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。
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
这是我的版本。它正在使用new Object。ES5中引入的keys特性以及+、+和+的想法/测试:
function objectEquals(x, y) { 'use strict'; if (x === null || x === undefined || y === null || y === undefined) { return x === y; } // after this just checking type of one would be enough if (x.constructor !== y.constructor) { return false; } // if they are functions, they should exactly refer to same one (because of closures) if (x instanceof Function) { return x === y; } // if they are regexps, they should exactly refer to same one (it is hard to better equality check on current ES) if (x instanceof RegExp) { return x === y; } if (x === y || x.valueOf() === y.valueOf()) { return true; } if (Array.isArray(x) && x.length !== y.length) { return false; } // if they are dates, they must had equal valueOf if (x instanceof Date) { return false; } // if they are strictly equal, they both need to be object at least if (!(x instanceof Object)) { return false; } if (!(y instanceof Object)) { return false; } // recursive object equality check var p = Object.keys(x); return Object.keys(y).every(function (i) { return p.indexOf(i) !== -1; }) && p.every(function (i) { return objectEquals(x[i], y[i]); }); } /////////////////////////////////////////////////////////////// /// The borrowed tests, run them by clicking "Run code snippet" /////////////////////////////////////////////////////////////// var printResult = function (x) { if (x) { document.write('<div style="color: green;">Passed</div>'); } else { document.write('<div style="color: red;">Failed</div>'); } }; var assert = { isTrue: function (x) { printResult(x); }, isFalse: function (x) { printResult(!x); } } assert.isTrue(objectEquals(null,null)); assert.isFalse(objectEquals(null,undefined)); assert.isFalse(objectEquals(/abc/, /abc/)); assert.isFalse(objectEquals(/abc/, /123/)); var r = /abc/; assert.isTrue(objectEquals(r, r)); assert.isTrue(objectEquals("hi","hi")); assert.isTrue(objectEquals(5,5)); assert.isFalse(objectEquals(5,10)); assert.isTrue(objectEquals([],[])); assert.isTrue(objectEquals([1,2],[1,2])); assert.isFalse(objectEquals([1,2],[2,1])); assert.isFalse(objectEquals([1,2],[1,2,3])); assert.isTrue(objectEquals({},{})); assert.isTrue(objectEquals({a:1,b:2},{a:1,b:2})); assert.isTrue(objectEquals({a:1,b:2},{b:2,a:1})); assert.isFalse(objectEquals({a:1,b:2},{a:1,b:3})); assert.isTrue(objectEquals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}},{1:{name:"mhc",age:28}, 2:{name:"arb",age:26}})); assert.isFalse(objectEquals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}},{1:{name:"mhc",age:28}, 2:{name:"arb",age:27}})); Object.prototype.equals = function (obj) { return objectEquals(this, obj); }; var assertFalse = assert.isFalse, assertTrue = assert.isTrue; assertFalse({}.equals(null)); assertFalse({}.equals(undefined)); assertTrue("hi".equals("hi")); assertTrue(new Number(5).equals(5)); assertFalse(new Number(5).equals(10)); assertFalse(new Number(1).equals("1")); assertTrue([].equals([])); assertTrue([1,2].equals([1,2])); assertFalse([1,2].equals([2,1])); assertFalse([1,2].equals([1,2,3])); assertTrue(new Date("2011-03-31").equals(new Date("2011-03-31"))); assertFalse(new Date("2011-03-31").equals(new Date("1970-01-01"))); assertTrue({}.equals({})); assertTrue({a:1,b:2}.equals({a:1,b:2})); assertTrue({a:1,b:2}.equals({b:2,a:1})); assertFalse({a:1,b:2}.equals({a:1,b:3})); assertTrue({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}}.equals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}})); assertFalse({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}}.equals({1:{name:"mhc",age:28}, 2:{name:"arb",age:27}})); var a = {a: 'text', b:[0,1]}; var b = {a: 'text', b:[0,1]}; var c = {a: 'text', b: 0}; var d = {a: 'text', b: false}; var e = {a: 'text', b:[1,0]}; var i = { a: 'text', c: { b: [1, 0] } }; var j = { a: 'text', c: { b: [1, 0] } }; var k = {a: 'text', b: null}; var l = {a: 'text', b: undefined}; assertTrue(a.equals(b)); assertFalse(a.equals(c)); assertFalse(c.equals(d)); assertFalse(a.equals(e)); assertTrue(i.equals(j)); assertFalse(d.equals(k)); assertFalse(k.equals(l)); // from comments on stackoverflow post assert.isFalse(objectEquals([1, 2, undefined], [1, 2])); assert.isFalse(objectEquals([1, 2, 3], { 0: 1, 1: 2, 2: 3 })); assert.isFalse(objectEquals(new Date(1234), 1234)); // no two different function is equal really, they capture their context variables // so even if they have same toString(), they won't have same functionality var func = function (x) { return true; }; var func2 = function (x) { return true; }; assert.isTrue(objectEquals(func, func)); assert.isFalse(objectEquals(func, func2)); assert.isTrue(objectEquals({ a: { b: func } }, { a: { b: func } })); assert.isFalse(objectEquals({ a: { b: func } }, { a: { b: func2 } }));
其他回答
简短的回答
简单的答案是:不,没有一般的方法来确定一个对象等于另一个你所指的意义。例外情况是当您严格地认为一个对象是无类型的。
长话短说
这个概念是一个Equals方法,它比较一个对象的两个不同实例,以指示它们在值级别上是否相等。但是,定义Equals方法应该如何实现取决于具体的类型。对具有基本值的属性进行迭代比较可能还不够:对象可能包含与相等无关的属性。例如,
function MyClass(a, b)
{
var c;
this.getCLazy = function() {
if (c === undefined) c = a * b // imagine * is really expensive
return c;
}
}
在上面的例子中,c对于确定MyClass的任何两个实例是否相等并不重要,只有a和b是重要的。在某些情况下,c可能在不同实例之间有所不同,但在比较中并不显著。
注意,当成员本身也可能是某个类型的实例,并且每个实例都需要有确定相等的方法时,这个问题就会出现。
更复杂的是,在JavaScript中,数据和方法之间的区别是模糊的。
一个对象可以引用一个作为事件处理程序调用的方法,这可能不被认为是其“值状态”的一部分。然而,另一个对象很可能被分配一个执行重要计算的函数,从而使这个实例与其他实例不同,仅仅因为它引用了不同的函数。
如果一个对象的现有原型方法被另一个函数覆盖,该怎么办?它还能被认为与另一个相同的实例相等吗?这个问题只能在每种类型的具体情况下回答。
如前所述,异常将是一个严格的无类型对象。在这种情况下,唯一明智的选择是对每个成员进行迭代和递归比较。即使这样,人们也要问一个函数的“值”是什么?
