JavaScript：将具有父键的对象数组转换为父/子树（包括没有父的对象）答案

【问题标题】：JavaScript: Convert array of objects with parent keys to parent / child tree (including objects with no parent)JavaScript：将具有父键的对象数组转换为父/子树（包括没有父的对象）
【发布时间】：2019-03-07 13:20:10
【问题描述】：

我有一个带有父键的对象列表，它描述了多个嵌套的父/子关系级别。

const table =[
    {
        "id": 791,
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {
        "id": 790,
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {
        "id": 845,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {   
        "id": 844,
        "sortOrder": 0,
        "parentCategoryId": 842
    },
    {   
        "id": 802,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {   
        "id": 788,
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {    
        "id": 863,
        "sortOrder": 0,
        "parentCategoryId": 863
    },
    {    
        "id": 858,
        "sortOrder": 0,
        "parentCategoryId": 858
    },
    {    
        "id": 867,
        "sortOrder": 0,
        "parentCategoryId": 867
    },
    {    
        "id": 871,
        "sortOrder": 0,
        "parentCategoryId": 867
    },
    {    
        "id": 801,
        "name": "Tickets",
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {    
        "id": 792,
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {    
        "id": 797,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {    
        "id": 789,
        "name": "Hot food",
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {    
        "id": 798,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {    
        "id": 671,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {    
        "id": 833,
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {    
        "id": 796,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {    
        "id": 843,
        "sortOrder": 0,
        "parentCategoryId": 842
    },
    {    
        "id": 840,
        "sortOrder": 0,
        "parentCategoryId": 793
    },
    {    
        "id": 868,
        "sortOrder": 0,
        "parentCategoryId": 868
    },
    {    
        "id": 851,
        "sortOrder": 0,
        "parentCategoryId": 851
    },
    {    
        "id": 839,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {    
        "id": 793,
        "sortOrder": 0,
        "parentCategoryId": 839
    },
    {    
        "id": 859,
        "sortOrder": 0,
        "parentCategoryId": 859
    },
    {   
        "id": 805,
        "sortOrder": 0,
        "parentCategoryId": 859
    },
    {    
        "id": 856,
        "name": "DRINKS",
        "sortOrder": 0,
        "parentCategoryId": 805
    },
    {    
        "id": 870,
        "sortOrder": 0,
        "parentCategoryId": 856
    },
    {    
        "id": 787,
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {    
        "id": 786,
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {    
        "id": 799,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {    
        "id": 852,
        "sortOrder": 0,
        "parentCategoryId": 852
    },
    {    
        "id": 795,
        "name": "Gents fragrance",
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {    
        "id": 864,
        "sortOrder": 0,
        "parentCategoryId": 864
    },
    {   
        "id": 854,
        "sortOrder": 0,
        "parentCategoryId": 854
    },
    {    
        "id": 865,
        "sortOrder": 0,
        "parentCategoryId": 865
    },
    {    
        "id": 869,
        "name": "GFI",
        "sortOrder": 0,
        "parentCategoryId": 869
    },
    {    
        "id": 785,
        "sortOrder": 0,
        "parentCategoryId": 833
    }
]

问题是我没有 0 的根父 ID。我想在一个数组中排序，该数组显示在第一级项目中，id 匹配 parentCategoryId，这意味着它们每个都是根并且比他们每个人都拥有孩子中的孩子。

这是我已经走了多远，但很难做到这一点：

var root = { cid: 0, parent_id: null, children: []};
var node_list = { 0 : root};

     for (var i = 0; i < table.length; i++) {

       console.log('updated list', node_list)
       console.log('item in cat', table[i])

       // check if parent ID exsits in the list
       if (!node_list[table[i].parentCategoryId]) {

         console.log('not in the list');
         console.log('node_list[table[i].parentCategoryId]', table[i].parentCategoryId)
         if (table[i].parentCategoryId === table[i].cid) {
           console.log('it is the root');
           node_list[table[i].cid] = table[i];
         }    

       } else {

         const item = table[i];
         console.log('item is ', item)

         node_list[table[i].parentCategoryId].children = {
           ...node_list[table[i].parentCategoryId].children,
           ...item
         };
       }
     }

