Javascript natural sort objects

Question

I have an object array like

array[0] = {id: 1, name: "First"}
array[1] = {id: 2, name: "Second"}

Which I've tried to sort by name field by using:

Array.prototype.alphanumSort = function (caseInsensitive) {
for (var z = 0, t; t = ((typeof this[z] == "string" || typeof this[z] == "undefined") ? this[z] : this[z].name); z++) {
    this[z] = new Array();
    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("");
}

but the list that I'm getting back from it contains an array with just the name field from it, without the corresponding id. Why is this happening? Is there a way to apply this algorithm to filter objects? I'm new in JavaScript, so please be gentle.

Answer 1

You could use sorting with map and String#localeCompare with options

sensitivity

Which differences in the strings should lead to non-zero result values. Possible values are:

• "base": Only strings that differ in base letters compare as unequal. Examples: a ≠ b, a = á, a = A.
• "accent": Only strings that differ in base letters or accents and other diacritic marks compare as unequal. Examples: a ≠ b, a ≠ á, a = A.
• "case": Only strings that differ in base letters or case compare as unequal. Examples: a ≠ b, a = á, a ≠ A.
• "variant": Strings that differ in base letters, accents and other diacritic marks, or case compare as unequal. Other differences may also be taken into consideration. Examples: a ≠ b, a ≠ á, a ≠ A.

The default is "variant" for usage "sort"; it's locale dependent for usage "search".

numeric

Whether numeric collation should be used, such that "1" < "2" < "10". Possible values are true and false; the default is false. This option can be set through an options property or through a Unicode extension key; if both are provided, the options property takes precedence. Implementations are not required to support this property.

var array = ['Ex 10', 'Ex 2', 'a 10.1', 'a 1.1', 'a 10.0', 'a 2.0'];

array.sort((a, b) => a.localeCompare(b, undefined, { numeric: true, sensitivity: 'base' }));

console.log(array)

Answer 2

You can use String.prototype.localeCompare

var arr = [
  {id: 1, name: "Third"},
  {id: 2, name: "Fifth"},
  {id: 3, name: "Second"},
  {id: 4, name: "Fourth"},
  {id: 5, name: "First"}
];

arr.sort((a, b) => a.name.localeCompare(b.name, undefined, { numeric: true, sensitivity: 'base' }));

console.log(arr);