虽然这个问题已经有很多答案了。我只是想提供另一种实现方法:
const primitveDataTypes = ['number', 'boolean', 'string', 'undefined']; const isDateOrRegExp = (value) => value instanceof Date || value instanceof RegExp; const compare = (first, second) => { let agg = true; if(typeof first === typeof second && primitveDataTypes.indexOf(typeof first) !== -1 && first !== second){ agg = false; } // adding support for Date and RegExp. else if(isDateOrRegExp(first) || isDateOrRegExp(second)){ if(first.toString() !== second.toString()){ agg = false; } } else { if(Array.isArray(first) && Array.isArray(second)){ if(first.length === second.length){ for(let i = 0; i < first.length; i++){ if(typeof first[i] === 'object' && typeof second[i] === 'object'){ agg = compare(first[i], second[i]); } else if(first[i] !== second[i]){ agg = false; } } } else { agg = false; } } else { const firstKeys = Object.keys(first); const secondKeys = Object.keys(second); if(firstKeys.length !== secondKeys.length){ agg = false; } for(let j = 0 ; j < firstKeys.length; j++){ if(firstKeys[j] !== secondKeys[j]){ agg = false; } if(first[firstKeys[j]] && second[secondKeys[j]] && typeof first[firstKeys[j]] === 'object' && typeof second[secondKeys[j]] === 'object'){ agg = compare(first[firstKeys[j]], second[secondKeys[j]]); } else if(first[firstKeys[j]] !== second[secondKeys[j]]){ agg = false; } } } } return agg; } console.log('result', compare({a: 1, b: { c: [4, {d:5}, {e:6}]}, r: null}, {a: 1, b: { c: [4, {d:5}, {e:6}]}, r: 'ffd'})); //returns false.
对象是否相等检查:JSON.stringify(array1.sort()) === JSON.stringify(array2.sort())
上面的测试还适用于对象数组,在这种情况下使用http://www.w3schools.com/jsref/jsref_sort.asp中记录的排序函数
对于具有平面JSON模式的小型数组可能足够了。
下面是一个使用ES6+的解决方案
// this comparison would not work for function and symbol comparisons
// this would only work best for compared objects that do not belong to same address in memory
// Returns true if there is no difference, and false otherwise
export const isObjSame = (obj1, obj2) => {
if (typeof obj1 !== "object" && obj1 !== obj2) {
return false;
}
if (typeof obj1 !== "object" && typeof obj2 !== "object" && obj1 === obj2) {
return true;
}
if (typeof obj1 === "object" && typeof obj2 === "object") {
if (Array.isArray(obj1) && Array.isArray(obj2)) {
if (obj1.length === obj2.length) {
if (obj1.length === 0) {
return true;
}
const firstElemType = typeof obj1[0];
if (typeof firstElemType !== "object") {
const confirmSameType = currentType =>
typeof currentType === firstElemType;
const checkObjOne = obj1.every(confirmSameType);
const checkObjTwo = obj2.every(confirmSameType);
if (checkObjOne && checkObjTwo) {
// they are primitves, we can therefore sort before and compare by index
// use number sort
// use alphabet sort
// use regular sort
if (firstElemType === "string") {
obj1.sort((a, b) => a.localeCompare(b));
obj2.sort((a, b) => a.localeCompare(b));
}
obj1.sort((a, b) => a - b);
obj2.sort((a, b) => a - b);
let equal = true;
obj1.map((element, index) => {
if (!isObjSame(element, obj2[index])) {
equal = false;
}
});
return equal;
}
if (
(checkObjOne && !checkObjTwo) ||
(!checkObjOne && checkObjTwo)
) {
return false;
}
if (!checkObjOne && !checkObjTwo) {
for (let i = 0; i <= obj1.length; i++) {
const compareIt = isObjSame(obj1[i], obj2[i]);
if (!compareIt) {
return false;
}
}
return true;
}
// if()
}
const newValue = isObjSame(obj1, obj2);
return newValue;
} else {
return false;
}
}
if (!Array.isArray(obj1) && !Array.isArray(obj2)) {
let equal = true;
if (obj1 && obj2) {
const allKeys1 = Array.from(Object.keys(obj1));
const allKeys2 = Array.from(Object.keys(obj2));
if (allKeys1.length === allKeys2.length) {
allKeys1.sort((a, b) => a - b);
allKeys2.sort((a, b) => a - b);
allKeys1.map((key, index) => {
if (
key.toLowerCase() !== allKeys2[index].toLowerCase()
) {
equal = false;
return;
}
const confirmEquality = isObjSame(obj1[key], obj2[key]);
if (!confirmEquality) {
equal = confirmEquality;
return;
}
});
}
}
return equal;
// return false;
}
}
};
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;
});