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

Stringify两个对象并进行比较

返回(JSON.stringify(obj1) === = JSON.stringify(obj2))

这将返回true或false

其他回答

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"})

Stringify两个对象并进行比较

返回(JSON.stringify(obj1) === = JSON.stringify(obj2))

这将返回true或false

判断两个对象是否相似的一个快速“破解”方法是使用它们的toString()方法。如果您正在检查对象A和B,请确保A和B具有有意义的toString()方法,并检查它们返回的字符串是否相同。

这不是灵丹妙药,但有时在正确的情况下是有用的。

const isEqual = function (var1, var2) {
    if (typeof var1 === 'object' && typeof var2 === 'object') {
        // Checking equality for each of the inner values of the objects
        const keys = [...new Set([...Object.keys(var1),...Object.keys(var2)])];
        return keys.every(key => isEqual(var1[key], var2[key]) && isEqual(var2[key], var1[key]));
    } else { // Primitive types (number, boolean etc..)
        return var1 === var2; // Normal equality
    }
}

我在这里看到了很多很好的答案,但我找不到一个简短的函数,所以我创建了一个。 这将检查两个变量是否相等,无论其类型如何。 希望你看得合适。

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 { ... }