?=^[0-9]{1,5}$)[0]*[1-9][0-9]*

Walter
----
Walter Gildersleeve
Freiburg, Germany

______________________________________________________
http://linkfrog.net
URL Shortening
Free and easy, small and green.

 

[loyal_caper]

a bit simpler:

(?=^[0-9]{1,5}$)0*[1-9]

Walter

----
Walter Gildersleeve
Freiburg, Germany

______________________________________________________
http://linkfrog.net
URL Shortening
Free and easy, small and green.

 

[Julian Turner]

I'm developing a web app and I want to validate a users input on a
form. I need a regular expression to validate a string which must be
from 1 to 5 characters in length - each character must be a number from
0 to 9 however the string can't be comprised entirely of 0's (i.e.
invalid entries would be 0, 00, 000, 0000, and 00000).

[/snip]

Another alternative:-

var r=/^(?!0+$)[0-9]{1,5}$/;

However, I think some older browsers (e.g. some older versions of IE)
do not support look aheads ?: ?= and ?!

Regards

Julian

 

[RobG]

Hi,

I'm developing a web app and I want to validate a users input on a
form. I need a regular expression to validate a string which must be
from 1 to 5 characters in length - each character must be a number from
0 to 9 however the string can't be comprised entirely of 0's (i.e.
invalid entries would be 0, 00, 000, 0000, and 00000).

Hope someone can help.

You've been given some great responses, here's one for browsers that
don't support look-ahead:

if (/^\d{1,5}$/.test(num) && num.replace(/0/g,'').length){
alert('OK');
} else {
alert('bzzzzt');
}


Where num is the string entered by the user.

 

[Dr John Stockton]

JRS: In article <43f5cb55$0$17107$5a62ac22@per-qv1-newsreader-
01.iinet.net.au>, dated Fri, 17 Feb 2006 23:10:03 remote, seen in

if (/^\d{1,5}$/.test(num) && num.replace(/0/g,'').length){

or ?

if (/^\d{1,5}$/.test(num) && /[1-9]/.test(num)) ) {

 

[paulsmith5]

Hi Walter,

Thanks for the reply. You're an expert in my books - it works
brilliantly.

 

[paulsmith5]

Hi Julian,

Thanks for the reply. It's great to know help is out there.

 

[paulsmith5]

Hi Rob,

Yes the responses have been great - thanks for the tip to overcome the
look-ahead thingy

 

[Randy Webb]

(e-mail address removed) said the following on 2/20/2006 7:19 AM:

Hi Rob,

Yes the responses have been great - thanks for the tip to overcome the
look-ahead thingy

How about another "tip" then?

Please quote what you are replying to.

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