严格相等运算符将告诉您两个对象类型是否相等。然而,是否有一种方法来判断两个对象是否相等,就像Java中的哈希码值一样?

堆栈溢出问题JavaScript中有hashCode函数吗?类似于这个问题,但需要一个更学术的答案。上面的场景说明了为什么有必要有一个,我想知道是否有等效的解决方案。


当前回答

我的版本,包括发现差异的地方,以及差异是什么。

function DeepObjectCompare(O1, O2)
{
    try {
        DOC_Val(O1, O2, ['O1->O2', O1, O2]);
        return DOC_Val(O2, O1, ['O2->O1', O1, O2]);
    } catch(e) {
        console.log(e.Chain);
        throw(e);
    }
}
function DOC_Error(Reason, Chain, Val1, Val2)
{
    this.Reason=Reason;
    this.Chain=Chain;
    this.Val1=Val1;
    this.Val2=Val2;
}

function DOC_Val(Val1, Val2, Chain)
{
    function DoThrow(Reason, NewChain) { throw(new DOC_Error(Reason, NewChain!==undefined ? NewChain : Chain, Val1, Val2)); }

    if(typeof(Val1)!==typeof(Val2))
        return DoThrow('Type Mismatch');
    if(Val1===null || Val1===undefined)
        return Val1!==Val2 ? DoThrow('Null/undefined mismatch') : true;
    if(Val1.constructor!==Val2.constructor)
        return DoThrow('Constructor mismatch');
    switch(typeof(Val1))
    {
        case 'object':
            for(var m in Val1)
            {
                if(!Val1.hasOwnProperty(m))
                    continue;
                var CurChain=Chain.concat([m]);
                if(!Val2.hasOwnProperty(m))
                    return DoThrow('Val2 missing property', CurChain);
                DOC_Val(Val1[m], Val2[m], CurChain);
            }
            return true;
        case 'number':
            if(Number.isNaN(Val1))
                return !Number.isNaN(Val2) ? DoThrow('NaN mismatch') : true;
        case 'string':
        case 'boolean':
            return Val1!==Val2 ? DoThrow('Value mismatch') : true;
        case 'function':
            if(Val1.prototype!==Val2.prototype)
                return DoThrow('Prototype mismatch');
            if(Val1!==Val2)
                return DoThrow('Function mismatch');
            return true;
        default:
            return DoThrow('Val1 is unknown type');
    }
}

其他回答

短功能deepEqual实现:

function deepEqual(x, y) {
  return (x && y && typeof x === 'object' && typeof y === 'object') ?
    (Object.keys(x).length === Object.keys(y).length) &&
      Object.keys(x).reduce(function(isEqual, key) {
        return isEqual && deepEqual(x[key], y[key]);
      }, true) : (x === y);
}

编辑:版本2,使用jib的建议和ES6的箭头函数:

function deepEqual(x, y) {
  const ok = Object.keys, tx = typeof x, ty = typeof y;
  return x && y && tx === 'object' && tx === ty ? (
    ok(x).length === ok(y).length &&
      ok(x).every(key => deepEqual(x[key], y[key]))
  ) : (x === y);
}

我需要模拟jQuery POST请求,因此对我来说重要的是两个对象具有相同的属性集(任何一个对象中都不缺少属性),并且每个属性值都是“相等的”(根据这个定义)。我不关心对象是否有不匹配的方法。

这是我将使用的,它应该足以满足我的特定要求:

function PostRequest() {
    for (var i = 0; i < arguments.length; i += 2) {
        this[arguments[i]] = arguments[i+1];
    }

    var compare = function(u, v) {
        if (typeof(u) != typeof(v)) {
            return false;
        }

        var allkeys = {};
        for (var i in u) {
            allkeys[i] = 1;
        }
        for (var i in v) {
            allkeys[i] = 1;
        }
        for (var i in allkeys) {
            if (u.hasOwnProperty(i) != v.hasOwnProperty(i)) {
                if ((u.hasOwnProperty(i) && typeof(u[i]) == 'function') ||
                    (v.hasOwnProperty(i) && typeof(v[i]) == 'function')) {
                    continue;
                } else {
                    return false;
                }
            }
            if (typeof(u[i]) != typeof(v[i])) {
                return false;
            }
            if (typeof(u[i]) == 'object') {
                if (!compare(u[i], v[i])) {
                    return false;
                }
            } else {
                if (u[i] !== v[i]) {
                    return false;
                }
            }
        }