预期结果：

const table =[
    {  
        "id": 791,
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {    
        "id": 790,
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {    
        "id": 845,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {    
        "id": 844,
        "sortOrder": 0,
        "parentCategoryId": 842
    },
    {    
        "id": 802,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {    
        "id": 788,
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {    
        "id": 863,
        "sortOrder": 0,
        "parentCategoryId": 863
    },
    {    
        "id": 858,
        "sortOrder": 0,
        "parentCategoryId": 858
    },
    {    
        "id": 867,
        "sortOrder": 0,
        "parentCategoryId": 867
    },
    {    
        "id": 871,
        "sortOrder": 0,
        "parentCategoryId": 867
    },
    {    
        "id": 801,
        "sortOrder": 0,
        "parentCategoryId": 847
    },    
    {    
        "id": 797,
        "sortOrder": 0,
        "parentCategoryId": 847,
        children:[
            {    
                "id": 792,
                "sortOrder": 0,
                "parentCategoryId": 797,
                children:[
                    {
                        "id": 671,
                        "sortOrder": 0,
                        "parentCategoryId": 792
                    },
                ]
            },
        ]
    },
    {    
        "id": 789,
        "name": "Hot food",
        "sortOrder": 0,
        "parentCategoryId": 833
    },
    {    
        "id": 798,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {    
        "id": 833,
        "sortOrder": 0,
        "parentCategoryId": 833,
        children:[
            {    
                "id": 785,
                "sortOrder": 0,
                "parentCategoryId": 833
            },
            {
                "id": 786,
                "sortOrder": 0,
                "parentCategoryId": 833
            },
            {
                "id": 787,
                "sortOrder": 0,
                "parentCategoryId": 833
            },
        ]
    },
    {   
        "id": 796,
        "sortOrder": 0,
        "parentCategoryId": 847
    },
    {   
        "id": 843,
        "sortOrder": 0,
        "parentCategoryId": 842
    },
    {   
        "id": 840,
        "sortOrder": 0,
        "parentCategoryId": 793
    },
    {    
        "id": 868,
        "sortOrder": 0,
        "parentCategoryId": 868
    },
    {    
        "id": 851,
        "sortOrder": 0,
        "parentCategoryId": 851
    },
    {    
        "id": 839,
        "sortOrder": 0,
        "parentCategoryId": 847,
        children:[
            {
                "id": 793,
                "sortOrder": 0,
                "parentCategoryId": 839,
                children:[
                    {    
                        "id": 870,
                        "sortOrder": 0,
                        "parentCategoryId": 856
                    },
                ]
            },
        ]
    },
    {    
        "id": 805,
        "sortOrder": 0,
        "parentCategoryId": 859,
        children:[
            {
                "id": 856,
                "sortOrder": 0,
                "parentCategoryId": 805
            },
            {    
                "id": 859,
                "sortOrder": 0,
                "parentCategoryId": 805
            },
        ]
    },      
]

【问题讨论】：

请同时添加想要的结果。
添加了想要的结果
表和结果不匹配，例如792和797。
我知道，因为我没有时间手动映射这个。但它显示了结构

标签： javascript arrays object spread-syntax

【解决方案1】：

看起来您想将带有一组值的 children 键添加到初始数组中的对象，其中 id 值对应于数组中其他对象的一个或多个 parentCategoryId 值 - 并且没有对象应作为嵌套对象数组中的父对象或子对象重复。

您可以map 数组来附加孩子，然后filter 只返回根父母（和孤儿）。例如（如果您想查看输出，请在示例下方使用 sn-p）：

const ids = table.map((x) => x.id);
const result = table.map((parent) => {
  const children = table.filter((child) => {
    if (child.id !== child.parentCategoryId && child.parentCategoryId === parent.id) {
      return true;
    }

    return false;
  });

  if (children.length) {
    parent.children = children;
  }

  return parent;
}).filter((obj) => {
  if (obj.id === obj.parentCategoryId || !ids.includes(obj.parentCategoryId)) {
    // include ultimate parents and orphans at root
    return true;
  }

  return false;
});

