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
var firstObject = {
key1 : 'value1',
key2 : 'value2'
};
var secondObject={
...firstObject,
key3 : 'value3',
key4 : 'value4',
key5 : 'value5'
}
console.log(firstObject);
console.log(secondObject);
其他回答
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;
}
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 }));
我的方式:
function mergeObjects(defaults, settings) {
Object.keys(defaults).forEach(function(key_default) {
if (typeof settings[key_default] == "undefined") {
settings[key_default] = defaults[key_default];
} else if (isObject(defaults[key_default]) && isObject(settings[key_default])) {
mergeObjects(defaults[key_default], settings[key_default]);
}
});
function isObject(object) {
return Object.prototype.toString.call(object) === '[object Object]';
}
return settings;
}
:)
以下两个可能是一个很好的起点。lodash还为那些特殊需求提供定制功能!
_.延伸(http://underscorejs.org/#extend) _.合并(https://lodash.com/docs#merge)
var obj1 = { food: 'pizza', car: 'ford' }
var obj2 = { animal: 'dog' }
// result
result: {food: "pizza", car: "ford", animal: "dog"}
使用jQuery.exde()-Link
// Merge obj1 & obj2 to result
var result1 = $.extend( {}, obj1, obj2 );
使用_.merge()-链接
// Merge obj1 & obj2 to result
var result2 = _.merge( {}, obj1, obj2 );
使用_.exde()-链接
// Merge obj1 & obj2 to result
var result3 = _.extend( {}, obj1, obj2 );
使用Object.assign()ECMAScript 2015(ES6)-Link
// Merge obj1 & obj2 to result
var result4 = Object.assign( {}, obj1, obj2 );
全部输出
obj1: { animal: 'dog' }
obj2: { food: 'pizza', car: 'ford' }
result1: {food: "pizza", car: "ford", animal: "dog"}
result2: {food: "pizza", car: "ford", animal: "dog"}
result3: {food: "pizza", car: "ford", animal: "dog"}
result4: {food: "pizza", car: "ford", animal: "dog"}