javascript regex when diffrent length of string

Question

I have a string of digits that's supposed to have 9 characters and I have a regex that replaces the string with the same string and some spaces; something like this:

TheString = '123456789';
TheSpacedString = TheString.replace(/(\d{1})(\d{2})(\d{2})(\d{2})(\d{2})/, '$1 $2 $3 $4 $5');
TheSpacedString format is now '1 23 45 67 89'

The problem is that when the length of the string is not 9, the formatting doesn't work: for instance, if we have this:

TheString = '12345';
TheSpacedString = TheString.replace(/(\d{1})(\d{2})(\d{2})(\d{2})(\d{2})/, '$1 $2 $3 $4 $5');
format should be '1 23 45'

But instead, the string is just '12345'. What's the problem with my regex? The jsFiddle is here

You should add validation before you perform your replacement - if you're expecting 9 digits but don't get 9, shouldn't that be an error first? — qJake
– qJake, Commented Dec 3, 2013 at 17:05

Matt Burland · Accepted Answer · 2013-12-03 16:58:19Z

4

Make the last two groups (or however many you think should be optional) optional with ?

http://jsfiddle.net/yWSR2/3/

TheSpacedString = TheString.replace(/(\d{1})(\d{2})(\d{2})(\d{2})?(\d{2})?/, '$1 $2 $3 $4 $5');

answered Dec 3, 2013 at 16:58

Matt Burland

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2 Comments

frenchie Over a year ago

Oh wow, thanks, fixed my day! Upvoted!! Actually, I added a question mark after each block and it works perfectly!!

anubhava Over a year ago

Though it will add some trailing spaces like: "1 23 45 " better to call .trim() after this.

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