如何从具有不同值的两个对象返回键和值

Question

我将如何遍历两个对象并仅返回不同的键和值的列表？

let list = [];

let previousObject = {

    key: 'key_1',

    name: 'Previous name',

    age: '30',

    location: '12345 Main St.',

    height: '77',

    weight: '215',

    ...

}

let newObject = {

    key: 'key_1',

    name: 'New name',

    age: '25',

    location: '54321 Main St.',

    height: '77',

    weight: '195',

    ...

}

我希望列表返回...

list = [{ name: 'New name', age: '25', location: '54321 Main St.' }]

需要考虑的事项：

我将如何遍历两个对象并仅返回不同的键和值的列表？

let list = [];

let previousObject = {

    key: 'key_1',

    name: 'Previous name',

    age: '30',

    location: '12345 Main St.',

    height: '77',

    weight: '215',

    ...

}

let newObject = {

    key: 'key_1',

    name: 'New name',

    age: '25',

    location: '54321 Main St.',

    height: '77',

    weight: '195',

    ...

}

我希望列表返回...

list = [{ name: 'New name', age: '25', location: '54321 Main St.' }]

需要考虑的事项：

• previousObject可能包含比 newObject 更多的键

• 每个用户会话可能不止一次检查这些差异

• 
  

Answer 1

在您的示例list中是一个数组，其中只有一个包含所有差异的对象。要获取该对象，您可以执行以下操作：let list = [];let previousObject = {    key: 'key_1',    name: 'Previous name',    age: '30',    location: '12345 Main St.',    height: '77',    weight: '215',}let newObject = {    key: 'key_1',    name: 'New name',    age: '25',    location: '54321 Main St.',    height: '77',    weight: '195',}let diff = {}for (const key in newObject) {  if (previousObject[key] != newObject[key]) {    diff[key] = newObject[key]  } }console.log(diff)