循环由于某种原因被卡住，然后最终返回未定义

Question

这就是我试图解决的问题：给定：一个包含名称哈希的数组Return：格式为用逗号分隔的名称列表的字符串，但最后两个名称除外，最后两个名称应以 & 符号分隔。例：list([ {name: 'Bart'}, {name: 'Lisa'}, {name: 'Maggie'} ])// returns 'Bart, Lisa & Maggie'list([ {name: 'Bart'}, {name: 'Lisa'} ])// returns 'Bart & Lisa'list([ {name: 'Bart'} ])// returns 'Bart'list([])// returns ''注意：所有哈希值都经过预先验证，并且仅包含 A-Z、a-z、“-”和 “。这是我的代码：var finalName;var notFinal;function list(names){  var finalNames = names.forEach(returnNames);        console.log(typeof finalNames);  function returnNames() {    for (var i = 0; i<names.length; i++) {      var nameValue = Object.keys(names[i]).map(key => names[i][key])    }  }  for(var i = 0; i<finalNames.length; i++) {    if (finalNames.length / i == 1) {      finalName = "& " + finalNames[i];     }    else {      notFinal = finalNames[i] + ", ";    }  }  console.log(notFinal + finalName);}list([{name: 'Bart'},{name: 'Lisa'},{name: 'Maggie'},{name: 'Homer'},{name: 'Marge'}])它卡在循环中，最后给出一个错误：TypeError: Cannot read property 'length' of undefined    at list    at /home/codewarrior/index.js:30:1    at Object.handleError        <anonymous>如何修复此问题？

Answer 1

这是因为 forEach 不会返回任何内容，请尝试改用映射函数。

var finalNames = names.map(returnNames);

Answer 2

正如他们已经指出的那样，返回 。相反，您可以使用此来修改您的函数Array.prototype.forEachundefined.mapreturnNames

var finalName;

var notFinal;

function list(names){

  // Changed .forEach with .map

  var finalNames = names.map(returnNames);

  console.log(typeof finalNames);

  function returnNames(person) {

    // If you only need to get the object values, use Object.values instead of Object.keys

    return Object.values(person);

  }

  for(var i = 0; i < finalNames.length; i++) {

    // Added + 1 because i could never be equal to the array length

    // Note that you'll need to make 1 or 2 more changes before this code works as expected

    if (finalNames.length / (i + 1) == 1) {

      finalName = "& " + finalNames[i]; 

    }

    else {

      notFinal = finalNames[i] + ", ";

    }

  }

  console.log(notFinal + finalName);

}

list([{name: 'Bart'},{name: 'Lisa'},{name: 'Maggie'},{name: 'Homer'},{name: 'Marge'}])