用mongoose包装数据

Question

我有3个不同模式的文件。用户有很多笔记本，笔记本有很多笔记。模式示例：

用户架构：

var mongoose    = require('mongoose'),
    Schema      = mongoose.Schema;

var User = new Schema({
    username:  { type: String, require: true, index: { unique: true }, trim: true},
    password:  { type: String, require: true, select: true },
    age: { type: Number, min: 0 },
    firstname: String,
    secondname: String,
    token: String,
    role: String,
    city: String, 
    rememberMe: Boolean
});

module.exports = mongoose.model('User', User);

笔记本架构：

var mongoose        = require('mongoose'),
    Schema          = mongoose.Schema,
    Note            = require('./note'),
    User            = require('./user');

var NoteBook = new Schema({
    creator: { type:Schema.ObjectId, ref:"User"},
    name: String,
    description: String
}); 

NoteBook.methods.getAllNotes = function(cb) {

    Note.find({notebook: this}, function(err, noteList){
        cb(noteList);
    });
};

module.exports = mongoose.model('NoteBook', NoteBook);

注意模式：

var mongoose=require（'mongoose'），Schema=mongoose。图式；

var NoteSchema = new Schema({
    notebook: { type: Schema.Types.ObjectId, ref: 'NoteBook'},
    name: String,
    description: String,
    content: String
});

module.exports = mongoose.model('Note', NoteSchema);

我得到了请求userId，我需要打包所有这行数据。使用笔记本电脑的用户，以及使用笔记本电脑的用户。我玩过这样的游戏：

函数getTree（req，res）{

    var data = [];

    User.findOne({_id: req.body.userId}, function(err, user) {

        NoteBook.find({creator: user._id}, function(err, notebookList) {

            for (var idx in notebookList) {

                Note.find({notebok: notebookList[idx]._id}, function(err, noteList) {

                    var children = [];

                    for (var noteIdx in noteList) {

                        children.push({
                            'text': noteList[idx].name,
                            'a_attr' : {
                                    'data-node-type': 'note',
                                    'data-node-id': noteList[idx]._id,
                                },
                            });         
                    }                       

                data.push({
                    'text': notebookList[idx].name,
                    'a_attr' : {
                            'data-node-type': 'notebook',
                            'data-node-id': notebookList[idx]._id,
                        },
                        'children': children
                    });
                });
            }


            res.json({ tree: data });
        });
    });
}

但它不起作用。

Answer 1

var async = require("async");

function getTree(req, res) {

    var data = [];

    User.findOne({_id: req.body.userId}, function(err, user) {

        NoteBook.find({creator: user._id}, function(err, notebookList) {

            async.forEach(notebookList, function(notebook, callback){


                Note.find({notebok: notebook._id}, function(err, noteList) {

                    var children = [];

                    for (var noteIdx in noteList) {

                        children.push({
                            'text': noteList[idx].name,
                            'a_attr' : {
                                'data-node-type': 'note',
                                'data-node-id': noteList[idx]._id,
                            },
                        });
                    }

                    data.push({
                        'text': notebookList[idx].name,
                        'a_attr' : {
                            'data-node-type': 'notebook',
                            'data-node-id': notebook._id,
                        },
                        'children': children
                    });
                });

            }, function(err){
                res.json({ tree: data });
            });



        });
    });
}