我需要能够在运行时合并两个(非常简单)JavaScript对象。例如,我想:
var obj1 = { food: 'pizza', car: 'ford' }
var obj2 = { animal: 'dog' }
obj1.merge(obj2);
//obj1 now has three properties: food, car, and animal
是否有一种内置的方法来实现这一点?我不需要递归,也不需要合并函数,只需要平面对象上的方法。
我需要能够在运行时合并两个(非常简单)JavaScript对象。例如,我想:
var obj1 = { food: 'pizza', car: 'ford' }
var obj2 = { animal: 'dog' }
obj1.merge(obj2);
//obj1 now has three properties: food, car, and animal
是否有一种内置的方法来实现这一点?我不需要递归,也不需要合并函数,只需要平面对象上的方法。
当前回答
基于Markus和vsync的回答,这是一个扩展版本。该函数接受任意数量的参数。它可以用于设置DOM节点上的财产,并对值进行深度复制。然而,第一个论点是通过引用给出的。
要检测DOM节点,使用isDOMNode()函数(请参阅堆栈溢出问题JavaScript isDOM-如何检查JavaScript对象是否为DOM对象?)
它在Opera 11、Firefox 6、Internet Explorer 8和Google Chrome 16中进行了测试。
Code
function mergeRecursive() {
// _mergeRecursive does the actual job with two arguments.
var _mergeRecursive = function (dst, src) {
if (isDOMNode(src) || typeof src !== 'object' || src === null) {
return dst;
}
for (var p in src) {
if (!src.hasOwnProperty(p))
continue;
if (src[p] === undefined)
continue;
if ( typeof src[p] !== 'object' || src[p] === null) {
dst[p] = src[p];
} else if (typeof dst[p]!=='object' || dst[p] === null) {
dst[p] = _mergeRecursive(src[p].constructor===Array ? [] : {}, src[p]);
} else {
_mergeRecursive(dst[p], src[p]);
}
}
return dst;
}
// Loop through arguments and merge them into the first argument.
var out = arguments[0];
if (typeof out !== 'object' || out === null)
return out;
for (var i = 1, il = arguments.length; i < il; i++) {
_mergeRecursive(out, arguments[i]);
}
return out;
}
一些示例
设置HTML元素的innerHTML和样式
mergeRecursive(
document.getElementById('mydiv'),
{style: {border: '5px solid green', color: 'red'}},
{innerHTML: 'Hello world!'});
合并数组和对象。请注意,undefined可用于保存左侧数组/对象中的值。
o = mergeRecursive({a:'a'}, [1,2,3], [undefined, null, [30,31]], {a:undefined, b:'b'});
// o = {0:1, 1:null, 2:[30,31], a:'a', b:'b'}
任何非JavaScript对象的参数(包括null)都将被忽略。除了第一个参数之外,也会丢弃DOM节点。注意,像new String()这样创建的字符串实际上是对象。
o = mergeRecursive({a:'a'}, 1, true, null, undefined, [1,2,3], 'bc', new String('de'));
// o = {0:'d', 1:'e', 2:3, a:'a'}
如果要将两个对象合并为一个新对象(不影响其中任何一个),请提供{}作为第一个参数
var a={}, b={b:'abc'}, c={c:'cde'}, o;
o = mergeRecursive(a, b, c);
// o===a is true, o===b is false, o===c is false
编辑(由收割者很快):
还要合并阵列
function mergeRecursive(obj1, obj2) {
if (Array.isArray(obj2)) { return obj1.concat(obj2); }
for (var p in obj2) {
try {
// Property in destination object set; update its value.
if ( obj2[p].constructor==Object ) {
obj1[p] = mergeRecursive(obj1[p], obj2[p]);
} else if (Array.isArray(obj2[p])) {
obj1[p] = obj1[p].concat(obj2[p]);
} else {
obj1[p] = obj2[p];
}
} catch(e) {
// Property in destination object not set; create it and set its value.
obj1[p] = obj2[p];
}
}
return obj1;
}
其他回答
function extend()
{
var o = {};
for (var i in arguments)
{
var s = arguments[i];
for (var i in s)
{
o[i] = s[i];
}
}
return o;
}
ECMAScript 2018标准方法
您可以使用对象扩散:
let merged = {...obj1, ...obj2};
merged现在是obj1和obj2的并集。obj2中的财产将覆盖obj1中的属性。
/** There's no limit to the number of objects you can merge.
* Later properties overwrite earlier properties with the same name. */
const allRules = {...obj1, ...obj2, ...obj3};
这里还有此语法的MDN文档。如果您正在使用babel,则需要@babel/plugin提议对象rest spread插件才能工作(该插件包含在ES2018中的@babel/preset-env中)。
ECMAScript 2015(ES6)标准方法
/* For the case in question, you would do: */
Object.assign(obj1, obj2);
/** There's no limit to the number of objects you can merge.
