获取数组中键的所有对象值[重复]

Question

我有一个这样的对象：

myObj = { 1:{name:'a'}, 2:{name:'v'}, 3:{name:'x'}, 10:{name:'t'} };

并且我想获取其键在以下列表中的值的列表：

myList = [1,2,10]

所以我想要的结果是：

[{name:'a'},{name:'v'}, {name:'t'}]

关于如何做到这一点的任何想法？

我尝试了以下方法：

const result = [];
        
for (const childId in state.frameObjects[id].childrenIds ) 
{
     if(state.frameObjects[childId]!==undefined)
     {
          result.push(state.frameObjects[childId]);
     }
}
        
return result;

但是 Vue 控制台给了我这个错误：InternalError: "too much recursion"

Answer 1

您可以将所需的密钥放入Set 并使用filter。如果数组包含对象中不存在的键，这也将起作用。

const myObj = { 1:{name:'a'}, 2:{name:'v'}, 3:{name:'x'}, 10:{name:'t'} };
const keys = new Set([1,2,10])
const res = Object.keys(myObj).filter(key=>keys.has(+key)).map(key=>myObj[key]);
console.log(res);

Answer 2

你可以这样做：

var objectArray = [];
myList.forEach(function(key){
    if(myObj[key]){ 
       objectArray.push(myObj[key]);
     }
});

Answer 3

您可以使用键迭代您的数组，并检查它们是否存在于您的myObj 中，如果它们存在，那么您可以将值推送到新的 arrya 中：

const myObj = { 1:{name:'a'}, 2:{name:'v'}, 3:{name:'x'}, 10:{name:'t'} };
const myList = [1,2,10]
const newArr = []
myList.forEach(key => {
  if (myObj[key]) {
    newArr.push(myObj[key])
  }
})


console.log(newArr)

Answer 4

您可以过滤它们对象的键，然后映射结果。

const valuesOf = (obj, keys) => Object.keys(obj)
  .filter(key => keys.includes(parseInt(key, 10)))
  .map(key => obj[key])

const myObj = { 1: { name: 'a'}, 2: { name: 'v'}, 3: { name: 'x'}, 10: { name: 't'} };

console.log(valuesOf(myObj, [1, 2, 10]))

.as-console-wrapper { top: 0; min-height: 100% }

或者，您可以稍后尝试映射和过滤。这允许您按所需键的顺序返回对象。

const valuesOf = (obj, keys) =>
  keys.map(key => obj[key]).filter(v => v !== undefined)

const myObj = { 1: { name: 'a'}, 2: { name: 'v'}, 3: { name: 'x'}, 10: { name: 't'} };

console.log(valuesOf(myObj, [1, 2, 10]))

.as-console-wrapper { top: 0; min-height: 100% }

Answer 5

更实用的方法是：myList.map(a => myObj[a]).filter(a => a)

注意：.filter(a => a) 将删除 undefined

Answer 6

这可以通过多种方式完成； 一种选择是循环遍历对象属性

const list = [];      
for(let prop in myObj){
       if(myList.includes(prop)){
           list.push(myObj[prop])
        }
    }

使用 Object.keys() 可以达到相同的结果

 const result = Object.keys(myObj).map((x) => {
        if (myList.includes(parseInt(x))) {
           return myObj[x];
        }
    })

Answer 7

试试这个

let myObj = { 1:{name:'a'}, 2:{name:'v'}, 3:{name:'x'}, 10:{name:'t'} };
    let objsArray = []
    let myList = [1,2,10];
    for (let index = 0; index < myList.length; index++) {
       const element = myList[index];
       objsArray.push(myObj[element])
  
    }
    
    console.log(objsArray)

Answer 8

只需要map

var myObj = { 1:{name:'a'}, 2:{name:'v'}, 3:{name:'x'}, 10:{name:'t'} };
var myList = [1,2,10]
// var result = myList.map( function(k){ return myObj[k]; })
var result = myList.map( k => myObj[k])

console.log(result);

如果人们说它可能不在列表中，请使用filter

var myObj = { 1:{name:'a'}, 2:{name:'v'}, 3:{name:'x'}, 10:{name:'t'} };
var myList = [1,2,10, 33, 44, 555]
// var result = myList.map( function(k){ return myObj[k]; }).filter(function(o) { return o; });
var result = myList.map( k => myObj[k]).filter(Boolean);

console.log(result);

或者在一个循环中使用reduce

var myObj = { 1:{name:'a'}, 2:{name:'v'}, 3:{name:'x'}, 10:{name:'t'} };
var myList = [1,2,10, 33, 44, 555]
var result = myList.reduce((a, k) =>
  (myObj[k] && a.push(myObj[k]), a), [])
// var result = myList.reduce(function(a, k) {
//  if(myObj[k]) { 
//     a.push(myObj[k]);
//   }
//   return a
// }, []);
console.log(result);

Answer 9

使用Map

 myObj = { 1:{name:'a'}, 2:{name:'v'}, 3:{name:'x'}, 10:{name:'t'} };
 myList = [1,2,10]
 map= new Map(),res=[]
 Object.entries(myObj).map(o=>map.set(o[0],o[1]))
 myList.forEach(o=>{ v=map.get(o.toString())
   if(v) res.push(v)})
 console.log(res)

Answer 10

const myObj = { 1:{name:'a'}, 2:{name:'v'}, 3:{name:'x'}, 10:{name:'t'} };
const myList = [1,2,10];

const result = myList?.map(key => (myObj == null ? null : myObj[key]), []) || [];
console.log(result);

Answer 11

这是你想要的：

const myObj = { 1:{name:'a'}, 2:{name:'v'}, 3:{name:'x'}, 10:{name:'t'} };

const myList = [1,2,10];

const res = Object.entries(myObj).filter(([k,v])=>{
  return myList.includes(+k);
}).map(([k,v])=>{
  return v;
})
console.log(res);//[{name:'a'},{name:'v'}, {name:'t'}]