我有一个对象x。我想将其复制为对象y,这样对y的更改不会修改x。我意识到复制从内置JavaScript对象派生的对象会导致额外的、不需要的财产。这不是问题,因为我正在复制我自己的一个文字构造对象。
如何正确克隆JavaScript对象?
我有一个对象x。我想将其复制为对象y,这样对y的更改不会修改x。我意识到复制从内置JavaScript对象派生的对象会导致额外的、不需要的财产。这不是问题,因为我正在复制我自己的一个文字构造对象。
如何正确克隆JavaScript对象?
当前回答
根据Apple JavaScript编码指南:
// Create an inner object with a variable x whose default
// value is 3.
function innerObj()
{
this.x = 3;
}
innerObj.prototype.clone = function() {
var temp = new innerObj();
for (myvar in this) {
// this object does not contain any objects, so
// use the lightweight copy code.
temp[myvar] = this[myvar];
}
return temp;
}
// Create an outer object with a variable y whose default
// value is 77.
function outerObj()
{
// The outer object contains an inner object. Allocate it here.
this.inner = new innerObj();
this.y = 77;
}
outerObj.prototype.clone = function() {
var temp = new outerObj();
for (myvar in this) {
if (this[myvar].clone) {
// This variable contains an object with a
// clone operator. Call it to create a copy.
temp[myvar] = this[myvar].clone();
} else {
// This variable contains a scalar value,
// a string value, or an object with no
// clone function. Assign it directly.
temp[myvar] = this[myvar];
}
}
return temp;
}
// Allocate an outer object and assign non-default values to variables in
// both the outer and inner objects.
outer = new outerObj;
outer.inner.x = 4;
outer.y = 16;
// Clone the outer object (which, in turn, clones the inner object).
newouter = outer.clone();
// Verify that both values were copied.
alert('inner x is '+newouter.inner.x); // prints 4
alert('y is '+newouter.y); // prints 16
史蒂夫
其他回答
使用lodash_.cloneDeep()。
浅拷贝:lodash_.clone()
只需复制参考即可进行浅层复制。
let obj1 = {
a: 0,
b: {
c: 0,
e: {
f: 0
}
}
};
let obj3 = _.clone(obj1);
obj1.a = 4;
obj1.b.c = 4;
obj1.b.e.f = 100;
console.log(JSON.stringify(obj1));
//{"a":4,"b":{"c":4,"e":{"f":100}}}
console.log(JSON.stringify(obj3));
//{"a":0,"b":{"c":4,"e":{"f":100}}}
深度复制:lodash_.cloneDeep()
取消引用字段:而不是复制对象的引用
let obj1 = {
a: 0,
b: {
c: 0,
e: {
f: 0
}
}
};
let obj3 = _.cloneDeep(obj1);
obj1.a = 100;
obj1.b.c = 100;
obj1.b.e.f = 100;
console.log(JSON.stringify(obj1));
{"a":100,"b":{"c":100,"e":{"f":100}}}
console.log(JSON.stringify(obj3));
{"a":0,"b":{"c":0,"e":{"f":0}}}
我找到了一种用函数克隆对象的方法(打断多行以便于理解):
const clone = Object.assign(
Object.create(
Object.getPrototypeOf(originalObject)
),
dataObject
);
//
// creates 'clone' method on context object
//
// var
// clon = Object.clone( anyValue );
//
!((function (propertyName, definition) {
this[propertyName] = definition();
}).call(
Object,
"clone",
function () {
function isfn(fn) {
return typeof fn === "function";
}
function isobj(o) {
return o === Object(o);
}
function isarray(o) {
return Object.prototype.toString.call(o) === "[object Array]";
}
function fnclon(fn) {
return function () {
fn.apply(this, arguments);
};
}
function owns(obj, p) {
return obj.hasOwnProperty(p);
}
function isemptyobj(obj) {
for (var p in obj) {
return false;
}
return true;
}
function isObject(o) {
return Object.prototype.toString.call(o) === "[object Object]";
}
return function (input) {
if (isfn(input)) {
return fnclon(input);
} else if (isobj(input)) {
var cloned = {};
for (var p in input) {
owns(Object.prototype, p)
|| (
isfn(input[p])
&& ( cloned[p] = function () { return input[p].apply(input, arguments); } )
|| ( cloned[p] = input[p] )
);
}
if (isarray(input)) {
cloned.length = input.length;
"concat every filter forEach indexOf join lastIndexOf map pop push reduce reduceRight reverse shift slice some sort splice toLocaleString toString unshift"
.split(" ")
.forEach(
function (methodName) {
isfn( Array.prototype[methodName] )
&& (
cloned[methodName] =
function () {
return Array.prototype[methodName].apply(cloned, arguments);
}
);
}
);
}
return isemptyobj(cloned)
? (
isObject(input)
? cloned
: input
)
: cloned;
} else {
return input;
}
};
}
));
//
在JavaScript中复制对象的方法
使用排列(…)语法使用Object.assign()方法使用JSON.stringify()和JSON.parse()方法
const person = {
firstName: 'John',
lastName: 'Doe'
};
// using spread ...
let p1 = {
...person
};
// using Object.assign() method
let p2 = Object.assign({}, person);
// using JSON
let p3 = JSON.parse(JSON.stringify(person));
如果对象中没有循环依赖关系,我建议使用其他答案之一或jQuery的复制方法,因为它们看起来都很有效。
如果存在循环依赖关系(即,两个子对象彼此链接),那么你就有点糟糕了,因为(从理论角度)没有办法优雅地解决这个问题。