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
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;
}
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));
这取决于你对平等的定义。因此,作为类的开发人员,要由您来定义它们的相等性。
有时会使用一种情况,如果两个实例指向内存中的相同位置,则认为它们是“相等的”,但这并不总是您想要的。例如,如果我有一个Person类,如果两个Person对象具有相同的Last Name、First Name和Social Security Number(即使它们指向内存中的不同位置),我可能会认为它们是“相等的”。
另一方面,我们不能简单地说两个对象是相等的,如果它们的每个成员的值都相同,因为,有时,你并不想这样。换句话说,对于每个类,由类开发人员定义组成对象“标识”的成员并开发适当的相等操作符(通过重载==操作符或Equals方法)。
Saying that two objects are equal if they have the same hash is one way out. However you then have to wonder how the hash is calculated for each instance. Going back to the Person example above, we could use this system if the hash was calculated by looking at the values of the First Name, Last Name, and Social Security Number fields. On top of that, we are then relying on the quality of the hashing method (that's a huge topic on its own, but suffice it to say that not all hashes are created equal, and bad hashing methods can lead to more collisions, which in this case would return false matches).
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;
};
下面是ES6/ES2015中使用函数式方法的解决方案:
const typeOf = x =>
({}).toString
.call(x)
.match(/\[object (\w+)\]/)[1]
function areSimilar(a, b) {
const everyKey = f => Object.keys(a).every(f)
switch(typeOf(a)) {
case 'Array':
return a.length === b.length &&
everyKey(k => areSimilar(a.sort()[k], b.sort()[k]));
case 'Object':
return Object.keys(a).length === Object.keys(b).length &&
everyKey(k => areSimilar(a[k], b[k]));
default:
return a === b;
}
}
这里有演示
当然,当我们在它的时候,我会抛出我自己对车轮的重新发明(我为辐条和使用的材料的数量感到自豪):
////////////////////////////////////////////////////////////////////////////////
var equals = function ( objectA, objectB ) {
var result = false,
keysA,
keysB;
// Check if they are pointing at the same variable. If they are, no need to test further.
if ( objectA === objectB ) {
return true;
}
// Check if they are the same type. If they are not, no need to test further.
if ( typeof objectA !== typeof objectB ) {
return false;
}
// Check what kind of variables they are to see what sort of comparison we should make.
if ( typeof objectA === "object" ) {
// Check if they have the same constructor, so that we are comparing apples with apples.
if ( objectA.constructor === objectA.constructor ) {
// If we are working with Arrays...
if ( objectA instanceof Array ) {
// Check the arrays are the same length. If not, they cannot be the same.
if ( objectA.length === objectB.length ) {
// Compare each element. They must be identical. If not, the comparison stops immediately and returns false.
return objectA.every(
function ( element, i ) {
return equals( element, objectB[ i ] );
}
);
}
// They are not the same length, and so are not identical.
else {
return false;
}
}
// If we are working with RegExps...
else if ( objectA instanceof RegExp ) {
// Return the results of a string comparison of the expression.
return ( objectA.toString() === objectB.toString() );
}
// Else we are working with other types of objects...
else {
// Get the keys as arrays from both objects. This uses Object.keys, so no old browsers here.
keysA = Object.keys( objectA );
keysB = Object.keys( objectB );
// Check the key arrays are the same length. If not, they cannot be the same.
if ( keysA.length === keysB.length ) {
// Compare each property. They must be identical. If not, the comparison stops immediately and returns false.
return keysA.every(
function ( element ) {
return equals( objectA[ element ], objectB[ element ] );
}
);
}
// They do not have the same number of keys, and so are not identical.
else {
return false;
}
}
}
// They don't have the same constructor.
else {
return false;
}
}
// If they are both functions, let us do a string comparison.
else if ( typeof objectA === "function" ) {
return ( objectA.toString() === objectB.toString() );
}
// If a simple variable type, compare directly without coercion.
else {
return ( objectA === objectB );
}
// Return a default if nothing has already been returned.
return result;
};
////////////////////////////////////////////////////////////////////////////////
它会尽可能快地返回false,但当然,对于一个差异嵌套很深的大对象,它可能不那么有效。在我自己的场景中,良好地处理嵌套数组非常重要。
希望它能帮助需要这种“轮子”的人。