创建mongodb对象的思维导图

Question

我想从模型分支中制作一些类似思维导图的东西，它具有“父”和“子”值。看起来是这样的：

Branches = [{
 children: {},
 _id: String,
 parent: { type: Schema.Types.ObjectId, ref: 'Branch' }
}]

我想要得到这样的东西：

[
 {
    "_id": "599c1f79f339dc3991d8250a",
    "name": "Robert",
    "children": [
        {
            "_id": "599c1f82f339dc3991d8250b",
            "name": "Robert 1",
            "parent": "599c1f79f339dc3991d8250a",
            "children": [
                {
                    "_id": "599c1f8ff339dc3991d8250c",
                    "name": "Robert 2",
                    "parent": "599c1f82f339dc3991d8250b",
                    "children": [
                        {
                            "_id": "599c2b7373a7d43e5205af1f",
                            "name": "Robert 4",
                            "parent": "599c1f8ff339dc3991d8250c",
                            "__v": 0,
                            "shareholder": [],
                            "createdDate": "2017-08-22T13:02:43.290Z",
                            "children": []
                        }
                    ],
                    "__v": 0,
                    "shareholder": [],
                    "createdDate": "2017-08-22T12:11:59.230Z"
                },
                {
                    "_id": "599c1f9df339dc3991d8250d",
                    "name": "Robert 3",
                    "parent": "599c1f82f339dc3991d8250b",
                    "children": [],
                    "__v": 0,
                    "shareholder": [],
                    "createdDate": "2017-08-22T12:12:13.156Z"
                }
            ],
            "__v": 0,
            "shareholder": [],
            "createdDate": "2017-08-22T12:11:46.938Z"
        }
    ],
    "__v": 0,
    "parent": null,
    "shareholder": [],
    "createdDate": "2017-08-22T12:11:37.005Z"
}

]“

如果一个对象具有值parent "1“，则具有_id：”1“分支必须将其包含在子对象中。我现在的代码是：

Branch.find((err, foundedBranches) => {

 function createTree(branch) {
  function inner(parent) {
      return branch.filter(x => x.parent === parent)
              .map(x => {
                  x.children = inner(x._id)
                  return x
              })
  }
  return inner(null)
}
 let formated = createTree(branches)
 res.json(formated)

如果您知道如何解决，请帮助:/

Answer 1

像这样怎么样--你将会得到每一个条目的childs

// All items from database
const dataFromDatabase = // ...

// For each item, go and look for childrens
const finalMap = dataFromDatabase.map((x) => {
  children: this.getChildren(dataFromDatabase, x._id),
  _id: x._id,
  // ...
});

// Get all children and for theses children go for children aswell (recursive)...
getChildren(all, id) {
  return all.reduce((tmp, x) => {
    if (x.parent === id) {
      return [
        ...tmp,
        {
          children: this.getChildren(all, x._id),
          _id: x._id,
          // ...
        },
      ];
    }

    return tmp;
  }, []);
}

如果您希望条目只出现一次，请在开始时过滤条目以仅获取标题(没有的条目，父项)

dataFromDatabase.filter(x => !x.parent).map((x) => {