如何获取密钥然后删除未关联到数组的密钥？ [复制]

Posted 2023-02-27

【中文标题】如何获取密钥然后删除未关联到数组的密钥？ [复制]

【英文标题】：How to get the keys and then remove the keys not associated to an array? [duplicate]

【发布时间】：2021-05-19 04:22:03

【问题描述】：

我想获取项目的值，然后删除与列无关的键

下面是我的代码

var columns =['title','code','notes','value','daysCode'];

var items =[
        title:'Marvel',
        code:'marvel-01',
        notes:'spider-man',
        value:'Value 2',
        daysCode:'Sept 1, 2020',
        daysAge:'Nov 1 , 2019',
        name:'Peter Parker',
        other:'comic'
    ,
    
        title:'DC',
        code:'dc-01',
        notes:'batman',
        value:'Value 1',
        daysCode:'Sept 1, 2020',
        daysAge:'Nov 1 , 2019',
        name:'Bruce Wayne',
        other:'comics'
    ,
    
        title:'Image',
        code:'image-02',
        notes:'spawn',
        value:'Value 3',
        daysCode:'Sept 1, 2020',
        daysAge:'Nov 1 , 2019',
        name:'Albert Simmons',
        other:'comics'
    
]

输出：

newItems = [
    title:'Marvel',
    code:'marvel-01',
    notes:'spider-man',
    value:'Value 2',
    daysCode:'Sept 1, 2020'
,

   title:'DC',
   code:'dc-01',
   notes:'batman',
   value:'Value 1',
   daysCode:'Sept 1, 2020'
,

   title:'Image',
   code:'image-02',
   notes:'spawn',
   value:'Value 3',
   daysCode:'Sept 1, 2020'
]

我该怎么做？

【问题讨论】：

或ecmascript 6 - Filter object properties by key in ES6 - Stack Overflow

你的问题有点不清楚。您应该发布您到目前为止编写的代码，以便网站用户可以看到您正在尝试做什么以及哪里出了问题。

@user202729 在这种情况下，您将问题标记为欺骗

@AnuragSrivastava 我知道这一点。但是由于不止一个，我只是在检查哪个更好等等......（除了我只有一个标志）

简单而简短的答案是 - const result = items.map(item => return ((title,code,notes,value,daysCode) => (title,code, notes,value,daysCode))(item) ) console.log("result----",result)

【参考方案1】：

使用map 和Object.fromEntries 来获得预期的输出：

var columns =['title','code','notes','value','daysCode'];

var items =[ title:'Marvel', code:'marvel-01', notes:'spider-man', value:'Value 2', daysCode:'Sept 1, 2020', daysAge:'Nov 1 , 2019', name:'Peter Parker', other:'comic' ,  title:'DC', code:'dc-01', notes:'batman', value:'Value 1', daysCode:'Sept 1, 2020', daysAge:'Nov 1 , 2019', name:'Bruce Wayne', other:'comics' ,  title:'Image', code:'image-02', notes:'spawn', value:'Value 3', daysCode:'Sept 1, 2020', daysAge:'Nov 1 , 2019', name:'Albert Simmons', other:'comics'  ]

var newItems = items.map(o=>Object.fromEntries(columns.map(n=>[n,o[n]])));

// using reduce method:
var newItems1 = items.map(o=>columns.reduce((a,e)=>(...a, [e]:o[e]),));

console.log(newItems);
console.log(newItems1);

【讨论】：

有没有办法不使用 Object.fromEntries？

@thePeculiarCoder 是的，然后使用reduce

@thePeculiarCoder 请检查更新的副本。您可以使用 reduce 来创建每个对象的子集***.com/a/32184094/3082296

@thePeculiarCoder 如果您有特定要求，您应该在问题中指定。