Question

我有一种非常天真的方式来比较两个对象属性的子集。当子集中有许多属性时，很难读取。可能有什么替代方案？

const obj1 = {a: 22, b: 33, c: 44};
const obj2 = {a: 22, b: 34, c: 44};

JSON.stringify([obj1.a, obj1.b]) === JSON.stringify([obj2.a, obj2.b])

Answer 1

子集比较：SubsetCompare函数对两个对象的子集进行平面比较。

＆＃13;

let obj1 = {a: 22, b: 34, c: 44};
let obj2 = {a: 22, b: 34, c: 10};

let subsetCompare = (obj1, obj2, keys) =>  keys.every(key =>  obj1[key] === obj2[key]);

console.log(subsetCompare(obj1, obj2, ['a', 'b']));

console.log(subsetCompare(obj1, obj2, ['a', 'b', 'c']))

＆＃13;

平面比较： flatCompare函数仅比较基本类型成员的值和引用类型成员的引用。

＆＃13;

let x = { a : 10};
let y = { a : 10};
let obj1 = {a: 22, b: 34, c: 44, d:y};
let obj2 = {a: 22, b: 34, c: 44, d:x};

let flatCompare = (obj1, obj2) => {
  let keys = [];
  if((keys = getKeys(obj1)).length != getKeys(obj2).length)
    return false;
  return keys.every(key => obj1[key] === obj2[key]);
}

let isObject = (obj) => obj === Object(obj);

let getKeys = (obj) => Object.getOwnPropertyNames(obj);

console.log(flatCompare(obj1, obj2));

＆＃13;

深度比较：deepCompare函数递归地比较嵌套对象的基本类型成员的值。

＆＃13;

let x = { a : 10};
let y = { a : 10};
let obj1 = {a: 22, b: 34, c: 44, d:y};
let obj2 = {a: 22, b: 34, c: 44, d:x};


let deepCompare = (obj1, obj2) => {
  let keys = [];
  if((keys = getKeys(obj1)).length != getKeys(obj2).length)
    return false;
  return keys.every(key => {
    if(isObject(obj1[key]) && isObject(obj2[key]))
      return deepCompare(obj1[key], obj2[key]);
    return obj1[key] === obj2[key];
  });
};

let isObject = (obj) => obj === Object(obj);

let getKeys = (obj) => Object.getOwnPropertyNames(obj);

console.log(deepCompare(obj1, obj2))

＆＃13;

Answer 2

为什么不使用功能？

＆＃13;

let obj1 = {a: 22, b: 33, c: 44};
let obj2 = {a: 22, b: 34, c: 44};
console.log(compareObjects(obj1, obj2, [`a`, `b`])); // false
console.log(compareObjects(obj1, obj2, [`a`, `c`])); // true
console.log(compareObjects(obj1, obj2, [`b`, `c`])); // false

function compareObjects(obj1, obj2, keys) {
    let i;
    for (i = keys.length - 1; i > -1 && obj1[keys[i]] === obj2[keys[i]]; i--);
    return i === -1; // if i equals -1 it went through the whole loop and did not fail for the second condition
}

＆＃13;

Answer 3

基于@ revilheart的回答，但是@ Ryan的every建议：

let obj1 = {a: 22, b: 33, c: 44};
let obj2 = {a: 22, b: 34, c: 44};
console.log(compareObjects(obj1, obj2, [`a`, `b`])); // false
console.log(compareObjects(obj1, obj2, [`a`, `c`])); // true
console.log(compareObjects(obj1, obj2, [`b`, `c`])); // false

function compareObjects(obj1, obj2, keys) {
  return keys.every(key => obj1[key] === obj2[key])
}

比较对象之间的属性子集

3 个答案: