什么是最有效的方式来克隆一个JavaScript对象?我已经看到obj = eval(uneval(o));被使用,但它是非标准的,仅支持Firefox.我做了事情,如obj = JSON.parse(JSON.stringify(o));但质疑效率。
当前回答
class Handler { static deepCopy (obj) { if (Object.prototype.toString.call(obj) === '[object Array]') { const result = []; for (let i = 0, len = obj.length; i < len; i++) { result[i] = Handler.deepCopy(obj[i]); } return result; } else if (Object.prototype.toString.call(obj) === '[object Object]') { const result = {}; for (let prop in obj) { result[prop] = Handler.deepCopy(obj[prop]); } return result; } return obj; } }
其他回答
代码:
// extends 'from' object with members from 'to'. If 'to' is null, a deep clone of 'from' is returned
function extend(from, to)
{
if (from == null || typeof from != "object") return from;
if (from.constructor != Object && from.constructor != Array) return from;
if (from.constructor == Date || from.constructor == RegExp || from.constructor == Function ||
from.constructor == String || from.constructor == Number || from.constructor == Boolean)
return new from.constructor(from);
to = to || new from.constructor();
for (var name in from)
{
to[name] = typeof to[name] == "undefined" ? extend(from[name], null) : to[name];
}
return to;
}
测试:
var obj =
{
date: new Date(),
func: function(q) { return 1 + q; },
num: 123,
text: "asdasd",
array: [1, "asd"],
regex: new RegExp(/aaa/i),
subobj:
{
num: 234,
text: "asdsaD"
}
}
var clone = extend(obj);
没有触摸原型遗产,你可以如下深孤独的物体和岩石;
函数 objectClone(o){ var ot = Array.isArray(o); return o!== null && typeof o === “object”? Object.keys(o).reduce((r,k) => o[k]!== null && typeof o[k] === “object”? (r[k] = objectClone(o[k]),r) : (r[k] = o[k],r), ot?
如果您正在使用它,UnderScore.js图书馆有一个克隆方法。
var newObject = _.clone(oldObject);
这是我的解决方案,没有使用任何图书馆或本地JavaScript功能。
function deepClone(obj) {
if (typeof obj !== "object") {
return obj;
} else {
let newObj =
typeof obj === "object" && obj.length !== undefined ? [] : {};
for (let key in obj) {
if (key) {
newObj[key] = deepClone(obj[key]);
}
}
return newObj;
}
}
A Recursive Deep Clone 比 JSON.parse(JSON.stringify(obj)) 提到的更快。
Jsperf 在这里排名第一: https://jsperf.com/deep-copy-vs-json-stringify-json-parse/5 Jsben 从上面的答案更新显示,一个重复的深度克隆打击所有其他提到的: http://jsben.ch/13YKQ
下面是快速参考的功能:
function cloneDeep (o) {
let newO
let i
if (typeof o !== 'object') return o
if (!o) return o
if (Object.prototype.toString.apply(o) === '[object Array]') {
newO = []
for (i = 0; i < o.length; i += 1) {
newO[i] = cloneDeep(o[i])
}
return newO
}
newO = {}
for (i in o) {
if (o.hasOwnProperty(i)) {
newO[i] = cloneDeep(o[i])
}
}
return newO
}