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

使用ES6方法。工作很有魅力

//数据集 //一个顶级注释 Const注释= [{ id: 1、 parent_id:零 }, { id: 2 parent_id: 1 }, { id: 3, parent_id: 1 }, { id: 4 parent_id: 2 }, { id: 5 parent_id: 4 }); Const nest = (items, id = null, link = 'parent_id') => 项目 .filter(item => item[link] === id) .map(item =>({…Item, children: nest(items, Item .id)})); console.log ( 巢(评论) )

其他回答

下面是Steven Harris的一个修改版本,它是普通的ES5,返回一个以id为键的对象,而不是返回顶层和子层的节点数组。

unflattenToObject = function(array, parent) {
  var tree = {};
  parent = typeof parent !== 'undefined' ? parent : {id: 0};

  var childrenArray = array.filter(function(child) {
    return child.parentid == parent.id;
  });

  if (childrenArray.length > 0) {
    var childrenObject = {};
    // Transform children into a hash/object keyed on token
    childrenArray.forEach(function(child) {
      childrenObject[child.id] = child;
    });
    if (parent.id == 0) {
      tree = childrenObject;
    } else {
      parent['children'] = childrenObject;
    }
    childrenArray.forEach(function(child) {
      unflattenToObject(array, child);
    })
  }

  return tree;
};

var arr = [
    {'id':1 ,'parentid': 0},
    {'id':2 ,'parentid': 1},
    {'id':3 ,'parentid': 1},
    {'id':4 ,'parentid': 2},
    {'id':5 ,'parentid': 0},
    {'id':6 ,'parentid': 0},
    {'id':7 ,'parentid': 4}
];
tree = unflattenToObject(arr);

正如@Sander提到的,@Halcyon的答案假设一个预先排序的数组,下面的不是。(然而,它假设你已经加载了underscore.js -尽管它可以用香草javascript编写):

Code

// Example usage var arr = [ {'id':1 ,'parentid' : 0}, {'id':2 ,'parentid' : 1}, {'id':3 ,'parentid' : 1}, {'id':4 ,'parentid' : 2}, {'id':5 ,'parentid' : 0}, {'id':6 ,'parentid' : 0}, {'id':7 ,'parentid' : 4} ]; unflatten = function( array, parent, tree ){ tree = typeof tree !== 'undefined' ? tree : []; parent = typeof parent !== 'undefined' ? parent : { id: 0 }; var children = _.filter( array, function(child){ return child.parentid == parent.id; }); if( !_.isEmpty( children ) ){ if( parent.id == 0 ){ tree = children; }else{ parent['children'] = children } _.each( children, function( child ){ unflatten( array, child ) } ); } return tree; } tree = unflatten( arr ); document.body.innerHTML = "<pre>" + (JSON.stringify(tree, null, " ")) <script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore-min.js"></script>

需求

它假设属性'id'和'parentid'分别表示id和父id。必须有父ID为0的元素,否则将返回一个空数组。孤儿元素及其后代“丢失”

http://jsfiddle.net/LkkwH/1/

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);

一个更简单的从列表到树的函数

NPM安装列表到树精简版

listToTree(列表)

来源:

function listToTree(data, options) {
    options = options || {};
    var ID_KEY = options.idKey || 'id';
    var PARENT_KEY = options.parentKey || 'parent';
    var CHILDREN_KEY = options.childrenKey || 'children';

    var tree = [],
        childrenOf = {};
    var item, id, parentId;

    for (var i = 0, length = data.length; i < length; i++) {
        item = data[i];
        id = item[ID_KEY];
        parentId = item[PARENT_KEY] || 0;
        // every item may have children
        childrenOf[id] = childrenOf[id] || [];
        // init its children
        item[CHILDREN_KEY] = childrenOf[id];
        if (parentId != 0) {
            // init its parent's children object
            childrenOf[parentId] = childrenOf[parentId] || [];
            // push it into its parent's children object
            childrenOf[parentId].push(item);
        } else {
            tree.push(item);
        }
    };

    return tree;
}

斯菲德尔

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