绕过多维数组

Question

我需要从复杂数组中获取一些数据到数组中。例如我有这个结构：

animals =
     /*...*/
         list: [
           {
             type: "fluffy"
             withCategories: false
             animal: [
               {
                 name: "Patrik"
                 description: "..."
                 price: 135
                 weight: 220
                 rating: 94
               }
               {
                 name: "Mike"
                 description: "..."
                 price: 135
                 weight: 235
                 rating: 97
               }
             ]
           }
           {
             imageUrl: "/img/borsh.jpg"
             type: "pets"
             withCategories: true
             categories: [
               {
                 name: "parrot"
                 imageUrl: "/img/parrot.jpg"
                 withCategories: false
                 animal: [
                   {
                     name: "Kesha"
                     description: "..."
                     price: 75
                     weight: 250
                     rating: 89
                   }
                 ]
               }
             ]
           }

因此我需要一个包含动物元素的数组： //result = [object1 -> name: Patrik, price: 135 ..], ..., [object3 -> name: Kesha, description: ...] 正如您在较低级别上看到的那样，如果参数 "withCategories" = true，我们将递归向下。我试着去实现它：

PlacesService.getAllAnimals = ->
    merged = []
    temp = (getAnimalByCategory(category) for category in animals.list)
    merged = merged.concat.apply(merged, temp)
    return merged

  getAnimalByCategory = (category) ->
    if category.withCategories == false
      return category.animal
    else
      (getAnimalByCategory(an) for an in category.categories)

但是出了点问题:(有人可以帮我吗？我找不到任何错误。

Answer 1

您希望您的 getAnimalByCategory 到 return 是一个平面列表，以便循环的结果可以被 concat 合并。但是，在递归情况下，您不需要 return。

PlacesService.getAllAnimals = ->
  [].concat.apply([], getAnimalByCategory(category) for category in animals.list)

getAnimalByCategory = (category) ->
  if not category.withCategories
    category.animal
  else
    [].concat.apply([], getAnimalByCategory(an) for an in category.categories)

绕过多维数组

Bypassing the multi-dimensional arrays

javascript

multidimensional-array

coffeescript