natural sort of text and numbers, JavaScript

Question

I'm looking for the easiest way to sort an array that consists of numbers and text, and a combination of these.

E.g.

'123asd'
'19asd'
'12345asd'
'asd123'
'asd12'

turns into

'19asd'
'123asd'
'12345asd'
'asd12'
'asd123'

This is going to be used in combination with the solution to another question I've asked here.

The sorting function in itself works, what I need is a function that can say that that '19asd' is smaller than '123asd'.

I'm writing this in JavaScript.

Edit: as adormitu pointed out, what I'm looking for is a function for natural sorting

Answer 1

So you need a natural sort? If so, than maybe this script would be what you need.

Answer 2

function a1Sort(a, b){
    var cnt= 0, tem;
    a= String(a).toLowerCase();
    b= String(b).toLowerCase();
    if(a== b) return 0;
    if(/\d/.test(a) ||  /\d/.test(b)){
        var Rx=  /^\d+(\.\d+)?/;
        while(a.charAt(cnt)=== b.charAt(cnt) && 
        !Rx.test(a.substring(cnt))){
            cnt++;
        }
        a= a.substring(cnt);
        b= b.substring(cnt);
        if(Rx.test(a) || Rx.test(b)){
            if(!Rx.test(a)) return a? 1: -1;
            if(!Rx.test(b)) return b? -1: 1;
            tem= parseFloat(a)-parseFloat(b);
            if(tem!= 0) return tem;
            a= a.replace(Rx,'');
            b= b.replace(Rx,'');
            if(/\d/.test(a) ||  /\d/.test(b)){
                return a1Sort(a, b);
            }
        }
    }
    if(a== b) return 0;
    return a> b? 1: -1;
}

var arr=['123asd','19asd','12345asd','asd123','asd12']; arr.sort(a1Sort)

/* returned value: (Array) 19asd,123asd,12345asd,asd12,asd123 */