如果使用JSON库，可以将每个对象编码为JSON，然后比较结果字符串是否相等。

var obj1={test:"value"};
var obj2={test:"value2"};

alert(JSON.encode(obj1)===JSON.encode(obj2));

注意:虽然这个答案在很多情况下都有效，但由于各种原因，一些人在评论中指出了它的问题。在几乎所有情况下，您都希望找到更健壮的解决方案。

我已经实现了一个方法，它接受两个json，并使用递归检查它们的键是否具有相同的值。 我用另一个问题来解决这个问题。

const arraysEqual = (a, b) => { if (a === b) return true; if (a === null || b === null) return false; if (a.length !== b.length) return false; // If you don't care about the order of the elements inside // the array, you should sort both arrays here. for (let i = 0; i < a.length; ++i) { if (a[i] !== b[i]) return false; } return true; }; const jsonsEqual = (a, b) => { if(typeof a !== 'object' || typeof b !== 'object') return false; if (Object.keys(a).length === Object.keys(b).length) { // if items have the same size let response = true; for (let key in a) { if (!b[key]) // if not key response = false; if (typeof a[key] !== typeof b[key]) // if typeof doesn't equals response = false; else { if (Array.isArray(a[key])) // if array response = arraysEqual(a[key], b[key]); else if (typeof a[key] === 'object') // if another json response = jsonsEqual(a[key], b[key]); else if (a[key] !== b[key]) // not equals response = false; } if (!response) // return if one item isn't equal return false; } } else return false; return true; }; const json1 = { a: 'a', b: 'asd', c: [ '1', 2, 2.5, '3', { d: 'asd', e: [ 1.6, { f: 'asdasd', g: '123' } ] } ], h: 1, i: 1.2, }; const json2 = { a: 'nops', b: 'asd' }; const json3 = { a: 'h', b: '484', c: [ 3, 4.5, '2ss', { e: [ { f: 'asdasd', g: '123' } ] } ], h: 1, i: 1.2, }; const result = jsonsEqual(json1,json2); //const result = jsonsEqual(json1,json3); //const result = jsonsEqual(json1,json1); if(result) // is equal $('#result').text("Jsons are the same") else $('#result').text("Jsons aren't equals") <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="result"></div>

如何确定部分对象(partial <T>)等于typescript中的原始对象(T)。

function compareTwoObjects<T>(original: T, partial: Partial<T>): boolean {
  return !Object.keys(partial).some((key) => partial[key] !== original[key]);
}

附注:最初我打算提出一个有答案的新问题。但这样的问题已经存在，并被标记为重复题。

我不是Javascript专家，但这里有一个简单的解决方法。我检查三件事:

它是一个对象，而且它不是null，因为typeof null是对象。 如果两个对象的属性计数相同?否则它们就不相等。 遍历一个对象的属性，并检查对应的属性在第二个对象中是否具有相同的值。

function deepEqual (first, second) { // Not equal if either is not an object or is null. if (!isObject(first) || !isObject(second) ) return false; // If properties count is different if (keys(first).length != keys(second).length) return false; // Return false if any property value is different. for(prop in first){ if (first[prop] != second[prop]) return false; } return true; } // Checks if argument is an object and is not null function isObject(obj) { return (typeof obj === "object" && obj != null); } // returns arrays of object keys function keys (obj) { result = []; for(var key in obj){ result.push(key); } return result; } // Some test code obj1 = { name: 'Singh', age: 20 } obj2 = { age: 20, name: 'Singh' } obj3 = { name: 'Kaur', age: 19 } console.log(deepEqual(obj1, obj2)); console.log(deepEqual(obj1, obj3));

这是一个非常干净的CoffeeScript版本，你可以这样做:

Object::equals = (other) ->
  typeOf = Object::toString

  return false if typeOf.call(this) isnt typeOf.call(other)
  return `this == other` unless typeOf.call(other) is '[object Object]' or
                                typeOf.call(other) is '[object Array]'

  (return false unless this[key].equals other[key]) for key, value of this
  (return false if typeof this[key] is 'undefined') for key of other

  true

下面是测试:

  describe "equals", ->

    it "should consider two numbers to be equal", ->
      assert 5.equals(5)

    it "should consider two empty objects to be equal", ->
      assert {}.equals({})

    it "should consider two objects with one key to be equal", ->
      assert {a: "banana"}.equals {a: "banana"}

    it "should consider two objects with keys in different orders to be equal", ->
      assert {a: "banana", kendall: "garrus"}.equals {kendall: "garrus", a: "banana"}

    it "should consider two objects with nested objects to be equal", ->
      assert {a: {fruit: "banana"}}.equals {a: {fruit: "banana"}}

    it "should consider two objects with nested objects that are jumbled to be equal", ->
      assert {a: {a: "banana", kendall: "garrus"}}.equals {a: {kendall: "garrus", a: "banana"}}

    it "should consider two objects with arrays as values to be equal", ->
      assert {a: ["apple", "banana"]}.equals {a: ["apple", "banana"]}



    it "should not consider an object to be equal to null", ->
      assert !({a: "banana"}.equals null)

    it "should not consider two objects with different keys to be equal", ->
      assert !({a: "banana"}.equals {})

    it "should not consider two objects with different values to be equal", ->
      assert !({a: "banana"}.equals {a: "grapefruit"})

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).