javascript - 类似对象 .map() 的 JavaScript 数组

Question

您好，我想使用 .map 来切换我的数组的排名和谷物，但是我在 console.log 中得到了一个未定义的结果。基于很棒的反馈，我能够让一切正常工作，但是我仍然对某些事情感到有些困惑。因为我不确定如何将谷物与排名倒序匹配？我完全被难住了。

var breakFastFood =[
  {
     cereal: "Captain Crunch",
     scale: "Yuck!"

  },
  {
    cereal: "Grape Nuts",
    scale: "Yum!"

 },
 {
  cereal: "Fruity Pebbles",
  scale: "Yuck!"

},
{
  cereal: "Oatmeal",
  scale: "Yum!"

}
];
var cereals = breakFastFood.map(function(bFood){
 return breakFastFood.cereal
});

var rank = breakFastFood.map(function(standing){
 return breakFastFood.scale
});

rank.forEach(function(rating){console.log(rating)});
cereals.forEach(function(food){console.log(food)});

标签： javascriptarraysarray.prototype.map

Answer 1

您没有在return语句中使用函数参数：

var breakFastFood =[
  {
     cereal: "Captain Crunch",
     scale: "Yuck!"

  },
  {
    cereal: "Grape Nuts",
    scale: "Yum!"

 },
 {
  cereal: "Fruity Pebbles",
  scale: "Yuck!"

},
{
  cereal: "Oatmeal",
  scale: "Yum!"

}
];
var cereals = breakFastFood.map(function(bFood){
 return bFood.cereal
});

var rank = breakFastFood.map(function(standing){
 return standing.scale
});

rank.forEach(function(rating){console.log(rating)});
cereals.forEach(function(food){console.log(food)});

您还可以使用简写属性：

var breakFastFood =[
  {
     cereal: "Captain Crunch",
     scale: "Yuck!"

  },
  {
    cereal: "Grape Nuts",
    scale: "Yum!"

 },
 {
  cereal: "Fruity Pebbles",
  scale: "Yuck!"

},
{
  cereal: "Oatmeal",
  scale: "Yum!"

}
];
var cereals = breakFastFood.map(({cereal}) => cereal);

var rank = breakFastFood.map(({scale}) => scale);

rank.forEach(function(rating){console.log(rating)});
cereals.forEach(function(food){console.log(food)});