…每个对象在同一个数组中也有对其他对象的引用?

当我第一次想到这个问题的时候,我就想到了

var clonedNodesArray = nodesArray.clone()

并搜索如何在JavaScript中克隆对象的信息。我确实在Stack Overflow上找到了一个问题(同样由@JohnResig回答),他指出用jQuery你可以做到

var clonedNodesArray = jQuery.extend({}, nodesArray);

克隆对象。虽然我尝试了这个,但这只复制了数组中对象的引用。如果我

nodesArray[0].value = "red"
clonedNodesArray[0].value = "green"

nodesArray[0]和clonedNodesArray[0]的值将显示为“绿色”。然后我尝试了

var clonedNodesArray = jQuery.extend(true, {}, nodesArray);

它深度复制了一个对象,但我分别从Firebug和Opera Dragonfly得到了“太多递归”和“控制堆栈溢出”的消息。

你会怎么做?这是不应该做的事情吗?在JavaScript中是否有可重用的方法来做到这一点?


当前回答

jQuery扩展工作正常。你只需要指定你克隆的是一个数组而不是一个对象(注意extend方法的参数是[]而不是{}):

var clonedNodesArray = jQuery.extend([], nodesArray);

其他回答

如果你只需要一个浅拷贝,一个非常简单的方法是:

new_array = old_array.slice(0);

在JavaScript中,数组和对象复制会改变原始值,因此深度复制是解决这个问题的解决方案。

深度复制实际上意味着创建一个新数组并复制值,因为无论它发生什么都不会影响原始数组。

JSON。解析和JSON。Stringify是深度复制的最佳和简单的方法。JSON.stringify()方法将JavaScript值转换为JSON字符串。JSON.parse()方法解析JSON字符串,构造由字符串描述的JavaScript值或对象。

深克隆

let a = [{ x:{z:1} , y: 2}];
let b = JSON.parse(JSON.stringify(a));
b[0].x.z=0

console.log(JSON.stringify(a)); //[{"x":{"z":1},"y":2}]
console.log(JSON.stringify(b)); // [{"x":{"z":0},"y":2}]

更多细节:阅读这里

$.evalJSON($.toJSON(origArray));
var game_popularity = [
     { game: "fruit ninja", popularity: 78 },
     { game: "road runner", popularity: 20 },
     { game: "maze runner", popularity: 40 },
     { game: "ludo", popularity: 75 },
     { game: "temple runner", popularity: 86 }
];
console.log("sorted original array before clonning");
game_popularity.sort((a, b) => a.popularity < b.popularity);
console.log(game_popularity);


console.log("clone using object assign");
const cl2 = game_popularity.map(a => Object.assign({}, a));
cl2[1].game = "clash of titan";
cl2.push({ game: "logan", popularity: 57 });
console.log(cl2);


// Adding new array element doesnt reflect in original array
console.log("clone using concat");
var ph = []
var cl = ph.concat(game_popularity);

// Copied by reference ?
cl[0].game = "rise of civilization";

game_popularity[0].game = 'ping me';
cl.push({ game: "angry bird", popularity: 67 });
console.log(cl);

console.log("clone using ellipses");
var cl3 = [...game_popularity];
cl3.push({ game: "blue whale", popularity: 67 });
cl3[2].game = "harry potter";
console.log(cl3);

console.log("clone using json.parse");
var cl4 = JSON.parse(JSON.stringify(game_popularity));
cl4.push({ game: "home alone", popularity: 87 });
cl4[3].game ="lockhead martin";
console.log(cl4);

console.log("clone using Object.create");
var cl5 = Array.from(Object.create(game_popularity));
cl5.push({ game: "fish ville", popularity: 87 });
cl5[3].game ="veto power";
console.log(cl5);


// Array function
console.log("sorted original array after clonning");
game_popularity.sort((a, b) => a.popularity < b.popularity);
console.log(game_popularity);


console.log("Object.assign deep clone object array");
console.log("json.parse deep clone object array");
console.log("concat does not deep clone object array");
console.log("ellipses does not deep clone object array");
console.log("Object.create does not deep clone object array");

输出

sorted original array before clonning
[ { game: 'temple runner', popularity: 86 },
{ game: 'fruit ninja', popularity: 78 },
{ game: 'ludo', popularity: 75 },
{ game: 'maze runner', popularity: 40 },
{ game: 'road runner', popularity: 20 } ]
clone using object assign
[ { game: 'temple runner', popularity: 86 },
{ game: 'clash of titan', popularity: 78 },
{ game: 'ludo', popularity: 75 },
{ game: 'maze runner', popularity: 40 },
{ game: 'road runner', popularity: 20 },
{ game: 'logan', popularity: 57 } ]
clone using concat
[ { game: 'ping me', popularity: 86 },
{ game: 'fruit ninja', popularity: 78 },
{ game: 'ludo', popularity: 75 },
{ game: 'maze runner', popularity: 40 },
{ game: 'road runner', popularity: 20 },
{ game: 'angry bird', popularity: 67 } ]
clone using ellipses
[ { game: 'ping me', popularity: 86 },
{ game: 'fruit ninja', popularity: 78 },
{ game: 'harry potter', popularity: 75 },
{ game: 'maze runner', popularity: 40 },
{ game: 'road runner', popularity: 20 },
{ game: 'blue whale', popularity: 67 } ]
clone using json.parse
[ { game: 'ping me', popularity: 86 },
{ game: 'fruit ninja', popularity: 78 },
{ game: 'harry potter', popularity: 75 },
{ game: 'lockhead martin', popularity: 40 },
{ game: 'road runner', popularity: 20 },
{ game: 'home alone', popularity: 87 } ]
clone using Object.create
[ { game: 'ping me', popularity: 86 },
{ game: 'fruit ninja', popularity: 78 },
{ game: 'harry potter', popularity: 75 },
{ game: 'veto power', popularity: 40 },
{ game: 'road runner', popularity: 20 },
{ game: 'fish ville', popularity: 87 } ]
sorted original array after clonning
[ { game: 'ping me', popularity: 86 },
{ game: 'fruit ninja', popularity: 78 },
{ game: 'harry potter', popularity: 75 },
{ game: 'veto power', popularity: 40 },
{ game: 'road runner', popularity: 20 } ]

Object.assign deep clone object array
json.parse deep clone object array
concat does not deep clone object array
ellipses does not deep clone object array
Object.create does not deep clone object array

最简洁的浅拷贝解决方案:

array = array.map(obj => {
    return { ...obj };
});

这将生成[{a: 1}, {b: 2}]等内容的完全独立副本,而不是[{a: {b: 2}}, {b: {a: 1}}]。