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 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"}

其他回答

如果有人正在使用Google闭包库:

goog.require('goog.object');
var a = {'a': 1, 'b': 2};
var b = {'b': 3, 'c': 4};
goog.object.extend(a, b);
// Now object a == {'a': 1, 'b': 3, 'c': 4};

数组存在类似的助手函数:

var a = [1, 2];
var b = [3, 4];
goog.array.extend(a, b); // Extends array 'a'
goog.array.concat(a, b); // Returns concatenation of array 'a' and 'b'

此解决方案创建一个新对象,并能够处理多个对象。

此外,它是递归的,您可以选择要覆盖值和对象的天气。

    function extendObjects() {

        var newObject        = {};
        var overwriteValues  = false;
        var overwriteObjects = false;

        for ( var indexArgument = 0; indexArgument < arguments.length; indexArgument++ ) {

            if ( typeof arguments[indexArgument] !== 'object' ) {

                if ( arguments[indexArgument] == 'overwriteValues_True' ) {

                    overwriteValues = true;            
                } else if ( arguments[indexArgument] == 'overwriteValues_False' ) {

                    overwriteValues = false;                             
                } else if ( arguments[indexArgument] == 'overwriteObjects_True' ) {

                    overwriteObjects = true;     
                } else if ( arguments[indexArgument] == 'overwriteObjects_False' ) {

                    overwriteObjects = false; 
                }

            } else {

                extendObject( arguments[indexArgument], newObject, overwriteValues, overwriteObjects );
            }

        }

        function extendObject( object, extendedObject, overwriteValues, overwriteObjects ) {

            for ( var indexObject in object ) {

                if ( typeof object[indexObject] === 'object' ) {

                    if ( typeof extendedObject[indexObject] === "undefined" || overwriteObjects ) {
                        extendedObject[indexObject] = object[indexObject];
                    }

                    extendObject( object[indexObject], extendedObject[indexObject], overwriteValues, overwriteObjects );

                } else {

                    if ( typeof extendedObject[indexObject] === "undefined" || overwriteValues ) {
                        extendedObject[indexObject] = object[indexObject];
                    }

                }

            }     

            return extendedObject;

        }

        return newObject;
    }

    var object1           = { a : 1, b : 2, testArr : [888, { innArr : 1 }, 777 ], data : { e : 12, c : { lol : 1 }, rofl : { O : 3 } } };
    var object2           = { a : 6, b : 9, data : { a : 17, b : 18, e : 13, rofl : { O : 99, copter : { mao : 1 } } }, hexa : { tetra : 66 } };
    var object3           = { f : 13, g : 666, a : 333, data : { c : { xD : 45 } }, testArr : [888, { innArr : 3 }, 555 ]  };

    var newExtendedObject = extendObjects( 'overwriteValues_False', 'overwriteObjects_False', object1, object2, object3 );

newExtendedObject的内容:

{"a":1,"b":2,"testArr":[888,{"innArr":1},777],"data":{"e":12,"c":{"lol":1,"xD":45},"rofl":{"O":3,"copter":{"mao":1}},"a":17,"b":18},"hexa":{"tetra":66},"f":13,"g":666}

小提琴:http://jsfiddle.net/o0gb2umb/

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;
}

请注意,underline.js的extend方法在一行中实现了这一点:

_.extend({name : 'moe'}, {age : 50});
=> {name : 'moe', age : 50}

浅的

var obj = { name : "Jacob" , address : ["America"] }
var obj2 = { name : "Shaun" , address : ["Honk Kong"] }

var merged = Object.assign({} , obj,obj2 ); //shallow merge 
obj2.address[0] = "new city"

result.地址[0]更改为“新城”,即合并对象也更改。这就是浅层合并的问题。

deep

var obj = { name : "Jacob" , address : ["America"] }
var obj2 = { name : "Shaun" , address : ["Honk Kong"] }

var result = Object.assign({} , JSON.parse(JSON.stringify(obj)),JSON.parse(JSON.stringify(obj2)) )

obj2.address[0] = "new city"

result.address[0]未更改