将 2 个对象合并为一个新对象，具有选择性条件答案

【问题标题】：Merging 2 objects into a new one, with selective condition将 2 个对象合并为一个新对象，具有选择性条件
【发布时间】：2021-09-06 00:46:42
【问题描述】：

我正在尝试根据条件创建一组新的对象，将新组件从第二个组件添加到第一个组件。这是一个例子：

拥有这 2 组对象：

const one = [{code: 'a'},{ code: 'b'}, {code: 'c'}]
const two = [{code: 'b', status: 'active'}]

并希望得到如下结果：

{ code: 'a', status: 'empty' }
{ code: 'b', status: 'active' }
{ code: 'c', status: 'empty' }

这是我编写的示例代码：

const one = [{code: 'a'},{ code: 'b'}, {code: 'c'}]
const two = [{code: 'b', status: 'active'}]



var final = []
var temp = {}

for(let holiday of one) {
    // here I started to create a temp object with components of object one
    temp.code = holiday.code
    
    // here I check if same code from object one exist in any object two
    var checksub= two.find(o => o.code == holiday.code)

    // then assigns proper value to temp object from object two, of a default instead
    if(checksub){temp.status = checksub.status}
    else {temp.status = 'empty'}

    // then push the object into new array of objects
    final.push(temp)
}
    
console.log(final)

但我得到的结果是：

[
  { code: 'c', status: 'empty' },
  { code: 'c', status: 'empty' },
  { code: 'c', status: 'empty' }
]

为什么我得到最后一个对象的 3 倍？我尝试了不同的方法（使用 forEach 和 for 循环）并且总是得到同样奇怪的结果......我做错了什么？谢谢！！！

【问题讨论】：

将这个var temp = {}移到for循环中。

标签： node.js arrays json object

【解决方案1】：

您通过引用传递对象，因此如果您在一个位置更改对象的值，则使用该值的所有位置都将被更新。由于在每个循环中， temp 的值都会更改，因此您的数组会使用更新后的值进行更新。因此，如果您只是传递临时对象的克隆，那么您的所有值都不会受到影响。

const one = [{code: 'a'},{ code: 'b'}, {code: 'c'}]
const two = [{code: 'b', status: 'active'}]



var final = []
var temp = {}

for(let holiday of one) {
    // here I started to create a temp object with components of object one
    temp.code = holiday.code
    
    // here I check if same code from object one exist in any object two
    var checksub= two.find(o => o.code == holiday.code)

    // then assigns proper value to temp object from object two, of a default instead
    if(checksub){temp.status = checksub.status}
    else {temp.status = 'empty'}

    // then push the clone of object into new array of objects
    final.push({...temp})
}
    
console.log(final)

您所要做的就是final.push({...temp})，它将解构临时对象并将其作为新对象传递。

决赛如下：

[
 {code: "a", status: "empty"},
 {code: "b", status: "active"},
 {code: "c", status: "empty"},
]

【讨论】：

欢迎 :) 您可以选择此答案作为正确答案，以便其他寻找类似内容的人可以立即找到答案

【解决方案2】：

const one = [{code: 'a'},{ code: 'b'}, {code: 'c'}]
const two = [{code: 'b', status: 'active'}]

const final = one.map((x) => {
    const twoContains = two.find(y => y.code === x.code);
    return {
        code: x.code,
        status: twoContains ? twoContains.status : 'empty'
    };
});

或简单（更短的版本）

const final = one.map(x => ({
    code: x.code,
    status: (two.find(y => y.code === x.code) || {status: 'empty'}).status
}));

console.log(final);

/* output
[
    { code: 'a', status: 'empty' },
    { code: 'b', status: 'active' },
    { code: 'c', status: 'empty' }
]    
*/

就这么简单

【讨论】：