提升

arr.sort((a, b) => a - b);

下行

arr.sort((a, b) => b - a);

只是为了好玩:

降序=升序+反转

arr.sort((a, b) => a - b).reverse();

sort_mixed

Object.defineProperty(Array.prototype,"sort_mixed",{
    value: function () { // do not use arrow function
        var N = [], L = [];
        this.forEach(e => {
            Number.isFinite(e) ? N.push(e) : L.push(e);
        });
        N.sort((a, b) => a - b);
        L.sort();
        [...N, ...L].forEach((v, i) => this[i] = v);
        return this;
    })

try a =[1，'u'，"V"，10,4，"c"，" a "].sort_mixed();console.log (a)

数组中。Sort默认使用字母排序，而不是数字排序。

要支持数字，请添加如下的like

var numArray = [140000, 104, 99];
numArray.sort((a, b) =>  a - b); // <-- Ascending
numArray.sort((a, b) =>  b - a); // <-- Descending
console.log(numArray);

输出:

Let grade =[80,100,50,90,40]; grade.sort ((x, y) = > x - y); grade.forEach(元素= > console.log(元素));

如果你需要从字符串列表中计算和排序最大的charCodeAt，这是正确的方法。

const arrayLines = '1.1.1.1\n1.0.1.1\n1.1.1.2\n1.1.1.0'.split('\n');

// Response: (4) ['1.0.1.1', '1.1.1.0', '1.1.1.1', '1.1.1.2']
arrayLines.sort((a, b) => {
    let a_charCodeSize = 0,
        b_charCodeSize = 0;

    // Loop true a & b characters and calculate the charCodeAt size.
    for (const aChar of a) a_charCodeSize += aChar.charCodeAt(0);
    for (const bChar of b) b_charCodeSize += bChar.charCodeAt(0);

    return a_charCodeSize - b_charCodeSize;
});

试试下面的代码:

HTML:

<div id="demo"></div>

JavaScript代码:

<script>
    (function(){
        var points = [40, 100, 1, 5, 25, 10];
        document.getElementById("demo").innerHTML = points;
        points.sort(function(a, b){return a-b});
        document.getElementById("demo").innerHTML = points;
    })();
</script>