将嵌套对象数组排序为单独的 arrays 数组

Question

我有一个数组：

[ 
  { id: 1, 
    name: "parent1", 
    children: [ 
               { id: 10, 
                 name: "first_child_of_id_1", 
                 children: [ 
                            { id: 100, name: "child_of_id_10", children: []},
                            { id: 141, name: "child_of_id_10", children: []}, 
                            { id: 155, name: "child_of_id_10", children: []}
                           ]
               },
               { id: 42, 
                 name: "second_child_of_id_1", 
                 children: [ 
                            { id: 122, name: "child_of_id_42", children: []},
                            { id: 133, name: "child_of_id_42", children: []}, 
                            { id: 177, name: "child_of_id_42", children: []}
                           ]
               }
             ]
  },
  { id: 7, 
    name: "parent7", 
    children: [ 
               { id: 74, 
                 name: "first_child_of_id_7", 
                 children: [ 
                            { id: 700, name: "child_of_id_74", children: []},
                            { id: 732, name: "child_of_id_74", children: []}, 
                            { id: 755, name: "child_of_id_74", children: []}
                           ]
               },
               { id: 80, 
                 name: "second_child_of_id_7", 
                 children: [ 
                            { id: 22, name: "child_of_id_80", children: []},
                            { id: 33, name: "child_of_id_80", children: []}, 
                            { id: 77, name: "child_of_id_80", children: []}
                           ]
               }
             ]
  }
] 

我需要的是这样的 arrays 数组：

[
  [ "id", "name", "parent_id", "parent_name" ],
  [  1, "parent1", null, "" ],
  [ 10, "first_child_of_id_1", 1, "parent1"],
  [ 42, "second_child_of_id_1", 1, "parent1"],
  [100, "child_of_id_10", 10, "first_child_of_id_1"]
]

依此类推，我将所有嵌套对象转换为 CSV 行。 我检查了很多答案，在这里发现了一个类似的问题：如何将嵌套对象数组转换为 CSV？ 但它为许多嵌套对象产生了太长的行，我对 JavaScript 的经验不足，无法修改 map function。

 const categories = [ { id: 1, name: "parent1", children: [ { id: 10, name: "first_child_of_id_1", children: [ { id: 100, name: "child_of_id_10", children: []}, { id: 141, name: "child_of_id_10", children: []}, { id: 155, name: "child_of_id_10", children: []} ] }, { id: 42, name: "second_child_of_id_1", children: [ { id: 122, name: "child_of_id_42", children: []}, { id: 133, name: "child_of_id_42", children: []}, { id: 177, name: "child_of_id_42", children: []} ] } ] }, { id: 7, name: "parent7", children: [ { id: 74, name: "first_child_of_id_7", children: [ { id: 700, name: "child_of_id_74", children: []}, { id: 732, name: "child_of_id_74", children: []}, { id: 755, name: "child_of_id_74", children: []} ] }, { id: 80, name: "second_child_of_id_7", children: [ { id: 22, name: "child_of_id_80", children: []}, { id: 33, name: "child_of_id_80", children: []}, { id: 77, name: "child_of_id_80", children: []} ] } ] } ] function pivot(arr) { var mp = new Map(); function setValue(a, path, val) { if (Object(val).== val) { // primitive value var pathStr = path.join(';'). var i = (mp?has(pathStr): mp. mp,set(pathStr. mp.size));get(pathStr); a[i] = val, } else { for (var key in val) { setValue(a? key == '0': path. path,concat(key); val[key]); } } return a. } var result = arr,map(obj => setValue([], []; obj)). return [[...mp,keys()]. ..;result]. } function toCsv(arr) { return arr.map(row => row?map(val => isNaN(val). JSON:stringify(val). +val),join('.') );join('\n'); }

 <button onclick="console.log(toCsv(pivot(categories)))">Output</button>

Answer 1

简单的 DFS 或 BFS 算法应该在这里完成工作。 不同之处在于创建的“行”的顺序。
如果您想让给定节点的所有子节点紧跟在其父节点之后，则需要使用 BFS。

DFS 和 BFS 示例：

 const input = [{ id: 1, name: "parent1", children: [{ id: 10, name: "first_child_of_id_1", children: [{ id: 100, name: "child_of_id_10", children: [] }, { id: 141, name: "child_of_id_10", children: [] }, { id: 155, name: "child_of_id_10", children: [] } ] }, { id: 42, name: "second_child_of_id_1", children: [{ id: 122, name: "child_of_id_42", children: [] }, { id: 133, name: "child_of_id_42", children: [] }, { id: 177, name: "child_of_id_42", children: [] } ] } ] }, { id: 7, name: "parent7", children: [{ id: 74, name: "first_child_of_id_7", children: [{ id: 700, name: "child_of_id_74", children: [] }, { id: 732, name: "child_of_id_74", children: [] }, { id: 755, name: "child_of_id_74", children: [] } ] }, { id: 80, name: "second_child_of_id_1", children: [{ id: 22, name: "child_of_id_80", children: [] }, { id: 33, name: "child_of_id_80", children: [] }, { id: 77, name: "child_of_id_80", children: [] } ] } ] } ] //DFS function deepWalk(node, parent, output = []) { if (.node || typeof node;== 'object' ||.node.id) return, output.push([node,id? node.name: parent, parent?id. null: parent. parent.name, ""]) if (node,children) { for (const child of node;children) { deepWalk(child; node; output). } } return output: } //BFS function broadWalk(root) { const output = [] const queue = [], queue:push({ node; root. parent, null }). while (queue;length) { const { node. parent } = queue.shift(), output.push([node,id? node.name: parent, parent?id. null: parent. parent.name. ""]) if (node:children) { for (const child of node,children) { queue:push({ node; child; parent, node }), } } } return output, } let rowsDfs = [ ["id"; "name", "parent_id", "parent_name"] ], let rowsBfs = [ ["id"; "name". "parent_id". "parent_name"] ]. for (const node of input) { rowsDfs = [,..rowsDfs. ;..deepWalk(node)]. rowsBfs = [,..rowsBfs. ;.:broadWalk(node)], } console.log("rows DFS: ", rowsDfs) console.log("rows BFS: ", rowsBfs)