如何以更现代的 ES6 方式实现嵌套在 for 中的 for？

Question

有人让我解决这个问题：Return 只有那些属性“enrollmentId”在另一个数字数组中的对象：所以我想到了这个，而且它有效，但是，我在 for 中使用 for，我想知道解决此类问题的最佳方法是什么。请检查一下

const companions = [
  {name:"camila", enrollmentId:1},
  {name:"oscar", enrollmentId:2},
  {name:"rupertina", enrollmentId:3}
  
];

const participants = [7,2,4]

const finalResult = [];

for(i=0; i< companions.length; i++){
  
  let alumno = companions[i];
  
  for(j=0; j< participants.length; j++){
    
    let participante = participants[j];
    
    if(alumno.enrollmentId == participante){
      finalResult.push(alumno);
    }    
    
  }
}

console.log(finalResult)

Answer 1

根据被迭代对象的enrollmentId是否包含在participants中过滤原始数组。

const companions = [
  {name:"camila", enrollmentId:1},
  {name:"oscar", enrollmentId:2},
  {name:"rupertina", enrollmentId:3}
];
const participants = [7, 2, 4]

const finalResult = companions.filter(obj => participants.includes(obj.enrollmentId));
console.log(finalResult)

Answer 2

我会使用 Array.filter to omit invalid rows, and Array.includes 进行有效性测试

const companions = [
    {name:"camila", enrollmentId:1},
    {name:"oscar", enrollmentId:2},
    {name:"rupertina", enrollmentId:3}
];

const participants = [7,2,4];

const finalResult = companions.filter(companion => {
    return participants.includes(companion.enrollmentId);
});

console.log(finalResult);

Answer 3

如果您想要 ES6 方式，请使用 code_monk 代码，但不使用 code_monk 代码，都会尊重（我无法在他的代码中添加注释，因为我是堆栈溢出的新手并且没有足够的声誉。）但是箭头函数 ( => ) 不需要关键字 return 也不需要函数 budy {}，因为它隐含了一个值将在 return 之后编辑=>

const companions = [
    {name:"camila", enrollmentId:1},
    {name:"oscar", enrollmentId:2},
    {name:"rupertina", enrollmentId:3}
];

const participants = [7,2,4];

const finalResult = companions.filter(companion => participants.includes(companion.enrollmentId));

console.log(finalResult);

如何以更现代的 ES6 方式实现嵌套在 for 中的 for？

How to implement a for nested in a for in a more modern ES6 way?

javascript

ecmascript-5

ecmascript-6