Question

我有这样的mongoose.Schema：

User.js：

const mongoose = require('mongoose');
const bcrypt = require('bcrypt-nodejs');
const Schema = mongoose.Schema;

const UserSchema = new Schema({

    username: String,
    password:  String,      
    email:  String,      
});

module.exports = mongoose.model('users', UserSchema, 'users');

product.js：

const mongoose = require('mongoose');
var mongoosePaginate = require('mongoose-paginate');
const Schema = mongoose.Schema;

const ProductSchema = new Schema({

    userid: {type:String, required: true},
    product_name: {type:String, required: true}     
});

ProductSchema.plugin(mongoosePaginate);
module.exports = mongoose.model('products', ProductSchema, 'products');

方式：

exports.selectUsersProductsCount = function (req, res, next) {

  Product.aggregate([
    {"$group" : {_id:{userid:"$userid" }, count:{$sum:1} }   }
],
function(err, result) {
    if(err){ res.status(400).json({ success: false, message:'Error processing request '+ err }); }
        res.status(201).json({
          success: true,
          data: result
    });
    console.log(result);
})

}

结果：

[ { _id: { userid: '5ab655bbb94733156c438112' }, count: 3 },
  { _id: { userid: '5ab425c6f5bff145304092f7' }, count: 1 } ]

我希望同时显示username字段。我不知道如何使用join。计算正确，但不显示用户名字段。

如何更正要显示的aggregate方法： 用户ID，用户名，计数

[ { _id: { userid: '5ab655bbb94733156c438112', username: 'Jon Alon' }, count: 3 },
  { _id: { userid: '5ab425c6f5bff145304092f7', username: 'Jonson Con' }, count: 1 } ]

示例数据：

产品

{
    _id:O bjectId(5ab7da972ade533790268f47),
    userid:"5ab655bbb94733156c438112",
    product_name:"gs",
    __v:0
},{
    _id: ObjectId(5ab7daa92ade533790268f48),
    userid:"5ab655bbb94733156c438112",
    product_name:"dg",
    __v:0 
}

用户

{
    _id: ObjectId(5ab655bbb94733156c438112),
    username: "rrrr",
    email:"rrrr",
    __v:0 
}

Answer 1

您必须使用$ lookup函数来实现这一目标，但您需要在两个集合之间具有一些共同属性。您需要在UserSchema或其他一些属性上添加userId，以便您可以加入＆＃34;

在您的示例中，您可以尝试

db.product.aggregate([ {$lookup:{ from: "user", localField: "ObjectId(userid)", foreignField: "ObjectId(_id)", as: "username" }}, {"$group" : {_id:{userid:"$userid", username : "$username" }, count:{$sum:1} } } ])

Answer 2

解决方案

products.aggregate([

  {$lookup:{
    from: "users",
    localField: "ObjectId(userid)",
    foreignField: "ObjectId(_id)",
    as: "users"
  }},


{ "$group" :
  {
    _id: { userid: "$userid" },
    name: { $last: "$users.username" } ,
    count: { $sum: 1 }
  }
}, {$sort: {"_id.userid":  1}}

  ], function(err, result) {
      console.log(result);
      if(err){ res.status(400).json({ success: false, message:'Error processing request '+ err }); }
          res.status(201).json({
            success: true,
            data: result
      });  
    }
  );
}

返回：

[ { _id: { userid: '5ab425c6f5bff145304092f7' },
    name: [ 'cccc', 'rrrr', 'zzzz' ],
    count: 3 },
  { _id: { userid: '5ab655bbb94733156c438112' },
    name: [ 'cccc', 'rrrr', 'zzzz' ],
    count: 1 },
  { _id: { userid: '5aba7e8c045115340496becd' },
    name: [ 'cccc', 'rrrr', 'zzzz' ],
    count: 2 } ]

Node.js - Angular - MongoDB - 用于`userid`，获取`username`

2 个答案: