如何在不更改当前架构的情况下在猫鼬中填充数据答案

【问题标题】：How to populate data in mongoose without changing current schema如何在不更改当前架构的情况下在猫鼬中填充数据
【发布时间】：2018-07-13 14:10:20
【问题描述】：

我已经为名为 associates 和 outlets 的 mongoose 创建了两个模式。
出口架构：-

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

var OutletSchema = new Schema(
{
  associate_id: {type: ObjectId, max: 100 },
  city_id:{type:ObjectId, max: 100},
  outlet_name: {type: String, max: 100},
  oulet_address: {type: String, max: 500},
  outlet_contact: {type: String, minlength: 9, maxlength: 12 },
  active: {type: Boolean}`enter code here`
}`
);

module.exports = mongoose.model('Outlet', OutletSchema);

关联架构：-

var mongoose = require('mongoose');

var Schema = mongoose.Schema;
ObjectId = Schema.ObjectId;

var AssociateSchema = new Schema(
{

  associate_name: {type: String, max: 100},
  associate_address: {type: String, max: 500},
  associate_contact: {type: String, minlength: 9, maxlength: 12 },
  associate_email: {type: String, max: 100},
  active: {type: Boolean}
}
);

module.exports = mongoose.model('Associate', AssociateSchema);

这些是我创建的架构。

我如何使用 relationship() 填充员工中的出口

我已经尝试过下面给出的代码。

router.get('/getUsingPopulate', function (req, res) {
    data = {}
    Associate.relationship({path:'outlet', ref:'Outlet',refPath:'associate_id'})
    Outlet.find({}).populate('Associate').exec()
    .then(function(alldata){
        console.log(alldata)

    })


})

合伙人数据库：-

_id: 5b40428088e5f0187dedb342
category_id: 5b34a99709d3ff3e2b5a399c
associate_name: "Cafe Coffee Day"
associate_address: "Coffee day kormangala"
associate_contact: "989545458645"
associate_email: "ccdkor@yahoo.com"
active: true
__v: 0

出口数据库：-

_id: 5b4471f4c5bf8018409c7c64
associate_id: 5b40428088e5f0187dedb342
city_id: 5b3c49fee0de9210a60a2266
outlet_name: "Coffee Day Whitefield"
oulet_address: " Coffee Day Whitefield"
outlet_contact: "7587454478"
active: true
__v:0

预期输出：-

  {
   "_id": "5b40428088e5f0187dedb342",
   "name": "Cafe Coffee Day",
   "outlets": [
     {
       "_id": "5b4471f4c5bf8018409c7c64",
       "associate_id": 5b40428088e5f0187dedb342,
       "name": "Coffee Day Whitefield",

     }]

【问题讨论】：

您可以使用$lookup 聚合来填充associates 模型中的outlet
在使用 $lookup 时，我得到的结果为 "{"_pipeline":[{"$project":{"deal_on":"$deal_on"}},{"$lookup":{" from":"Outlet","localField":"outlet_id","foreignField":"_id","as":"OutletData"}}],"options":{}}"
你必须把你的样本集合和输出看看$lookup是如何工作的
我正在尝试使用 associate_id 获取所有出口详细信息和关联详细信息。我怎样才能加入并将所有详细信息放在一起..请参考：以上模式并帮助我。提前致谢。
请发布您从 robomongo 或 mongo shell 复制的样本集合，这样我会更容易向您展示 $lookup 的工作原理

标签： javascript mongodb mongoose

【解决方案1】：

我用这段代码得到了答案：-

出口架构更改为：-

var OutletSchema = new Schema(
{
  associate_id: {type: Schema.Types.ObjectId, ref: 'Associate'  },
  city_id:{type:ObjectId, max: 100},
  outlet_name: {type: String, max: 100},
  oulet_address: {type: String, max: 500},
  outlet_contact: {type: String, minlength: 9, maxlength: 12 },
  active: {type: Boolean}
})

我给出的填充代码是：-

Deal.find({..}).populate({
                                path: 'outlet_id',
                                model: 'Outlet',
                                populate: {
                                  path: 'associate_id',
                                  model: 'Associate'
                                }
                              }).exec()

感谢大家的支持..

【讨论】：