Question

我已经做了这个jsfiddle，我无法解释为什么以下内容不会导致5个对象的数组都具有不同的id属性：

var arr = ["1", "2", "3", "4", "5"];
var clone = {"id": "0", "name":"Matthew"};
var arrObj = [];

var idArr = [];

while((a=arr.pop()) != null){ 
    clone.id = a;
    console.log(clone);
    arrObj.push(clone);
}

console.log(arrObj);

我最终得到的是我的控制台中的以下内容：

Object {id: "5", name: "Matthew"} (index):28
Object {id: "4", name: "Matthew"} (index):28
Object {id: "3", name: "Matthew"} (index):28
Object {id: "2", name: "Matthew"} (index):28
Object {id: "1", name: "Matthew"} (index):28

[Object, Object, Object, Object, Object]

当我打开5个克隆对象中的每一个时，它们都具有“id”值“1”

为什么会这样？

Answer 1

clone是一个对象。在javascript中，对象通过引用传递，因此您不会在每个索引中传递不同的对象。

这是可以工作的东西

while((a=arr.pop()) != null){
    var clone = {"id": a, "name":"Matthew"};
    clone.id = a;
    arrObj.push(clone);
}

console.log(arrObj);

结果：

Array [
    Object {id: "5", name: "Matthew"}
    Object {id: "4", name: "Matthew"}
    Object {id: "3", name: "Matthew"}
    Object {id: "2", name: "Matthew"}
    Object {id: "1", name: "Matthew"}
]

Answer 2

JS中的对象是通过引用分配的。您需要复制它，例如jQuery.extend。

var arr = ["1", "2", "3", "4", "5"];
var clone = {"id": "0", "name":"Matthew"};
var arrObj = [];

var idArr = [];

while((a=arr.pop()) != null){ 
    clone = $.extend({}, clone);
    clone.id = a;
    console.log(clone);
    arrObj.push(clone);
}

console.log(arrObj);

Answer 3

如果你的克隆不包含任何功能，你可以这样做

while((a=arr.pop()) != null){
    clone = JSON.parse(JSON.stringify(clone));
    clone.id = a;
    console.log(clone);
    arrObj.push(clone);
}

流行循环时Javascript数组推送内部

3 个答案: