我有一个复杂的json文件,我必须处理javascript使其分层,以便稍后构建树。 json的每个条目都有: Id:唯一的Id, parentId:父节点的id(如果节点是树的根,则为0) Level:树的深度级别

json数据已经“有序”。我的意思是,一个条目在它上面有一个父节点或兄弟节点,在它下面有一个子节点或兄弟节点。

输入:

{
    "People": [
        {
            "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
        }
    ],
    "Animals": [
        {
            "id": "5",
            "parentId": "0",
            "text": "Dog",
            "level": "1",
            "children": null
        },
        {
            "id": "8",
            "parentId": "5",
            "text": "Puppy",
            "level": "2",
            "children": null
        },
        {
            "id": "10",
            "parentId": "13",
            "text": "Cat",
            "level": "1",
            "children": null
        },
        {
            "id": "14",
            "parentId": "13",
            "text": "Kitten",
            "level": "2",
            "children": null
        },
    ]
}

预期产量:

{
    "People": [
        {
            "id": "12",
            "parentId": "0",
            "text": "Man",
            "level": "1",
            "children": [
                {
                    "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":
            {

                "id": "11",
                "parentId": "9",
                "text": "Girl",
                "level": "2",
                "children": null
            }
        }

    ],    

    "Animals": [
        {
            "id": "5",
            "parentId": "0",
            "text": "Dog",
            "level": "1",
            "children": 
                {
                    "id": "8",
                    "parentId": "5",
                    "text": "Puppy",
                    "level": "2",
                    "children": null
                }
        },
        {
            "id": "10",
            "parentId": "13",
            "text": "Cat",
            "level": "1",
            "children": 
            {
                "id": "14",
                "parentId": "13",
                "text": "Kitten",
                "level": "2",
                "children": null
            }
        }

    ]
}

当前回答

更新2022

这是一个针对无序项的建议。该函数使用单个循环和哈希表,并收集所有带有id的项。如果找到根节点,则将该对象添加到结果数组中。

const getTree = (data, root) => { const t = {}; data.forEach(o => ((t[o.parentId] ??= {}).children ??= []).push(Object.assign(t[o.id] ??= {}, o))); return t[root].children; }, data = { People: [{ 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 }], Animals: [{ id: "5", parentId: "0", text: "Dog", level: "1", children: null }, { id: "8", parentId: "5", text: "Puppy", level: "2", children: null }, { id: "10", parentId: "13", text: "Cat", level: "1", children: null }, { id: "14", parentId: "13", text: "Kitten", level: "2", children: null }] }, result = Object.fromEntries(Object .entries(data) .map(([k, v]) => [k, getTree(v, '0')]) ); console.log(result); .as-console-wrapper { max-height: 100% !important; top: 0; }

其他回答

我有类似的问题,几天前必须从平面数组显示文件夹树。我在TypeScript中没有看到任何解决方案,所以我希望它会有帮助。

在我的情况下,主父只有一个,rawData数组也不需要排序。解决方案基于准备临时对象 {parentId: [child1, child2,…]]}

示例原始数据

const flatData: any[] = Folder.ofCollection([
  {id: '1', title: 'some title' },
  {id: '2', title: 'some title', parentId: 1 },
  {id: '3', title: 'some title', parentId: 7 },
  {id: '4', title: 'some title', parentId: 1 },
  {id: '5', title: 'some title', parentId: 2 },
  {id: '6', title: 'some title', parentId: 5 },
  {id: '7', title: 'some title', parentId: 5 },

]);

文件夹的定义

export default class Folder {
    public static of(data: any): Folder {
        return new Folder(data);
    }

    public static ofCollection(objects: any[] = []): Folder[] {
        return objects.map((obj) => new Folder(obj));
    }

    public id: string;
    public parentId: string | null;
    public title: string;
    public children: Folder[];

    constructor(data: any = {}) {
        this.id = data.id;
        this.parentId = data.parentId || null;
        this.title = data.title;
        this.children = data.children || [];
    }
}

解决方案:返回扁平参数的树结构的函数

    public getTree(flatData: any[]): Folder[] {
        const addChildren = (item: Folder) => {
            item.children = tempChild[item.id] || [];
            if (item.children.length) {
                item.children.forEach((child: Folder) => {
                    addChildren(child);
                });
            }
        };

        const tempChild: any = {};
        flatData.forEach((item: Folder) => {
            const parentId = item.parentId || 0;
            Array.isArray(tempChild[parentId]) ? tempChild[parentId].push(item) : (tempChild[parentId] = [item]);
        });

        const tree: Folder[] = tempChild[0];
        tree.forEach((base: Folder) => {
            addChildren(base);
        });
        return tree;
    }

