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

EDIT: This method is quite flawed, and is rife with its own issues. I don't recommend it, and would appreciate some down-votes! It is problematic because 1) Some things can not be compared (i.e. functions) because they can not be serialized, 2) It isn't a very fast method of comparing, 3) It has ordering issues, 4) It can have collision issues/false positives if not properly implemented, 5) It can't check for "exactness" (===), and instead is based of value equality, which is oftentimes not what is desired in a comparison method.

这个问题的一个简单解决方案是对JSON字符串进行排序(每个字符),但很多人没有意识到这一点。这通常也比这里提到的其他解决方案更快:

function areEqual(obj1, obj2) {
    var a = JSON.stringify(obj1), b = JSON.stringify(obj2);
    if (!a) a = '';
    if (!b) b = '';
    return (a.split('').sort().join('') == b.split('').sort().join(''));
}

关于这个方法的另一个有用的事情是,您可以通过向JSON传递一个“replace”函数来筛选比较。stringify函数(https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/JSON/stringify#Example_of_using_replacer_parameter)。下面只比较所有命名为“derp”的对象键:

function areEqual(obj1, obj2, filter) {
    var a = JSON.stringify(obj1, filter), b = JSON.stringify(obj2, filter);
    if (!a) a = '';
    if (!b) b = '';
    return (a.split('').sort().join('') == b.split('').sort().join(''));
}
var equal = areEqual(obj1, obj2, function(key, value) {
    return (key === 'derp') ? value : undefined;
});

其他回答

let user1 = { name: "John", address: { line1: "55 Green Park Road", line2: { a:[1,2,3] } }, email:null } let user2 = { name: "John", address: { line1: "55 Green Park Road", line2: { a:[1,2,3] } }, email:null } // Method 1 function isEqual(a, b) { return JSON.stringify(a) === JSON.stringify(b); } // Method 2 function isEqual(a, b) { // checking type of a And b if(typeof a !== 'object' || typeof b !== 'object') { return false; } // Both are NULL if(!a && !b ) { return true; } else if(!a || !b) { return false; } let keysA = Object.keys(a); let keysB = Object.keys(b); if(keysA.length !== keysB.length) { return false; } for(let key in a) { if(!(key in b)) { return false; } if(typeof a[key] === 'object') { if(!isEqual(a[key], b[key])) { return false; } } else { if(a[key] !== b[key]) { return false; } } } return true; } console.log(isEqual(user1,user2));

纯JS方法:我的答案是基于生成一个返回相同值的字符串,无论属性顺序是否相同。设置对象可用于切换大小写和空白是否重要。(为了避免失去焦点,我没有包括那些支持函数,或者我猜应该在任何实用程序集中的isObject。)

这里也没有显示,但为了减少字符串比较时间,如果对象很大,你想加快比较,你也可以散列字符串和比较子字符串;这只适用于非常大的对象(当然也有很小的机会出现错误的相等)。

然后你可以比较genObjStr(obj1) ?= genObjStr(obj2)

function genObjStr(obj, settings) {
// Generate a string that corresponds to an object guarenteed to be the same str even if
// the object have different ordering. The string would largely be used for comparison purposes

var settings = settings||{};
var doStripWhiteSpace = defTrue(settings.doStripWhiteSpace);
var doSetLowerCase = settings.doSetLowerCase||false;

if(isArray(obj)) {
    var vals = [];
    for(var i = 0; i < obj.length; ++i) {
        vals.push(genObjStr(obj[i], settings));
    }
    vals = arraySort(vals);
    return vals.join(`,`);

} else if(isObject(obj)) {

    var keys = Object.keys(obj);
    keys = arraySort(keys);

    var vals = [];
    for(var key of keys) {
        
        var value = obj[key];
        
        value = genObjStr(value, settings);

        if(doStripWhiteSpace) {
            key = removeWhitespace(key);
            var value = removeWhitespace(value);
        };
        if(doSetLowerCase) {
            key = key.toLowerCase();
            value = value.toLowerCase();
        }

        vals.push(value);
    }
    var str = JSON.stringify({keys: keys, vals: vals});
    return str
} else {
    if(doStripWhiteSpace) {
        obj = removeWhitespace(obj);
    };
    if(doSetLowerCase) {
        obj = obj.toLowerCase();
    }
    return obj
}

}

var obj1 = {foo: 123, bar: `Test`};
var obj2 = {bar: `Test`, foo: 123};

console.log(genObjStr(obj1) == genObjStr(obj1))

添加这个版本,因为它处理日期,并有一个流程图显示它是如何工作的。

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"})

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。