RegExp sorting

[Kor]

I am not (yet ) very good on RegExp, so can anyone help me sorting an array using RegExp?

a = new Array()
a[0] = 'b';
a[1] = '1';
a[2] = 'a';
a[3] = '2';

I need to change the order in array so, after the sorting, I'd like to have the array in normal sorting (number-alphabetical)

a[0] = '1';
a[1] = '2';
a[2] = 'a';
a[3] = 'b';

[Danne]

Hmm, I'm not sure I understood, but why do u need RegExp?

compare=function(a,b){
    if(a<b)return -1;
    if(a>b)return 1;
    return 0;
}

a = new Array() 
a[0] = 'b'; 
a[1] = '1'; 
a[2] = 'a'; 
a[3] = '2'; 

a.sort(compare);

alert(a.join("\n"));

[liorean]

That's a bit overkill - the sort function by default sorts lexically, alphanumerically, case grouping. The only thing you need to do is to run the sort method of the array object in question. See:

['b','ab','aa',2,10,1,'a','A'].sort(); => [1,10,2,'A','a','aa','ab','b']

[fredmv]

David is of couse completely correct; answered here as well: http://forums.webdeveloper.com/showt...threadid=28702.

[glenngv]

But if you want it to be case-insensitive...

function compare(a,b){
    a=a.toLowerCase();
    b=b.toLowerCase();
    if(a<b) return -1;
    if(a>b) return 1;
    return 0;
}

or if you're comparing date strings...

function compare(a,b) {
  if (Date.parse(a) < Date.parse(b)) return -1;
  if (Date.parse(a) > Date.parse(b)) return 1;
  return 0;
}

[Kor]

Thank you all.

Actually I might need either RegExp or split() before sorting, as I have long strings, 2 or 3 words (name, surname) and intend to sort them by first, second or third word in string.

Anyway, this is the kind of work I can do it by myself. I just never use the sort() method before.