* All objects get merged into the first object.
* Only the object in the first argument is mutated and returned.
* Later properties overwrite earlier properties with the same name. */
const allRules = Object.assign({}, obj1, obj2, obj3, etc);
(参见MDN JavaScript参考)
ES5及更早版本的方法
for (var attrname in obj2) { obj1[attrname] = obj2[attrname]; }
请注意,这将简单地将obj2的所有属性添加到obj1中,如果您仍然希望使用未修改的obj1,那么这可能不是您想要的。
如果你使用的是一个在你的原型上到处都是垃圾的框架,那么你必须通过hasOwnProperty这样的检查来获得更高的效率,但这段代码在99%的情况下都是有效的。
示例函数:
/**
* Overwrites obj1's values with obj2's and adds obj2's if non existent in obj1
* @param obj1
* @param obj2
* @returns obj3 a new object based on obj1 and obj2
*/
function merge_options(obj1,obj2){
var obj3 = {};
for (var attrname in obj1) { obj3[attrname] = obj1[attrname]; }
for (var attrname in obj2) { obj3[attrname] = obj2[attrname]; }
return obj3;
}
Use:
//Takes any number of objects and returns one merged object
var objectMerge = function(){
var out = {};
if(!arguments.length)
return out;
for(var i=0; i<arguments.length; i++) {
for(var key in arguments[i]){
out[key] = arguments[i][key];
}
}
return out;
}
测试方法:
console.log(objectMerge({a:1, b:2}, {a:2, c:4}));
结果是:
{ a: 2, b: 2, c: 4 }
这是我的刺
支持深度合并不改变参数采用任意数量的参数不扩展对象原型不依赖于其他库(jQuery、MooTools、Undercore.js等)包括检查hasOwnProperty短:)/*递归合并财产并返回新对象对象1<;-对象2[<;-…]*/函数合并(){变量dst={},srcp,args=[].splice.call(参数,0);while(参数长度>0){src=参数拼接(0,1)[0];if(toString.call(src)=='[object object]'){for(src中的p){if(src.hasOwnProperty(p)){if(toString.call(src[p])=='[object object]'){dst[p]=合并(dst[p]||{},src[p]);}其他{dst[p]=src[p];}}}}}返回dst;}
例子:
a = {
"p1": "p1a",
"p2": [
"a",
"b",
"c"
],
"p3": true,
"p5": null,
"p6": {
"p61": "p61a",
"p62": "p62a",
"p63": [
"aa",
"bb",
"cc"
],
"p64": {
"p641": "p641a"
}
}
};
b = {
"p1": "p1b",
"p2": [
"d",
"e",
"f"
],
"p3": false,
"p4": true,
"p6": {
"p61": "p61b",
"p64": {
"p642": "p642b"
}
}
};
c = {
"p1": "p1c",
"p3": null,
"p6": {
"p62": "p62c",
"p64": {
"p643": "p641c"
}
}
};
d = merge(a, b, c);
/*
d = {
"p1": "p1c",
"p2": [
"d",
"e",
"f"
],
"p3": null,
"p5": null,
"p6": {
"p61": "p61b",
"p62": "p62c",
"p63": [
"aa",
"bb",
"cc"
],
"p64": {
"p641": "p641a",
"p642": "p642b",
"p643": "p641c"
}
},
"p4": true
};
*/
我今天需要合并对象,这个问题(和答案)对我帮助很大。我尝试了一些答案,但没有一个符合我的需要,所以我组合了一些答案并自己添加了一些东西,并提出了一个新的合并函数。这里是:
var merge = function() {
var obj = {},
i = 0,
il = arguments.length,
key;
for (; i < il; i++) {
for (key in arguments[i]) {
if (arguments[i].hasOwnProperty(key)) {
obj[key] = arguments[i][key];
}
}
}
return obj;
};
一些示例用法:
var t1 = {
key1: 1,
key2: "test",
key3: [5, 2, 76, 21]
};
var t2 = {
key1: {
ik1: "hello",
ik2: "world",
ik3: 3
}
};
var t3 = {
key2: 3,
key3: {
t1: 1,
t2: 2,
t3: {
a1: 1,
a2: 3,
a4: [21, 3, 42, "asd"]
}
}
};
console.log(merge(t1, t2));
console.log(merge(t1, t3));
console.log(merge(t2, t3));
console.log(merge(t1, t2, t3));
console.log(merge({}, t1, { key1: 1 }));