如何确定相等的两个JavaScript对象?

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

我对这个函数做如下假设:

你控制你比较的对象，你只有基本的值(例如。而不是嵌套的对象、函数等)。您的浏览器支持Object.keys。

这应该被视为一个简单策略的示范。

/**
 * Checks the equality of two objects that contain primitive values. (ie. no nested objects, functions, etc.)
 * @param {Object} object1
 * @param {Object} object2
 * @param {Boolean} [order_matters] Affects the return value of unordered objects. (ex. {a:1, b:2} and {b:2, a:1}).
 * @returns {Boolean}
 */
function isEqual( object1, object2, order_matters ) {
    var keys1 = Object.keys(object1),
        keys2 = Object.keys(object2),
        i, key;

    // Test 1: Same number of elements
    if( keys1.length != keys2.length ) {
        return false;
    }

    // If order doesn't matter isEqual({a:2, b:1}, {b:1, a:2}) should return true.
    // keys1 = Object.keys({a:2, b:1}) = ["a","b"];
    // keys2 = Object.keys({b:1, a:2}) = ["b","a"];
    // This is why we are sorting keys1 and keys2.
    if( !order_matters ) {
        keys1.sort();
        keys2.sort();
    }

    // Test 2: Same keys
    for( i = 0; i < keys1.length; i++ ) {
        if( keys1[i] != keys2[i] ) {
            return false;
        }
    }

    // Test 3: Values
    for( i = 0; i < keys1.length; i++ ) {
        key = keys1[i];
        if( object1[key] != object2[key] ) {
            return false;
        }
    }

    return true;
}

2013-09-27 01:07:00

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"})

2013-05-28 15:28:43

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;
};

2014-09-24 12:26:05

我有一个更短的函数，它将深入到所有子对象或数组。它和JSON.stringify(obj1) === JSON.stringify(obj2)一样高效，但是JSON.stringify(obj2)。如果顺序不相同(如此处所述)，Stringify将无法工作。

var obj1 = { a : 1, b : 2 };
var obj2 = { b : 2, a : 1 };

console.log(JSON.stringify(obj1) === JSON.stringify(obj2)); // false

这个函数也是一个很好的开始如果你想处理不相等的值。

如果你想增加对Function, Date和RegExp的支持，你可以在deep_equal的开头添加这个(未测试):

if ((typeof obj1 === 'function' && typeof obj2 === 'function') ||
(obj1 instanceof Date && obj2 instanceof Date) ||
(obj1 instanceof RegExp && obj2 instanceof RegExp))
{
    obj1 = obj1.toString();
    obj2 = obj2.toString();
}

2016-07-29 14:24:13

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;
}

2010-05-27 12:00:59

ES6:我能做到的最小代码是这样的。它通过对代表对象的所有键值数组进行字符串化来进行递归的深度比较，唯一的限制是没有方法或符号进行比较。

const compareObjects = (a, b) => { let s = (o) => Object.entries(o).sort()。映射(i => { if(i[1]实例对象)i[1] = s(i[1]); 返回我 }) 返回JSON.stringify(s(a)) === JSON.stringify(s(b)) ｝ console.log (compareObjects ({b: 4,答:{b: 1}}, {} {b: 1, b: 4}));

重要提示:这个函数正在执行JSON。stringfy在数组中，并将键排序，而不是在对象本身中:

["a" ["b"， 1]] [" b ", 4]

2019-09-24 16:47:44

如何确定相等的两个JavaScript对象?

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