我扩展了David Coallier的方法：

增加了合并多个对象的可能性支持深层对象override参数（如果最后一个参数是布尔值，则检测到）

如果覆盖为false，则不会覆盖任何属性，但会添加新的财产。

用法：obj.merge（合并…[，覆盖]）；

这是我的代码：

Object.defineProperty(Object.prototype, "merge", {
    enumerable: false,
    value: function () {
        var override = true,
            dest = this,
            len = arguments.length,
            props, merge, i, from;

        if (typeof(arguments[arguments.length - 1]) === "boolean") {
            override = arguments[arguments.length - 1];
            len = arguments.length - 1;
        }

        for (i = 0; i < len; i++) {
            from = arguments[i];
            if (from != null) {
                Object.getOwnPropertyNames(from).forEach(function (name) {
                    var descriptor;

                    // nesting
                    if ((typeof(dest[name]) === "object" || typeof(dest[name]) === "undefined")
                            && typeof(from[name]) === "object") {

                        // ensure proper types (Array rsp Object)
                        if (typeof(dest[name]) === "undefined") {
                            dest[name] = Array.isArray(from[name]) ? [] : {};
                        }
                        if (override) {
                            if (!Array.isArray(dest[name]) && Array.isArray(from[name])) {
                                dest[name] = [];
                            }
                            else if (Array.isArray(dest[name]) && !Array.isArray(from[name])) {
                                dest[name] = {};
                            }
                        }
                        dest[name].merge(from[name], override);
                    } 

                    // flat properties
                    else if ((name in dest && override) || !(name in dest)) {
                        descriptor = Object.getOwnPropertyDescriptor(from, name);
                        if (descriptor.configurable) {
                            Object.defineProperty(dest, name, descriptor);
                        }
                    }
                });
            }
        }
        return this;
    }
});

示例和测试用例：

function clone (obj) {
    return JSON.parse(JSON.stringify(obj));
}
var obj = {
    name : "trick",
    value : "value"
};

var mergeObj = {
    name : "truck",
    value2 : "value2"
};

var mergeObj2 = {
    name : "track",
    value : "mergeObj2",
    value2 : "value2-mergeObj2",
    value3 : "value3"
};

assertTrue("Standard", clone(obj).merge(mergeObj).equals({
    name : "truck",
    value : "value",
    value2 : "value2"
}));

assertTrue("Standard no Override", clone(obj).merge(mergeObj, false).equals({
    name : "trick",
    value : "value",
    value2 : "value2"
}));

assertTrue("Multiple", clone(obj).merge(mergeObj, mergeObj2).equals({
    name : "track",
    value : "mergeObj2",
    value2 : "value2-mergeObj2",
    value3 : "value3"
}));

assertTrue("Multiple no Override", clone(obj).merge(mergeObj, mergeObj2, false).equals({
    name : "trick",
    value : "value",
    value2 : "value2",
    value3 : "value3"
}));

var deep = {
    first : {
        name : "trick",
        val : "value"
    },
    second : {
        foo : "bar"
    }
};

var deepMerge = {
    first : {
        name : "track",
        anotherVal : "wohoo"
    },
    second : {
        foo : "baz",
        bar : "bam"
    },
    v : "on first layer"
};

assertTrue("Deep merges", clone(deep).merge(deepMerge).equals({
    first : {
        name : "track",
        val : "value",
        anotherVal : "wohoo"
    },
    second : {
        foo : "baz",
        bar : "bam"
    },
    v : "on first layer"
}));

assertTrue("Deep merges no override", clone(deep).merge(deepMerge, false).equals({
    first : {
        name : "trick",
        val : "value",
        anotherVal : "wohoo"
    },
    second : {
        foo : "bar",
        bar : "bam"
    },
    v : "on first layer"
}));

var obj1 = {a: 1, b: "hello"};
obj1.merge({c: 3});
assertTrue(obj1.equals({a: 1, b: "hello", c: 3}));

obj1.merge({a: 2, b: "mom", d: "new property"}, false);
assertTrue(obj1.equals({a: 1, b: "hello", c: 3, d: "new property"}));

