假设对象中属性的顺序没有改变。

JSON.stringify()适用于深度和非深度这两种类型的对象，不太确定性能方面:

Var object1 = { 关键:“价值” }; Var object2 = { 关键:“价值” }; Var object3 = { 键:“无值” }; console.log('object1和object2相等:'，JSON.stringify(object1) === JSON.stringify(object2)); console.log('object2和object3相等:'，JSON.stringify(object2) === JSON.stringify(object3));

下面的一些解决方案在性能、功能和风格方面存在问题……它们没有经过充分的考虑，其中一些在不同的情况下失败了。我试图在自己的解决方案中解决这个问题，我非常感谢您的反馈:

http://stamat.wordpress.com/javascript-object-comparison/

//Returns the object's class, Array, Date, RegExp, Object are of interest to us
var getClass = function(val) {
    return Object.prototype.toString.call(val)
        .match(/^\[object\s(.*)\]$/)[1];
};

//Defines the type of the value, extended typeof
var whatis = function(val) {

    if (val === undefined)
        return 'undefined';
    if (val === null)
        return 'null';

    var type = typeof val;

    if (type === 'object')
        type = getClass(val).toLowerCase();

    if (type === 'number') {
        if (val.toString().indexOf('.') > 0)
            return 'float';
        else
        return 'integer';
    }

    return type;
   };

var compareObjects = function(a, b) {
    if (a === b)
        return true;
    for (var i in a) {
        if (b.hasOwnProperty(i)) {
            if (!equal(a[i],b[i])) return false;
        } else {
            return false;
        }
    }

    for (var i in b) {
        if (!a.hasOwnProperty(i)) {
            return false;
        }
    }
    return true;
};

var compareArrays = function(a, b) {
    if (a === b)
        return true;
    if (a.length !== b.length)
        return false;
    for (var i = 0; i < a.length; i++){
        if(!equal(a[i], b[i])) return false;
    };
    return true;
};

var _equal = {};
_equal.array = compareArrays;
_equal.object = compareObjects;
_equal.date = function(a, b) {
    return a.getTime() === b.getTime();
};
_equal.regexp = function(a, b) {
    return a.toString() === b.toString();
};
//  uncoment to support function as string compare
//  _equal.fucntion =  _equal.regexp;



/*
 * Are two values equal, deep compare for objects and arrays.
 * @param a {any}
 * @param b {any}
 * @return {boolean} Are equal?
 */
var equal = function(a, b) {
    if (a !== b) {
        var atype = whatis(a), btype = whatis(b);

        if (atype === btype)
            return _equal.hasOwnProperty(atype) ? _equal[atype](a, b) : a==b;

        return false;
    }

    return true;
};

需要一个比已经发布的更通用的对象比较函数，我炮制了以下。批判赞赏……

Object.prototype.equals = function(iObj) {
  if (this.constructor !== iObj.constructor)
    return false;
  var aMemberCount = 0;
  for (var a in this) {
    if (!this.hasOwnProperty(a))
      continue;
    if (typeof this[a] === 'object' && typeof iObj[a] === 'object' ? !this[a].equals(iObj[a]) : this[a] !== iObj[a])
      return false;
    ++aMemberCount;
  }
  for (var a in iObj)
    if (iObj.hasOwnProperty(a))
      --aMemberCount;
  return aMemberCount ? false : true;
}

在Node.js中，你可以使用它的原生require("assert"). deepstrictequal。更多信息: http://nodejs.org/api/assert.html

例如:

var assert = require("assert");
assert.deepStrictEqual({a:1, b:2}, {a:1, b:3}); // will throw AssertionError

另一个返回true / false而不是返回错误的例子:

var assert = require("assert");

function deepEqual(a, b) {
    try {
      assert.deepEqual(a, b);
    } catch (error) {
      if (error.name === "AssertionError") {
        return false;
      }
      throw error;
    }
    return true;
};

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');
    }
}

这取决于你对平等的定义。因此，作为类的开发人员，要由您来定义它们的相等性。

有时会使用一种情况，如果两个实例指向内存中的相同位置，则认为它们是“相等的”，但这并不总是您想要的。例如，如果我有一个Person类，如果两个Person对象具有相同的Last Name、First Name和Social Security Number(即使它们指向内存中的不同位置)，我可能会认为它们是“相等的”。

另一方面，我们不能简单地说两个对象是相等的，如果它们的每个成员的值都相同，因为，有时，你并不想这样。换句话说，对于每个类，由类开发人员定义组成对象“标识”的成员并开发适当的相等操作符(通过重载==操作符或Equals方法)。

Saying that two objects are equal if they have the same hash is one way out. However you then have to wonder how the hash is calculated for each instance. Going back to the Person example above, we could use this system if the hash was calculated by looking at the values of the First Name, Last Name, and Social Security Number fields. On top of that, we are then relying on the quality of the hashing method (that's a huge topic on its own, but suffice it to say that not all hashes are created equal, and bad hashing methods can lead to more collisions, which in this case would return false matches).