它可能是有用的包列表到树 安装:

bower install list-to-tree --save

or

npm install list-to-tree --save

例如,有列表:

var list = [
  {
    id: 1,
    parent: 0
  }, {
    id: 2,
    parent: 1
  }, {
    id: 3,
    parent: 1
  }, {
    id: 4,
    parent: 2
  }, {
    id: 5,
    parent: 2
  }, {
    id: 6,
    parent: 0
  }, {
    id: 7,
    parent: 0
  }, {
    id: 8,
    parent: 7
  }, {
    id: 9,
    parent: 8
  }, {
    id: 10,
    parent: 0
  }
];

使用包列表到树:

var ltt = new LTT(list, {
  key_id: 'id',
  key_parent: 'parent'
});
var tree = ltt.GetTree();

结果:

[{
  "id": 1,
  "parent": 0,
  "child": [
    {
      "id": 2,
      "parent": 1,
      "child": [
        {
          "id": 4,
          "parent": 2
        }, {
          "id": 5, "parent": 2
        }
      ]
    },
    {
      "id": 3,
      "parent": 1
    }
  ]
}, {
  "id": 6,
  "parent": 0
}, {
  "id": 7,
  "parent": 0,
  "child": [
    {
      "id": 8,
      "parent": 7,
      "child": [
        {
          "id": 9,
          "parent": 8
        }
      ]
    }
  ]
}, {
  "id": 10,
  "parent": 0
}];

我使用@FurkanO answer并创建了一个可以用于任何对象类型的泛型函数,我还用TypeScript写了这个函数,我更喜欢它,因为它有自动补全功能。

实现:

1. Javascript:

export const flatListToTree = (flatList, idPath, parentIdPath, childListPath, isParent) => {
  const rootParents = [];
  const map = {};
  for (const item of flatList) {
    if (!item[childListPath]) item[childListPath] = [];
    map[item[idPath]] = item;
  }
  for (const item of flatList) {
    const parentId = item[parentIdPath];
    if (isParent(item)) {
      rootParents.push(item);
    } else {
      const parentItem = map[parentId];
      parentItem[childListPath].push(item);
    }
  }
  return rootParents;
};

2. TypeScript:我假设“T”类型有一个属性的孩子列表,你可以改变“childListPath”是一个字符串而不是“keyof T”如果你有不同的用例。

export const flatListToTree = <T>(
  flatList: T[],
  idPath: keyof T,
  parentIdPath: keyof T,
  childListPath: keyof T,
  isParent: (t: T) => boolean,
) => {
  const rootParents: T[] = [];
  const map: any = {};
  for (const item of flatList) {
    if (!(item as any)[childListPath]) (item as any)[childListPath] = [];
    map[item[idPath]] = item;
  }
  for (const item of flatList) {
    const parentId = item[parentIdPath];
    if (isParent(item)) {
      rootParents.push(item);
    } else {
      const parentItem = map[parentId];
      parentItem[childListPath].push(item);
    }
  }
  return rootParents;
};

使用方法:

  const nodes = [
    { id: 2, pid: undefined, children: [] },
    { id: 3, pid: 2 },
    { id: 4, pid: 2 },
    { id: 5, pid: 4 },
    { id: 6, pid: 5 },
    { id: 7, pid: undefined },
    { id: 8, pid: 7 },
  ];
  
  const result = flatListToTree(nodes, "id", "pid", "children", node => node.pid === undefined);

你可以使用这个来自Github或NPM的“treeify”包。

安装:

$ NPM install——save-dev treeify-js

var data = [{"country":"india","gender":"male","type":"lower","class":"X"}, {"country":"china","gender":"female","type":"upper"}, {"country":"india","gender":"female","type":"lower"}, {"country":"india","gender":"female","type":"upper"}]; var seq = ["country","type","gender","class"]; var treeData = createHieArr(data,seq); console.log(treeData) function createHieArr(data,seq){ var hieObj = createHieobj(data,seq,0), hieArr = convertToHieArr(hieObj,"Top Level"); return [{"name": "Top Level", "parent": "null", "children" : hieArr}] function convertToHieArr(eachObj,parent){ var arr = []; for(var i in eachObj){ arr.push({"name":i,"parent":parent,"children":convertToHieArr(eachObj[i],i)}) } return arr; } function createHieobj(data,seq,ind){ var s = seq[ind]; if(s == undefined){ return []; } var childObj = {}; for(var ele of data){ if(ele[s] != undefined){ if(childObj[ele[s]] == undefined){ childObj[ele[s]] = []; } childObj[ele[s]].push(ele); } } ind = ind+1; for(var ch in childObj){ childObj[ch] = createHieobj(childObj[ch],seq,ind) } return childObj; } }