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
如下图所示,John Resig的jQuery克隆器将非数字属性的<unk>转化为非数字属性的对象,而RegDwight的JSON克隆器将非数字属性放下。
function jQueryClone(obj) {
return jQuery.extend(true, {}, obj)
}
function JSONClone(obj) {
return JSON.parse(JSON.stringify(obj))
}
var arrayLikeObj = [[1, "a", "b"], [2, "b", "a"]];
arrayLikeObj.names = ["m", "n", "o"];
var JSONCopy = JSONClone(arrayLikeObj);
var jQueryCopy = jQueryClone(arrayLikeObj);
alert("Is arrayLikeObj an array instance?" + (arrayLikeObj instanceof Array) +
"\nIs the jQueryClone an array instance? " + (jQueryCopy instanceof Array) +
"\nWhat are the arrayLikeObj names? " + arrayLikeObj.names +
"\nAnd what are the JSONClone names? " + JSONCopy.names)
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);
通过提出新的方法 Object.fromEntries(),它支持一些浏览器的更新的版本(参考)。
const obj = { key1: {key11: “key11”, key12: “key12”, key13: {key131: 22}}, key2: {key21: “key21”, key22: “key22”}, key3: “key3”, key4: [1,2,3, {key: “value”} } const cloneObj = (obj) => { if (Object(obj)!== obj) return obj; other if (Array.isArray(obj)) return obj.map(cloneObj); return Object.fromEntries(Object.entries(obj).map(([k,v])
对于未来的参考,可以使用此代码
第6章:
_clone: function(obj){
let newObj = {};
for(let i in obj){
if(typeof(obj[i]) === 'object' && Object.keys(obj[i]).length){
newObj[i] = clone(obj[i]);
} else{
newObj[i] = obj[i];
}
}
return Object.assign({},newObj);
}
第5章:
function clone(obj){
let newObj = {};
for(let i in obj){
if(typeof(obj[i]) === 'object' && Object.keys(obj[i]).length){
newObj[i] = clone(obj[i]);
} else{
newObj[i] = obj[i];
}
}
return Object.assign({},newObj);
) )
E.G
var obj ={a:{b:1,c:3},d:4,e:{f:6}}
var xc = clone(obj);
console.log(obj); //{a:{b:1,c:3},d:4,e:{f:6}}
console.log(xc); //{a:{b:1,c:3},d:4,e:{f:6}}
xc.a.b = 90;
console.log(obj); //{a:{b:1,c:3},d:4,e:{f:6}}
console.log(xc); //{a:{b:90,c:3},d:4,e:{f:6}}
对象的克隆一直是JS的担忧,但在ES6之前,我列出了在下面的JavaScript中复制对象的不同方式,想象你有下面的对象,我希望有一个深刻的副本:
var obj = {a:1, b:2, c:3, d:4};
有几种方法可以复制这个对象,而不会改变其起源:
ES5+,使用一个简单的函数来为您进行复制:函数 deepCopyObj(obj) {如果(null == obj <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk>
希望这些帮助......
在我的经验中,一个重复版本大大超越了JSON.parse(JSON.stringify(obj))。这里是一个现代化的重复深对象复制功能,可以在单一线上匹配:
function deepCopy(obj) {
return Object.keys(obj).reduce((v, d) => Object.assign(v, {
[d]: (obj[d].constructor === Object) ? deepCopy(obj[d]) : obj[d]
}), {});
}
这比JSON.parse方法快40倍。