        return true;
    };

    this.equals = function(o) {
        return compare(this, o);
    };

    return this;
}

像这样使用:

foo = new PostRequest('text', 'hello', 'html', '<p>hello</p>');
foo.equals({ html: '<p>hello</p>', text: 'hello' });

let user1 = { name: "John", address: { line1: "55 Green Park Road", line2: { a:[1,2,3] } }, email:null } let user2 = { name: "John", address: { line1: "55 Green Park Road", line2: { a:[1,2,3] } }, email:null } // Method 1 function isEqual(a, b) { return JSON.stringify(a) === JSON.stringify(b); } // Method 2 function isEqual(a, b) { // checking type of a And b if(typeof a !== 'object' || typeof b !== 'object') { return false; } // Both are NULL if(!a && !b ) { return true; } else if(!a || !b) { return false; } let keysA = Object.keys(a); let keysB = Object.keys(b); if(keysA.length !== keysB.length) { return false; } for(let key in a) { if(!(key in b)) { return false; } if(typeof a[key] === 'object') { if(!isEqual(a[key], b[key])) { return false; } } else { if(a[key] !== b[key]) { return false; } } } return true; } console.log(isEqual(user1,user2));

简短的回答

简单的答案是:不,没有一般的方法来确定一个对象等于另一个你所指的意义。例外情况是当您严格地认为一个对象是无类型的。

长话短说

这个概念是一个Equals方法,它比较一个对象的两个不同实例,以指示它们在值级别上是否相等。但是,定义Equals方法应该如何实现取决于具体的类型。对具有基本值的属性进行迭代比较可能还不够:对象可能包含与相等无关的属性。例如,

 function MyClass(a, b)
 {
     var c;
     this.getCLazy = function() {
         if (c === undefined) c = a * b // imagine * is really expensive
         return c;
     }
  }

在上面的例子中,c对于确定MyClass的任何两个实例是否相等并不重要,只有a和b是重要的。在某些情况下,c可能在不同实例之间有所不同,但在比较中并不显著。

注意,当成员本身也可能是某个类型的实例,并且每个实例都需要有确定相等的方法时,这个问题就会出现。

更复杂的是,在JavaScript中,数据和方法之间的区别是模糊的。

一个对象可以引用一个作为事件处理程序调用的方法,这可能不被认为是其“值状态”的一部分。然而,另一个对象很可能被分配一个执行重要计算的函数,从而使这个实例与其他实例不同,仅仅因为它引用了不同的函数。

如果一个对象的现有原型方法被另一个函数覆盖,该怎么办?它还能被认为与另一个相同的实例相等吗?这个问题只能在每种类型的具体情况下回答。

如前所述,异常将是一个严格的无类型对象。在这种情况下,唯一明智的选择是对每个成员进行迭代和递归比较。即使这样,人们也要问一个函数的“值”是什么?

这是我的版本。它正在使用new Object。ES5中引入的keys特性以及+、+和+的想法/测试:

