从不同的 collection 获取数据，在 mongoose 中没有任何关系

Question

从不同的 collection 获取数据，与 mongoose

没有任何关系

Collection 架构

mongoose.Schema({
    skillid:{type:Number},
    name:{type:String},
});

技能Collection

mongoose.Schema({
  userid: {type: mongoose.Types.ObjectId},
  OverView:{type:String},
  location:{type:String},
  skills:[{Type:Number}] 
});

 { 
_id: 5f48d5d1b98bffee67b49917
 skillid: 1
 name: 'HTML' 
}

{ 
_id: 5f48d612b98bffee67b49919
skillid: 2
 name: 'PHP' 
}


User Collection

{
  _id: 5f425bdb311b791670d60de6,
  userid: 5f41115fbd904134883ae2d8,
  OverView: 'sdsdssdsd',
  skills: [1,2],   // skill id
  Education: [ 5f453e7f53895727f0e39d82, 5f453fb963d4ab181c115982 ],
  location: 'India',
}

如何从技能 Collection - mongoose

中获取技能名称

我想要这样的结果

 { 
  _id: 5f425bdb311b791670d60de6, 
   userid: 5f41115fbd904134883ae2d8, 
   OverView: 'sdsdssdsd', 
   skills: ['HTML','PHP'], // skill id
   Education: [ 5f453e7f53895727f0e39d82, 5f453fb963d4ab181c115982 ],
   location: 'India'
 } ```

Answer 1

您可以使用 aggregate(),

$lookup加入skill合集，在技能合集中匹配技能
$addFields to add skill name in array format using $reduce

db.user.aggregate([
  {
    $lookup: {
      from: "skill",
      localField: "skills",
      foreignField: "skillid",
      as: "skills"
    }
  },
  {
    $addFields: {
      skills: {
        $reduce: {
          input: "$skills",
          initialValue: [],
          in: {
            $concatArrays: [["$$this.name"], "$$value"]
          }
        }
      }
    }
  }
])

Playground

您可以使用猫鼬的另一个选项Virtual and also look at this

// SKILL SCHEMA
const SkillSchema = mongoose.Schema({
    skillid:{type:Number},
    name:{type:String},
});

// USER SCHEMA
const UserSchema = mongoose.Schema({
  userid: {type: mongoose.Types.ObjectId},
  OverView:{type:String},
  location:{type:String},
  skills:[{Type:Number}] 
});

// CREATES VIRTUAL CONNECTION WITH SKILL COLLECTION
UserSchema.virtual('skills', {
  ref: 'Skill', // The model to use
  localField: 'skills', // Find people where `localField`
  foreignField: 'skillid', // is equal to `foreignField`
  // If `justOne` is true, 'members' will be a single doc as opposed to
  // an array. `justOne` is false by default.
  justOne: false,
  // you can use other options
  // options: { sort: { name: -1 }, limit: 5 } 
});

// CREATE MODELS
const User = mongoose.model('User', UserSchema);
const Skill = mongoose.model('Skill', SkillSchema);

// QUERY TO FIND USERS AND POPULATE SKILLS
User.find({}).populate('skills').exec(function(error, user) {
  console.log(user);
});

Note: please refer documentation, if you are getting any problem, This is not tested code!

从不同的 collection 获取数据，在 mongoose 中没有任何关系

Get Data from the different collection without any relation in mongoose

node.js

mongodb

express

mongoose

mean-stack