var obj2 = {};
obj2.merge({a: 1}, {b: 2}, {a: 3});
assertTrue(obj2.equals({a: 3, b: 2}));

var a = [];
var b = [1, [2, 3], 4];
a.merge(b);
assertEquals(1, a[0]);
assertEquals([2, 3], a[1]);
assertEquals(4, a[2]);


var o1 = {};
var o2 = {a: 1, b: {c: 2}};
var o3 = {d: 3};
o1.merge(o2, o3);
assertTrue(o1.equals({a: 1, b: {c: 2}, d: 3}));
o1.b.c = 99;
assertTrue(o2.equals({a: 1, b: {c: 2}}));

// checking types with arrays and objects
var bo;
a = [];
bo = [1, {0:2, 1:3}, 4];
b = [1, [2, 3], 4];

a.merge(b);
assertTrue("Array stays Array?", Array.isArray(a[1]));

a = [];
a.merge(bo);
assertTrue("Object stays Object?", !Array.isArray(a[1]));

a = [];
a.merge(b);
a.merge(bo);
assertTrue("Object overrides Array", !Array.isArray(a[1]));

a = [];
a.merge(b);
a.merge(bo, false);
assertTrue("Object does not override Array", Array.isArray(a[1]));

a = [];
a.merge(bo);
a.merge(b);
assertTrue("Array overrides Object", Array.isArray(a[1]));

a = [];
a.merge(bo);
a.merge(b, false);
assertTrue("Array does not override Object", !Array.isArray(a[1]));

我的equals方法可以在这里找到：JavaScript中的对象比较

我刚开始使用JavaScript，所以如果我错了，请纠正我。

但如果可以合并任意数量的对象，不是更好吗？下面是我如何使用本机Arguments对象实现的。

关键在于，实际上可以向JavaScript函数传递任意数量的参数，而无需在函数声明中定义它们。如果不使用Arguments对象，就无法访问它们。

function mergeObjects() (
    var tmpObj = {};

    for(var o in arguments) {
        for(var m in arguments[o]) {
            tmpObj[m] = arguments[o][m];
        }
    }
    return tmpObj;
}

在Ext JS 4中，可以如下所示：

var mergedObject = Ext.Object.merge(object1, object2)

// Or shorter:
var mergedObject2 = Ext.merge(object1, object2)

请参见合并（对象）：对象。

使用object.assign和spread运算符合并两个对象。

错误的方式（修改原始对象，因为目标是o1）

var o1 = { X: 10 };
var o2 = { Y: 20 };
var o3 = { Z: 30 };
var merge = Object.assign(o1, o2, o3);
console.log(merge)  // {X:10, Y:20, Z:30}
console.log(o1)     // {X:10, Y:20, Z:30}

正确的方式

Object.assign（{}，o1，o2，o3）==>以新对象为目标｛…o1，…o2，…o3｝==>扩展对象

变量o1＝{X:10}；var o2＝{Y:20}；var o3＝{Z:30}；console.log（'不修改原始对象，因为目标｛｝'）；var merge=对象赋值（｛｝，o1，o2，o3）；console.log（合并）；//{X:10，Y:20，Z:30}控制台日志（o1）console.log（'不修改原始对象'）var spreadMerge＝{…o1，…o2，…o3}；console.log（spreadMerge）；控制台日志（o1）；

这似乎就是你所需要的：

var obj1 = { food: 'pizza', car: 'ford' }
var obj2 = { animal: 'dog' }

var obj3 = { ...obj1, ...obj2 }

之后，obj3现在应该具有以下值：

{food: "pizza", car: "ford", animal: "dog"}

在这里试试：

var obj1={food:“pizza”，car:“ford”}var obj2=｛动物：“狗”｝var obj3=｛…obj1，…obj2｝console.log（obj3）；

使用此

var t=merge（｛test:123｝，｛mysecondTest：｛blub:｛test2:‘string‘｝，args：｛‘test‘：2｝｝）控制台日志（t）；函数合并（…args）{return Object.assign（｛｝，…args）；}