有关js的双向绑定解除方法

最近碰到了一个bug

var persons = [{ number: 1, age: 11, name: "wanghaha", money: -1 }, { number: 2, age: 12, name: "王一一", money: -3 }, { number: 3, age: 13, name: "王三十", money: 4 }];
        var per_con = [];

        for (var i = persons.length; i > 0; i--) {
            if (persons[i-1].money > 0) {
                per_con.push(persons[i-1]);
            }
        }
        console.log("111"+persons);
        console.log("222"+per_con);
        var pp = { number: 0, age: 0, name: "", money: 0 };
        persons.push(pp);
        for (var i = persons.length; i > 0; i--) {
            if (persons[i - 1].number > 2) {
                persons[i].number += 1;
                persons[i].age = persons[i - 1].age;
                persons[i].name = persons[i - 1].name;
                persons[i].money = persons[i - 1].money;
            }
            if (persons[i - 1].number == 2) {
                persons[i].number = 2;
                persons[i].age = 100;
                persons[i].name = "王插队";
                persons[i].money = -90;
            }
        }
        console.log("333"+persons);
        console.log("444"+per_con);

　　结果：

console.log("111"+persons);
0: {number: 1, age: 11, name: "wanghaha", money: -1}
1: {number: 2, age: 12, name: "王一一", money: -3}
2: {number: 3, age: 13, name: "王三十", money: 4}

console.log("222"+per_con);
0: {number: 3, age: 13, name: "王三十", money: 4}

console.log("333"+persons);
0: {number: 1, age: 11, name: "wanghaha", money: -1}
1: {number: 2, age: 12, name: "王一一", money: -3}
2: {number: 2, age: 100, name: "王插队", money: -90}
3: {number: 1, age: 13, name: "王三十", money: 4}

console.log("444"+per_con);
0: {number: 2, age: 100, name: "王插队", money: -90}

看得出来：per_con的值因为绑定数据发生了变化，相当于只保留了persons[2]

问题出在per_con.push(persons[i-1]);

解决方法1：解除绑定

               

 per_con.push(persons[i-1]);替换成

 per_con.push(JSON.parse(JSON.stringify(persons[i-1])));

console.log("444"+per_con);

0: {number: 3, age: 13, name: "王三十", money: 4}

解决方法1：替换绑定

 function extend(o,p){
            for(index in p){
                o[index] = p[index];
            }
            return(o);
        }
        var persons = [{ number: 1, age: 11, name: "wanghaha", money: -1 }, { number: 2, age: 12, name: "王一一", money: -3 }, { number: 3, age: 13, name: "王三十", money: 4 }];
        var per_con = [];

        for (var i = persons.length; i > 0; i--) {
            if (persons[i-1].money > 0) {
                var p = {};
                extend(p , persons[i-1]);
                per_con.push(p);
            }
        }
        console.log("111"+persons);
        console.log("222"+per_con);
        var pp = { number: 0, age: 0, name: "", money: 0 };
        persons.push(pp);
        for (var i = persons.length; i > 0; i--) {
            if (persons[i - 1].number > 2) {
                persons[i].number += 1;
                persons[i].age = persons[i - 1].age;
                persons[i].name = persons[i - 1].name;
                persons[i].money = persons[i - 1].money;
            }
            if (persons[i - 1].number == 2) {
                persons[i].number = 2;
                persons[i].age = 100;
                persons[i].name = "王插队";
                persons[i].money = -90;
            }
        }
        console.log("333"+persons);
        console.log("444"+per_con);

添加替换函数

function extend(o,p){
　　for(index in p){
　　　　o[index] = p[index];
　　}
　　return(o);
}

per_con.push(persons[i-1]);替换成 

var p = {};
extend(p , persons[i-1]);
per_con.push(p);
也成功解决

console.log("444"+per_con);

0: {number: 3, age: 13, name: "王三十", money: 4}