如何从具有不同值的两个对象返回键和值

Question

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',
    ...
}

我希望返回列表...

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

在您的示例中，list 是一个数组，其中只有一个对象包含所有差异。要获取该对象，您可以这样做：

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)