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

这是一个旧线程,但我认为更新永远不会伤害,与ES6你可以做到:

const data = [{ id: 1, parent_id: 0 }, { id: 2, parent_id: 1 }, { id: 3, parent_id: 1 }, { id: 4, parent_id: 2 }, { id: 5, parent_id: 4 }, { id: 8, parent_id: 7 }, { id: 9, parent_id: 8 }, { id: 10, parent_id: 9 }]; const arrayToTree = (items=[], id = null, link = 'parent_id') => items.filter(item => id==null ? !items.some(ele=>ele.id===item[link]) : item[link] === id ).map(item => ({ ...item, children: arrayToTree(items, item.id) })) const temp1=arrayToTree(data) console.log(temp1) const treeToArray = (items=[], key = 'children') => items.reduce((acc, curr) => [...acc, ...treeToArray(curr[key])].map(({ [`${key}`]: child, ...ele }) => ele), items); const temp2=treeToArray(temp1) console.log(temp2)

希望它能帮助到别人

其他回答

我的解决方案:

允许双向映射(根到叶,叶到根) 返回所有节点、根节点和叶节点 一次数据传递和非常快的性能 香草Javascript

/**
 * 
 * @param data items array
 * @param idKey item's id key (e.g., item.id)
 * @param parentIdKey item's key that points to parent (e.g., item.parentId)
 * @param noParentValue item's parent value when root (e.g., item.parentId === noParentValue => item is root)
 * @param bidirectional should parent reference be added
 */
function flatToTree(data, idKey, parentIdKey, noParentValue = null, bidirectional = true) {
  const nodes = {}, roots = {}, leaves = {};

  // iterate over all data items
  for (const i of data) {

    // add item as a node and possibly as a leaf
    if (nodes[i[idKey]]) { // already seen this item when child was found first
      // add all of the item's data and found children
      nodes[i[idKey]] = Object.assign(nodes[i[idKey]], i);
    } else { // never seen this item
      // add to the nodes map
      nodes[i[idKey]] = Object.assign({ $children: []}, i);
      // assume it's a leaf for now
      leaves[i[idKey]] = nodes[i[idKey]];
    }

    // put the item as a child in parent item and possibly as a root
    if (i[parentIdKey] !== noParentValue) { // item has a parent
      if (nodes[i[parentIdKey]]) { // parent already exist as a node
        // add as a child
        (nodes[i[parentIdKey]].$children || []).push( nodes[i[idKey]] );
      } else { // parent wasn't seen yet
        // add a "dummy" parent to the nodes map and put the item as its child
        nodes[i[parentIdKey]] = { $children: [ nodes[i[idKey]] ] };
      }
      if (bidirectional) {
        // link to the parent
        nodes[i[idKey]].$parent = nodes[i[parentIdKey]];
      }
      // item is definitely not a leaf
      delete leaves[i[parentIdKey]];
    } else { // this is a root item
      roots[i[idKey]] = nodes[i[idKey]];
    }
  }
  return {roots, nodes, leaves};
}

使用的例子:

const data = [{id: 2, parentId: 0}, {id: 1, parentId: 2} /*, ... */];
const { nodes, roots, leaves } = flatToTree(data, 'id', 'parentId', 0);

有同样的问题,但我不能确定数据是否已排序。我不能使用第三方库,所以这只是香草Js;输入数据可以从@Stephen的例子中获取;

var arr = [ {'id':1 ,'parentid' : 0}, {'id':4 ,'parentid' : 2}, {'id':3 ,'parentid' : 1}, {'id':5 ,'parentid' : 0}, {'id':6 ,'parentid' : 0}, {'id':2 ,'parentid' : 1}, {'id':7 ,'parentid' : 4}, {'id':8 ,'parentid' : 1} ]; function unflatten(arr) { var tree = [], mappedArr = {}, arrElem, mappedElem; // First map the nodes of the array to an object -> create a hash table. for(var i = 0, len = arr.length; i < len; i++) { arrElem = arr[i]; mappedArr[arrElem.id] = arrElem; mappedArr[arrElem.id]['children'] = []; } for (var id in mappedArr) { if (mappedArr.hasOwnProperty(id)) { mappedElem = mappedArr[id]; // If the element is not at the root level, add it to its parent array of children. if (mappedElem.parentid) { mappedArr[mappedElem['parentid']]['children'].push(mappedElem); } // If the element is at the root level, add it to first level elements array. else { tree.push(mappedElem); } } } return tree; } var tree = unflatten(arr); document.body.innerHTML = "<pre>" + (JSON.stringify(tree, null, " "))

JS小提琴

平面阵列到树

(奖励1:节点可以排序,也可以不排序)

(bonus2:不需要第三方库,纯js)

(BONUS3:用户“Elias Rabl”说这是最高效的解决方案,见下面他的回答)

下面就是:

const createDataTree = dataset => {
  const hashTable = Object.create(null);
  dataset.forEach(aData => hashTable[aData.ID] = {...aData, childNodes: []});
  const dataTree = [];
  dataset.forEach(aData => {
    if(aData.parentID) hashTable[aData.parentID].childNodes.push(hashTable[aData.ID])
    else dataTree.push(hashTable[aData.ID])
  });
  return dataTree;
};

下面是一个测试,它可能会帮助你理解解决方案是如何工作的:

it('creates a correct shape of dataTree', () => {
  const dataSet = [{
    "ID": 1,
    "Phone": "(403) 125-2552",
    "City": "Coevorden",
    "Name": "Grady"
  }, {
    "ID": 2,
    "parentID": 1,
    "Phone": "(979) 486-1932",
    "City": "Chełm",
    "Name": "Scarlet"
  }];

  const expectedDataTree = [{
    "ID": 1,
    "Phone": "(403) 125-2552",
    "City": "Coevorden",
    "Name": "Grady",
    childNodes: [{
      "ID": 2,
      "parentID": 1,
      "Phone": "(979) 486-1932",
      "City": "Chełm",
      "Name": "Scarlet",
      childNodes : []
    }]
  }];

  expect(createDataTree(dataSet)).toEqual(expectedDataTree);
});

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;
    }

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; }