const table = [{ "id": 791, "sortOrder": 0, "parentCategoryId": 833 }, { "id": 790, "sortOrder": 0, "parentCategoryId": 833 }, { "id": 845, "sortOrder": 0, "parentCategoryId": 847 }, { "id": 844, "sortOrder": 0, "parentCategoryId": 842 }, { "id": 802, "sortOrder": 0, "parentCategoryId": 847 }, { "id": 788, "sortOrder": 0, "parentCategoryId": 833 }, { "id": 863, "sortOrder": 0, "parentCategoryId": 863 }, { "id": 858, "sortOrder": 0, "parentCategoryId": 858 }, { "id": 867, "sortOrder": 0, "parentCategoryId": 867 }, { "id": 871, "sortOrder": 0, "parentCategoryId": 867 }, { "id": 801, "name": "Tickets", "sortOrder": 0, "parentCategoryId": 847 }, { "id": 792, "sortOrder": 0, "parentCategoryId": 833 }, { "id": 797, "sortOrder": 0, "parentCategoryId": 847 }, { "id": 789, "name": "Hot food", "sortOrder": 0, "parentCategoryId": 833 }, { "id": 798, "sortOrder": 0, "parentCategoryId": 847 }, { "id": 671, "sortOrder": 0, "parentCategoryId": 847 }, { "id": 833, "sortOrder": 0, "parentCategoryId": 833 }, { "id": 796, "sortOrder": 0, "parentCategoryId": 847 }, { "id": 843, "sortOrder": 0, "parentCategoryId": 842 }, { "id": 840, "sortOrder": 0, "parentCategoryId": 793 }, { "id": 868, "sortOrder": 0, "parentCategoryId": 868 }, { "id": 851, "sortOrder": 0, "parentCategoryId": 851 }, { "id": 839, "sortOrder": 0, "parentCategoryId": 847 }, { "id": 793, "sortOrder": 0, "parentCategoryId": 839 }, { "id": 859, "sortOrder": 0, "parentCategoryId": 859 }, { "id": 805, "sortOrder": 0, "parentCategoryId": 859 }, { "id": 856, "name": "DRINKS", "sortOrder": 0, "parentCategoryId": 805 }, { "id": 870, "sortOrder": 0, "parentCategoryId": 856 }, { "id": 787, "sortOrder": 0, "parentCategoryId": 833 }, { "id": 786, "sortOrder": 0, "parentCategoryId": 833 }, { "id": 799, "sortOrder": 0, "parentCategoryId": 847 }, { "id": 852, "sortOrder": 0, "parentCategoryId": 852 }, { "id": 795, "name": "Gents fragrance", "sortOrder": 0, "parentCategoryId": 847 }, { "id": 864, "sortOrder": 0, "parentCategoryId": 864 }, { "id": 854, "sortOrder": 0, "parentCategoryId": 854 }, { "id": 865, "sortOrder": 0, "parentCategoryId": 865 }, { "id": 869, "name": "GFI", "sortOrder": 0, "parentCategoryId": 869 }, { "id": 785, "sortOrder": 0, "parentCategoryId": 833 }];
const ids = table.map((x) => x.id);
const result = table.map((parent) => {
  const children = table.filter((child) => {
    if (child.id !== child.parentCategoryId && child.parentCategoryId === parent.id) {
      return true;
    }
    
    return false;
  });
  
  if (children.length) {
    parent.children = children;
  }
  
  return parent;
}).filter((obj) => {
  if (obj.id === obj.parentCategoryId || !ids.includes(obj.parentCategoryId)) {
    // include ultimate parents and orphans at root
    return true;
  }
  
  return false;
});

// stringify just to flatten out SO console result for easier result scanning
console.log(JSON.stringify(result));

【讨论】：

谢谢@benvc，我看到的唯一问题是该对象在添加为子项后仍在根列表中。所以结果中的元素是重复的。
@ChrisTarasovs 我明白了，在这种情况下，包装函数需要是 filter 而不是 map - 编辑答案以反映澄清。
@ChrisTarasovs - 我改变了从过滤器开始的想法，我认为先映射然后过滤更简单（并且更具可读性） - 相应地编辑了答案。

【解决方案2】：

我们有一个复杂的 json 文件，我们必须用 JavaScript 处理它以使其分层，以便以后构建一棵树。 JSON数组的每个条目都有 - id - 一个唯一的 id， parentId - 父节点的 id（如果节点是树的根，则为 0） level - 树中的深度级别。

JSON 数据已经“有序”，这意味着条目将在其自身之上有一个父节点或兄弟节点，在其自身之下有一个子节点或兄弟节点。

    const arr = [
   {
      "id": "12",
      "parentId": "0",
      "text": "Man",
      "level": "1",
      "children": null
   },
   {
      "id": "6",
      "parentId": "12",
      "text": "Boy",
      "level": "2",
      "children": null
   },
   {
      "id": "7",
      "parentId": "12",
      "text": "Other",
      "level": "2",
      "children": null
   },
   {
      "id": "9",
      "parentId": "0",
      "text": "Woman",
      "level": "1",
      "children": null
   },
   {
      "id": "11",
      "parentId": "9",
      "text": "Girl",
      "level": "2",
      "children": null
   }
];
const listToTree = (arr = []) => {
   let map = {}, node, res = [], i;
   for (i = 0; i < arr.length; i += 1) {
      map[arr[i].id] = i;
      arr[i].children = [];
   };
   for (i = 0; i < arr.length; i += 1) {
      node = arr[i];
      if (node.parentId !== "0") {
         arr[map[node.parentId]].children.push(node);
      }
      else {
         res.push(node);
      };
   };
   return res;
};
console.log(JSON.stringify(listToTree(arr), undefined, 4));

【讨论】：