Question

我有一系列任务，这些任务通过parentId彼此链接。但是，任何任务都可以有多个父母。因此，不一定是简单的单树分层结构。我想要实现的是代替父母[]，我想要一个孩子[]。但是我似乎无法理解如何递归地解决这个问题。任何帮助将不胜感激。

这是Flat JSON数组

[
  {
    "_id": "4b04e450-06d5-4453-8d50-d3b2a70d9b2d",
    "task_name": "Parent2",
    "parents": []
  },
  {
    "_id": "a15ca08e-f13b-4d73-a496-ba23832ea233",
    "task_name": "Endpoints",
    "parents": [
      {
        "_id": "97bbf892-8a2a-4f45-befd-4fdbebded04b",
        "task_name": "Parent1"
      },
      {
        "_id": "4b04e450-06d5-4453-8d50-d3b2a70d9b2d",
        "task_name": "Parent2"
      }
    ]
  },
  {
    "_id": "ee78316a-491e-4db5-8f82-13b12b5b86fc",
    "task_name": "Mapping",
    "parents": [
      {
        "_id": "97bbf892-8a2a-4f45-befd-4fdbebded04b",
        "task_name": "Parent1"
      }
    ]
  },
  {
    "_id": "97bbf892-8a2a-4f45-befd-4fdbebded04b",
    "task_name": "Parent1",
    "parents": []
  }
]

我想要实现的是以下

[
  {
    "_id": "97bbf892-8a2a-4f45-befd-4fdbebded04b",
    "task_name": "Parent1",
    "children": [
      {
        "_id": "ee78316a-491e-4db5-8f82-13b12b5b86fc",
        "task_name": "Mapping",
        "children": []
      },
      {
        "_id": "a15ca08e-f13b-4d73-a496-ba23832ea233",
        "task_name": "Endpoints",
        "children": []
      }
    ]
  },
  {
    "_id": "4b04e450-06d5-4453-8d50-d3b2a70d9b2d",
    "task_name": "Parent2",
    "children": [
      {
        "_id": "a15ca08e-f13b-4d73-a496-ba23832ea233",
        "task_name": "Endpoints",
        "children": []
      }
    ]
  }
]

我尝试过的

function transform(list, idAttr, parentAttr, childrenAttr) {
    if (!idAttr) idAttr = '_id';
    if (!parentAttr) parentAttr = 'parents';
    if (!childrenAttr) childrenAttr = 'children';

    var newArr = [];
    var lookup = {};
    list.forEach(function(obj) {
        lookup[obj[idAttr]] = obj;
        obj[childrenAttr] = [];
    });
    list.forEach(function(obj) {
        if (obj[parentAttr] != null) {
            lookup[obj[parentAttr]][childrenAttr].push(obj);
        } else {
            newArr.push(obj);
        }
    });
    return newArr;
};

如果原始数组中的父键等于任务的_id，则此方法很好。但是，我不确定如何将其用于具有对象数组作为值的父键。

Answer 1

您可以使用_id作为键的对象，并存储父母及其关系。最后带走没有父母的物品的孩子。

var data = [{ _id: "4b04e450-06d5-4453-8d50-d3b2a70d9b2d", task_name: "Parent2", parents: [] }, { _id: "a15ca08e-f13b-4d73-a496-ba23832ea233", task_name: "Endpoints", parents: [{ _id: "97bbf892-8a2a-4f45-befd-4fdbebded04b", task_name: "Parent1" }, { _id: "4b04e450-06d5-4453-8d50-d3b2a70d9b2d", task_name: "Parent2" }] }, { _id: "ee78316a-491e-4db5-8f82-13b12b5b86fc", task_name: "Mapping", parents: [{ _id: "97bbf892-8a2a-4f45-befd-4fdbebded04b", task_name: "Parent1" }] }, { _id: "97bbf892-8a2a-4f45-befd-4fdbebded04b", task_name: "Parent1", parents: [] }],
    tree = function (data, root) {
        var t = {};
        data.forEach(({ parents, ...o }) => {
            Object.assign(t[o._id] = t[o._id] || {}, o);
            if (!parents.length) {
                t[root] = t[root] || {};
                t[root].children = t[root].children || [];
                t[root].children.push(t[o._id]);
                return;
            }
            parents.forEach(p => {
                Object.assign(t[p._id] = t[p._id] || {}, p);
                t[p._id].children = t[p._id].children || [];
                t[p._id].children.push(t[o._id]);
            });
        });
        return t[root].children;
    }(data);

