有条件地从javascript中的嵌套数组中提取值

Question

如何从嵌套数组中提取值列表？符合条件？

const tree = {
   "menu_id":"root",
   "open":true,
   "items":[
      {
        "menu_id":"Brussels",
        "open":false,
         "items":[
            {
               "menu_id":"Brussels/CBD",
               "open":true,
               "items":[
                  {
                     "menu_id":"Brussels/CBD/Centre",
                     "open":false
                  },
                  {
                     "menu_id":"Brussels/CBD/Louise",
                     "open":true
                  },
                  {
                     "menu_id":"Brussels/CBD/Léopold",
                     "open":false
                  },
                  {
                     "menu_id":"Brussels/CBD/Midi",
                     "open":true
                  },
                  {
                     "menu_id":"Brussels/CBD/North",
                     "open":false
                  }
               ]
            }
        ]
      }
    ]
}

我想获取一个 menu_id 值列表，其中 open 属性为 false，因此会给出：

var closed = ['Brussels','Brussels/CBD/Centre','Brussels/CBD/Léopold','Brussels/CBD/North']

我该怎么做？

谢谢！

Answer 1

编写此代码的一种有用方法是构建一个通用函数来处理各种树和任意谓词，然后在其上构建您的函数。

此版本执行此操作，然后部分应用描述树结构的函数以获取新函数，部分应用谓词以获取另一个函数，该函数接受树并返回与谓词匹配的所有节点。最后，我们的 main 函数接受一棵树，调用前面的函数并从每个结果中提取 id 节点。

const filterDeep = (getKids) => (pred) => (tree) => [
  ... (pred (tree) ? [tree] : []),
  ... (getKids (tree)) .flatMap (filterDeep (getKids) (pred))
]

const filterItems = 
  filterDeep (({items = []}) => items)

const closed =
  filterItems (({open}) => !open)

const closedIds = (tree) =>
  closed (tree) .map (({menu_id}) => menu_id)

const tree = {menu_id: "root", open: true, items: [{menu_id: "Brussels", open: false, items: [{menu_id: "Brussels/CBD", open: true, items: [{menu_id: "Brussels/CBD/Centre", open: false}, {menu_id: "Brussels/CBD/Louise", open: true}, {menu_id: "Brussels/CBD/Léopold", open: false}, {menu_id: "Brussels/CBD/Midi", open: true}, {menu_id: "Brussels/CBD/North", open: false}]}]}]}

console .log (closedIds (tree))

我们显然可以在filterDeep 上直接写closedIds，如下所示：

const closedIds = (tree) =>
  filterDeep (({items}) => items) (({open}) => open == false) (tree) 
    .map (({menu_id}) => menu_id)

而且，虽然这并没有什么问题，但这些中间函数可能是可重用的，我认为这有助于使问题更加清晰。例如，filterItems 是一个函数，它接受一个谓词和一个树，其中一个节点的子节点位于一个名为 items 的数组中，并返回与谓词匹配的所有节点（无论嵌套级别如何）。看起来很有用。

---

但是，如果您只想要一个专用功能，我们可以将所有这些合并成一个相当简单的东西：

const closedIds = (tree) => [
  ... (tree.open ? [] : [tree.menu_id]),
  ... (tree.items || []) .flatMap (closedIds)
]

const tree = {menu_id: "root", open: true, items: [{menu_id: "Brussels", open: false, items: [{menu_id: "Brussels/CBD", open: true, items: [{menu_id: "Brussels/CBD/Centre", open: false}, {menu_id: "Brussels/CBD/Louise", open: true}, {menu_id: "Brussels/CBD/Léopold", open: false}, {menu_id: "Brussels/CBD/Midi", open: true}, {menu_id: "Brussels/CBD/North", open: false}]}]}]}

console .log (closedIds (tree))