我想知道这条线上发生了什么，我用粗体。我知道他们的工作，但我不知道为什么。 [英] I'd like to know what's happening in the line's I've bolded. I know they 'work, ' but I'm not sure why.

问题描述

我正在锻炼，我必须编写一个名为select的函数。给定一个数组和对象，我不得不返回一个新对象，其属性仅包括数组和原始对象中常见的对象。例如：

I was doing in exercise in which I had to write a function called select. Given an array and object, I had to return a new object whose properties included only those which were common among the array and the original object. For example:

var arr = ['a', 'c', 'e'];
var obj = {
  a: 1,
  b: 2,
  c: 3,
  d: 4
};
var output = select(arr, obj);
console.log(output); // --> { a: 1, c: 3 }

下面的代码有效，但我正在努力理解newObject [search] = obj [search]如何在newObject中产生正确的结果。

The code below 'works,' but I'm trying to understand how newObject[search] = obj[search] yields the right results in newObject.

function select(arr, obj) {
  var newObject = {};
  var keys = Object.keys(obj);
  for (var i = 0; i < arr.length; i++) {
    var search = arr[i];
      if (keys.indexOf(search) !== - 1) {
        newObject[search] = obj[search];
    }
  }
  return newObject;
}

我的尝试：



我已经多次通过Google开发工具中的调试器运行此代码。我最好的理解是obj [search]一直回到'keys'变量，以找出它与'search'变量的关系。



真诚地感谢您的帮助！

What I have tried:

I've run this code through the debugger in Google dev tools multiple times. My best understanding is that obj[search] reaches all the way back up to the 'keys' variable to find out how it relates to the 'search' variable.

Thanks sincerely for your help!

推荐答案

function select(arr, obj) {
  var newObject = {};
  var keys = Object.keys(obj);
  for (var i = 0; i < arr.length; i++) {
    var search = arr[i];
    if (keys.indexOf(search) !== - 1) {
      newObject[search] = obj[search];
    }
  }
  return newObject;
}

您正在设置的变量包含 obj 参数的所有键 - 键。然后迭代遍历 arr 参数，检查每个元素（ search ）是否在键中 list（ indexOf 如果未找到密钥则返回-1）。如果元素在列表中，则将值从 obj 移动到 newObject 。



给出你的例子的赋值语句将执行 newObject ['a'] = obj ['a']; 和 newObject ['c'] = obj ['c']; 。



MDN [ ^ ]有一个很棒的JS指南和参考资料。

What's happening is you setup a variable which holds all the keys of the obj parameter - keys. You then iterate through the arr parameter checking whether each element (search) is in the keys list (indexOf returns -1 if no key is found). If the element is in the list you then move the value from obj into newObject.

The assignment statement given your example will execute newObject['a'] = obj['a']; and newObject['c'] = obj['c'];.

The MDN[^] has a great JS guide and reference materials.

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