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

在JavaScript for Objects中,当它们引用内存中的相同位置时,默认的相等运算符将产生true。

var x = {};
var y = {};
var z = x;

x === y; // => false
x === z; // => true

如果你需要一个不同的相等操作符,你需要添加一个equals(other)方法,或者类似的东西到你的类中,你的问题领域的细节将决定它的确切含义。

这里有一个扑克牌的例子:

function Card(rank, suit) {
  this.rank = rank;
  this.suit = suit;
  this.equals = function(other) {
     return other.rank == this.rank && other.suit == this.suit;
  };
}

var queenOfClubs = new Card(12, "C");
var kingOfSpades = new Card(13, "S");

queenOfClubs.equals(kingOfSpades); // => false
kingOfSpades.equals(new Card(13, "S")); // => true

其他回答

function isEqual(obj1, obj2){
    type1 = typeof(obj1);
    type2 = typeof(obj2);
    if(type1===type2){
        switch (type1){
            case "object": return JSON.stringify(obj1)===JSON.stringify(obj2);
            case "function": return eval(obj1).toString()===eval(obj2).toString();
            default: return obj1==obj2;
        }
    }
    return false;
}//have not tried but should work.

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);
}

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"})

最简单和逻辑的解决方案,比较一切像对象,数组,字符串,Int…

JSON。stringify({a: val1}) == JSON。stringify ({a: val2})

注意:

你需要用你的Object替换val1和val2 对于对象,必须对两侧对象进行递归排序(按键)

下面是stringify技巧的一个版本,它的输入较少,在很多情况下适用于简单的JSON数据比较。

var obj1Fingerprint = JSON.stringify(obj1).replace(/\{|\}/g,'').split(',').sort().join(',');
var obj2Fingerprint = JSON.stringify(obj2).replace(/\{|\}/g,'').split(',').sort().join(',');
if ( obj1Fingerprint === obj2Fingerprint) { ... } else { ... }