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

这是上面的一个修改版本,适用于多个根项,我使用guid为我的id和parentid,所以在创建它们的UI中,我硬编码根项为0000000-00000-00000-TREE-ROOT-ITEM

var树= unflatten(记录," tree - root - item ");

function unflatten(records, rootCategoryId, parent, tree){
    if(!_.isArray(tree)){
        tree = [];
        _.each(records, function(rec){
            if(rec.parentId.indexOf(rootCategoryId)>=0){        // change this line to compare a root id
            //if(rec.parentId == 0 || rec.parentId == null){    // example for 0 or null
                var tmp = angular.copy(rec);
                tmp.children = _.filter(records, function(r){
                    return r.parentId == tmp.id;
                });
                tree.push(tmp);
                //console.log(tree);
                _.each(tmp.children, function(child){
                    return unflatten(records, rootCategoryId, child, tree);
                });
            }
        });
    }
    else{
        if(parent){
            parent.children = _.filter(records, function(r){
                return r.parentId == parent.id;
            });
            _.each(parent.children, function(child){
                return unflatten(records, rootCategoryId, child, tree);
            });
        }
    }
    return tree;
}

其他回答

将节点数组转换为树

ES6函数转换数组节点(由父ID相关)到树结构:

/**
 * Convert nodes list related by parent ID - to tree.
 * @syntax getTree(nodesArray [, rootID [, propertyName]])
 *
 * @param {Array} arr   Array of nodes
 * @param {integer} id  Defaults to 0
 * @param {string} p    Property name. Defaults to "parent_id"
 * @returns {Object}    Nodes tree
 */

const getTree = (arr, p = "parent_id") => arr.reduce((o, n) => {

  if (!o[n.id]) o[n.id] = {};
  if (!o[n[p]]) o[n[p]] = {};
  if (!o[n[p]].nodes) o[n[p]].nodes= [];
  if (o[n.id].nodes) n.nodes= o[n.id].nodes;

  o[n[p]].nodes.push(n);
  o[n.id] = n;

  return o;
}, {});

从节点树生成HTML列表

有了我们的树,这里有一个递归函数来构建UL > LI元素:

/**
 * Convert Tree structure to UL>LI and append to Element
 * @syntax getTree(treeArray [, TargetElement [, onLICreatedCallback ]])
 *
 * @param {Array} tree Tree array of nodes
 * @param {Element} el HTMLElement to insert into
 * @param {function} cb Callback function called on every LI creation
 */

const treeToHTML = (tree, el, cb) => el.append(tree.reduce((ul, n) => {
  const li = document.createElement('li');

  if (cb) cb.call(li, n);
  if (n.nodes?.length) treeToHTML(n.nodes, li, cb);

  ul.append(li);
  return ul;
}, document.createElement('ul')));

演示时间

下面是一个使用上述两个函数的线性节点数组的例子:

const getTree = (arr, p = "parent_id") => arr.reduce((o, n) => { if (!o[n.id]) o[n.id] = {}; if (!o[n[p]]) o[n[p]] = {}; if (!o[n[p]].nodes) o[n[p]].nodes = []; if (o[n.id].nodes) n.nodes = o[n.id].nodes; o[n[p]].nodes.push(n); o[n.id] = n; return o; }, {}); const treeToHTML = (tree, el, cb) => el.append(tree.reduce((ul, n) => { const li = document.createElement('li'); if (cb) cb.call(li, n); if (n.nodes?.length) treeToHTML(n.nodes, li, cb); ul.append(li); return ul; }, document.createElement('ul'))); // DEMO TIME: const nodesList = [ {id: 10, parent_id: 4, text: "Item 10"}, // PS: Order does not matters {id: 1, parent_id: 0, text: "Item 1"}, {id: 4, parent_id: 0, text: "Item 4"}, {id: 3, parent_id: 5, text: "Item 3"}, {id: 5, parent_id: 4, text: "Item 5"}, {id: 2, parent_id: 1, text: "Item 2"}, ]; const myTree = getTree(nodesList)[0].nodes; // Get nodes of Root (0) treeToHTML(myTree, document.querySelector("#tree"), function(node) { this.textContent = `(${node.parent_id} ${node.id}) ${node.text}`; this._node = node; this.addEventListener('click', clickHandler); }); function clickHandler(ev) { if (ev.target !== this) return; console.clear(); console.log(this._node.id); }; <div id="tree"></div>

你可以使用npm包数组到树https://github.com/alferov/array-to-tree。 它将普通的节点数组(带有指向父节点的指针)转换为嵌套的数据结构。

解决了从数据库数据集检索到嵌套数据结构(即导航树)的转换问题。

用法:

var arrayToTree = require('array-to-tree');

var dataOne = [
  {
    id: 1,
    name: 'Portfolio',
    parent_id: undefined
  },
  {
    id: 2,
    name: 'Web Development',
    parent_id: 1
  },
  {
    id: 3,
    name: 'Recent Works',
    parent_id: 2
  },
  {
    id: 4,
    name: 'About Me',
    parent_id: undefined
  }
];

arrayToTree(dataOne);

/*
 * Output:
 *
 * Portfolio
 *   Web Development
 *     Recent Works
 * About Me
 */

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);

ES6地图版本:

getTreeData = (items) => {
  if (items && items.length > 0) {
    const data = [];
    const map = {};
    items.map((item) => {
      const id = item.id; // custom id selector !!!
      if (!map.hasOwnProperty(id)) {
        // in case of duplicates
        map[id] = {
          ...item,
          children: [],
        };
      }
    });
    for (const id in map) {
      if (map.hasOwnProperty(id)) {
        let mappedElem = [];
        mappedElem = map[id];
        /// parentId : use custom id selector for parent
        if (
          mappedElem.parentId &&
          typeof map[mappedElem.parentId] !== "undefined"
        ) {
          map[mappedElem.parentId].children.push(mappedElem);
        } else {
          data.push(mappedElem);
        }
      }
    }
    return data;
  }
  return [];
};

/// use like this :

const treeData = getTreeData(flatList);

你可以用两行代码来解决这个问题:

_(flatArray).forEach(f=>
           {f.nodes=_(flatArray).filter(g=>g.parentId==f.id).value();});

var resultArray=_(flatArray).filter(f=>f.parentId==null).value();

在线测试(查看浏览器控制台以获得创建的树)

要求:

1-安装lodash 4(一个Javascript库,用于操作对象和集合,使用性能方法=>,就像c#中的Linq)

2-如下所示的flatArray:

    var flatArray=
    [{
      id:1,parentId:null,text:"parent1",nodes:[]
    }
   ,{
      id:2,parentId:null,text:"parent2",nodes:[]
    }
    ,
    {
      id:3,parentId:1,text:"childId3Parent1",nodes:[]
    }
    ,
    {
      id:4,parentId:1,text:"childId4Parent1",nodes:[]
    }
    ,
    {
      id:5,parentId:2,text:"childId5Parent2",nodes:[]
    }
    ,
    {
      id:6,parentId:2,text:"childId6Parent2",nodes:[]
    }
    ,
    {
      id:7,parentId:3,text:"childId7Parent3",nodes:[]
    }
    ,
    {
      id:8,parentId:5,text:"childId8Parent5",nodes:[]
    }];

谢谢Bakhshabadi先生

祝你好运