console.log(tree);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

您可以保留平面JSON数组结构，并仅存储ID以进行交叉引用（id只是示例）

    [
        {
            "_id": "97bbf892-8a2a-4f45-befd-4fdbebded04b",
            "task_name": "Task1",
            "children": [
                "ee78316a-491e-4db5-8f82-13b12b5b86fc",
                "a15ca08e-f13b-4d73-a496-ba23832ea233"
            ],
            "parents": [
                "ee78316a-491e-4db5-8f82-13b12b5b86fc",
                "a15ca08e-f13b-4d73-a496-ba23832ea233"
            ]
        },
        {
            "_id": "97bbf892-8a2a-4f45-befd-4fdbebded04b",
            "task_name": "Task2",
            "children": [
                "ee78316a-491e-4db5-8f82-13b12b5b86fc",
                "a15ca08e-f13b-4d73-a496-ba23832ea233"
            ],
            "parents": [
                "ee78316a-491e-4db5-8f82-13b12b5b86fc",
                "a15ca08e-f13b-4d73-a496-ba23832ea233"
            ]
        },
        {
            "_id": "97bbf892-8a2a-4f45-befd-4fdbebded04b",
            "task_name": "Task3",
            "children": [
                "ee78316a-491e-4db5-8f82-13b12b5b86fc",
                "a15ca08e-f13b-4d73-a496-ba23832ea233"
            ],
            "parents": [
                "ee78316a-491e-4db5-8f82-13b12b5b86fc",
                "a15ca08e-f13b-4d73-a496-ba23832ea233"
            ]
        }
    ]

或者如果您想以当前希望的方式实现它，则可以使用此功能。没有经过完全测试，但您明白了

var tasks =     [
  {
    "_id": "4b04e450-06d5-4453-8d50-d3b2a70d9b2d",
    "task_name": "Parent2",
    "parents": []
  },
  {
    "_id": "a15ca08e-f13b-4d73-a496-ba23832ea233",
    "task_name": "Endpoints",
    "parents": [
      {
        "_id": "97bbf892-8a2a-4f45-befd-4fdbebded04b",
        "task_name": "Parent1"
      },
      {
        "_id": "4b04e450-06d5-4453-8d50-d3b2a70d9b2d",
        "task_name": "Parent2"
      }
    ]
  },
  {
    "_id": "ee78316a-491e-4db5-8f82-13b12b5b86fc",
    "task_name": "Mapping",
    "parents": [
      {
        "_id": "97bbf892-8a2a-4f45-befd-4fdbebded04b",
        "task_name": "Parent1"
      }
    ]
  },
  {
    "_id": "97bbf892-8a2a-4f45-befd-4fdbebded04b",
    "task_name": "Parent1",
    "parents": []
  }
];

function transformTasks(tasks){
    var transformed = {};
    var transformedAsArray = [];
    for(var taskIndex in tasks){
        var task = tasks[taskIndex];
        if(transformed[task["_id"]] === undefined){
            transformed[task["_id"]] = {_id:task["_id"],task_name:task["task_name"],children:[]};
            transformedAsArray.push(transformed[task["_id"]]);
        }
        for(var parentIndex in task["parents"]){
            var parentTask = task["parents"][parentIndex];
            if(transformed[parentTask["_id"]] === undefined){
                transformed[parentTask["_id"]] = {_id:parentTask["_id"],task_name:parentTask["task_name"],children:[]};
                transformedAsArray.push(transformed[parentTask["_id"]]);
            }
            transformed[parentTask["_id"]].children.push(transformed[task["_id"]]);

        }
    }

    return transformedAsArray;
}

console.log(transformTasks(tasks));

如何将对象的平面数组（可能有多个父对象）转换为嵌套的对象数组

2 个答案: