只有在函数内部时，循环才会返回额外的时间答案

【问题标题】：A loop is returning an extra time more only when inside a function只有在函数内部时，循环才会返回额外的时间
【发布时间】：2018-08-07 06:27:32
【问题描述】：

如果我错过了一些非常简单的东西，非常抱歉。

我一直在尝试写一些东西，它将返回一个属性编号、属性名称和对象的所有属性值的列表，格式如下：

"n property - property value"

我确信我已经以一种完全过于复杂的方式解决了这个问题，但这里是代码：

var person = {
    names: ["A", "B", "C"],
    dob: "D",
    age: 21,
    job: "E",
    car: ["F", 2, "G", "H", "I", "J", "K"],
    phone: "L",
    instrument: "M",
    parents: ["N", "O", "P"], 
    born: "Q",
    lives: "R",
    university: "S",
};

function lister (name) {
  for (i=0; i<(Object.keys(name).length); i++) {
  if (((Object.values(name)[i].constructor) == Array) === true) {
    for (x=0; x<(Object.keys(Object.values(name)[i]).length); x++) {
      console.log(i + " " + Object.keys(name)[i] + " - " + Object.values(Object.values(name)[i])[x]);
    }
  }else{
    console.log(i + " " + Object.keys(name)[i] + " - " + Object.values(name)[i]);
  }
}
};

console.log(lister(person));

for (i=0; i<(Object.keys(person).length); i++) {
  if (((Object.values(person)[i].constructor) == Array) === true) {
    for (x=0; x<(Object.keys(Object.values(person)[i]).length); x++) {
      console.log(i + " " + Object.keys(person)[i] + " - " + Object.values(Object.values(person)[i])[x]);
    }
  }else{
    console.log(i + " " + Object.keys(person)[i] + " - " + Object.values(person)[i]);
  }
};

问题是，当我在 function lister (name) 中调用 for 循环时，它似乎额外运行了一次，给了我：

"0 names - A"

"0 names - B"

"0 names - C"

"1 dob - D"

...

"9 lives - R"

"10 university - S"

undefined

但是，当从函数中取出 for 循环时（正如我在底部所做的那样），它会正常返回，而不是返回额外的“未定义”。我做错了什么？

【问题讨论】：

(Object.values(name)[i].constructor) == Array) === truesrsly?

标签： javascript list function for-loop undefined

【解决方案1】：

我会怎么做

 function lister(obj){
   let i = 0;
   for(const [key, value] of Object.entries(obj)){
      if(Array.isArray(value)){
        for(const entry of value)
           console.log(i + " " + key + "-" + entry);
      } else {
        console.log(i + " " + key + "-" + value);
      }
      i++;
   }
}

【讨论】：

【解决方案2】：

extra 不是来自循环，而是来自函数调用。如果函数没有显式返回任何内容，则返回 undefined，在下面的 sn-p 中，函数 lister 正在返回 x，因此您将看到 x 而不是 undefined

var person = {
  names: ["A", "B", "C"],
  dob: "D",
  age: 21,
  job: "E",
  car: ["F", 2, "G", "H", "I", "J", "K"],
  phone: "L",
  instrument: "M",
  parents: ["N", "O", "P"],
  born: "Q",
  lives: "R",
  university: "S",
};

function lister(name) {
  for (i = 0; i < (Object.keys(name).length); i++) {
    if (((Object.values(name)[i].constructor) == Array) === true) {
      for (x = 0; x < (Object.keys(Object.values(name)[i]).length); x++) {
        console.log(i + " " + Object.keys(name)[i] + " - " + Object.values(Object.values(name)[i])[x]);
      }
    } else {
      console.log(i + " " + Object.keys(name)[i] + " - " + Object.values(name)[i]);
    }
  }
  return 'x'
};

console.log(lister(person));

for (i = 0; i < (Object.keys(person).length); i++) {
  if (((Object.values(person)[i].constructor) == Array) === true) {
    for (x = 0; x < (Object.keys(Object.values(person)[i]).length); x++) {
      console.log(i + " " + Object.keys(person)[i] + " - " + Object.values(Object.values(person)[i])[x]);
    }
  } else {
    console.log(i + " " + Object.keys(person)[i] + " - " + Object.values(person)[i]);
  }
};

【讨论】：