javascript - Sort Array of numeric & alphabetical elements (Natural Sort)

Question

Suppose I have an array

var arr = [1,5,"ahsldk",10,55,3,2,7,8,1,2,75,"abc","huds"];

and I try sorting it, I get something like ...

[1, 1, 10, 2, 2, 3, 5, 55, 7, 75, 8, "abc", "ahsldk", "huds"]

notice 10 is before 2, how can I have something more like

[1,1,2,2,3,5 ..., "abc", "ahs...",...]

Answer 1

From http://snipplr.com/view/36012/javascript-natural-sort/ by mrhoo:

Array.prototype.naturalSort= function(){
    var a, b, a1, b1, rx=/(d+)|(D+)/g, rd=/d+/;
    return this.sort(function(as, bs){
        a= String(as).toLowerCase().match(rx);
        b= String(bs).toLowerCase().match(rx);
        while(a.length && b.length){
            a1= a.shift();
            b1= b.shift();
            if(rd.test(a1) || rd.test(b1)){
                if(!rd.test(a1)) return 1;
                if(!rd.test(b1)) return -1;
                if(a1!= b1) return a1-b1;
            }
            else if(a1!= b1) return a1> b1? 1: -1;
        }
        return a.length- b.length;
    });
}

Or, from Alphanum: Javascript Natural Sorting Algorithm by Brian Huisman:

Array.prototype.alphanumSort = function(caseInsensitive) {
  for (var z = 0, t; t = this[z]; z++) {
    this[z] = [];
    var x = 0, y = -1, n = 0, i, j;

    while (i = (j = t.charAt(x++)).charCodeAt(0)) {
      var m = (i == 46 || (i >=48 && i <= 57));
      if (m !== n) {
        this[z][++y] = "";
        n = m;
      }
      this[z][y] += j;
    }
  }

  this.sort(function(a, b) {
    for (var x = 0, aa, bb; (aa = a[x]) && (bb = b[x]); x++) {
      if (caseInsensitive) {
        aa = aa.toLowerCase();
        bb = bb.toLowerCase();
      }
      if (aa !== bb) {
        var c = Number(aa), d = Number(bb);
        if (c == aa && d == bb) {
          return c - d;
        } else return (aa > bb) ? 1 : -1;
      }
    }
    return a.length - b.length;
  });

  for (var z = 0; z < this.length; z++)
    this[z] = this[z].join("");
}