function objectEquals(x, y) { 'use strict'; if (x === null || x === undefined || y === null || y === undefined) { return x === y; } // after this just checking type of one would be enough if (x.constructor !== y.constructor) { return false; } // if they are functions, they should exactly refer to same one (because of closures) if (x instanceof Function) { return x === y; } // if they are regexps, they should exactly refer to same one (it is hard to better equality check on current ES) if (x instanceof RegExp) { return x === y; } if (x === y || x.valueOf() === y.valueOf()) { return true; } if (Array.isArray(x) && x.length !== y.length) { return false; } // if they are dates, they must had equal valueOf if (x instanceof Date) { return false; } // if they are strictly equal, they both need to be object at least if (!(x instanceof Object)) { return false; } if (!(y instanceof Object)) { return false; } // recursive object equality check var p = Object.keys(x); return Object.keys(y).every(function (i) { return p.indexOf(i) !== -1; }) && p.every(function (i) { return objectEquals(x[i], y[i]); }); } /////////////////////////////////////////////////////////////// /// The borrowed tests, run them by clicking "Run code snippet" /////////////////////////////////////////////////////////////// var printResult = function (x) { if (x) { document.write('<div style="color: green;">Passed</div>'); } else { document.write('<div style="color: red;">Failed</div>'); } }; var assert = { isTrue: function (x) { printResult(x); }, isFalse: function (x) { printResult(!x); } } assert.isTrue(objectEquals(null,null)); assert.isFalse(objectEquals(null,undefined)); assert.isFalse(objectEquals(/abc/, /abc/)); assert.isFalse(objectEquals(/abc/, /123/)); var r = /abc/; assert.isTrue(objectEquals(r, r)); assert.isTrue(objectEquals("hi","hi")); assert.isTrue(objectEquals(5,5)); assert.isFalse(objectEquals(5,10)); assert.isTrue(objectEquals([],[])); assert.isTrue(objectEquals([1,2],[1,2])); assert.isFalse(objectEquals([1,2],[2,1])); assert.isFalse(objectEquals([1,2],[1,2,3])); assert.isTrue(objectEquals({},{})); assert.isTrue(objectEquals({a:1,b:2},{a:1,b:2})); assert.isTrue(objectEquals({a:1,b:2},{b:2,a:1})); assert.isFalse(objectEquals({a:1,b:2},{a:1,b:3})); assert.isTrue(objectEquals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}},{1:{name:"mhc",age:28}, 2:{name:"arb",age:26}})); assert.isFalse(objectEquals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}},{1:{name:"mhc",age:28}, 2:{name:"arb",age:27}})); Object.prototype.equals = function (obj) { return objectEquals(this, obj); }; var assertFalse = assert.isFalse, assertTrue = assert.isTrue; assertFalse({}.equals(null)); assertFalse({}.equals(undefined)); assertTrue("hi".equals("hi")); assertTrue(new Number(5).equals(5)); assertFalse(new Number(5).equals(10)); assertFalse(new Number(1).equals("1")); assertTrue([].equals([])); assertTrue([1,2].equals([1,2])); assertFalse([1,2].equals([2,1])); assertFalse([1,2].equals([1,2,3])); assertTrue(new Date("2011-03-31").equals(new Date("2011-03-31"))); assertFalse(new Date("2011-03-31").equals(new Date("1970-01-01"))); assertTrue({}.equals({})); assertTrue({a:1,b:2}.equals({a:1,b:2})); assertTrue({a:1,b:2}.equals({b:2,a:1})); assertFalse({a:1,b:2}.equals({a:1,b:3})); assertTrue({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}}.equals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}})); assertFalse({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}}.equals({1:{name:"mhc",age:28}, 2:{name:"arb",age:27}})); var a = {a: 'text', b:[0,1]}; var b = {a: 'text', b:[0,1]}; var c = {a: 'text', b: 0}; var d = {a: 'text', b: false}; var e = {a: 'text', b:[1,0]}; var i = { a: 'text', c: { b: [1, 0] } }; var j = { a: 'text', c: { b: [1, 0] } }; var k = {a: 'text', b: null}; var l = {a: 'text', b: undefined}; assertTrue(a.equals(b)); assertFalse(a.equals(c)); assertFalse(c.equals(d)); assertFalse(a.equals(e)); assertTrue(i.equals(j)); assertFalse(d.equals(k)); assertFalse(k.equals(l)); // from comments on stackoverflow post assert.isFalse(objectEquals([1, 2, undefined], [1, 2])); assert.isFalse(objectEquals([1, 2, 3], { 0: 1, 1: 2, 2: 3 })); assert.isFalse(objectEquals(new Date(1234), 1234)); // no two different function is equal really, they capture their context variables // so even if they have same toString(), they won't have same functionality var func = function (x) { return true; }; var func2 = function (x) { return true; }; assert.isTrue(objectEquals(func, func)); assert.isFalse(objectEquals(func, func2)); assert.isTrue(objectEquals({ a: { b: func } }, { a: { b: func } })); assert.isFalse(objectEquals({ a: { b: func } }, { a: { b: func2 } }));