猫鼬的关系一对多

Question

I have a problem, more precisely i don't understand one thing with monggose and relationship.

I have one collection "Athletes" :

var athletesSchema = mongoose.Schema({
name : String,
regionName : String,
age : Number,
overallScore : Number,
scores : {
    ordinnal : String,
    ordinnal2 : String,
    ordinnal3: String,
    ordinnal4: String,
    ordinnal5:String
}
})

I have another collection "Regions"

var regionsSchema = mongoose.Schema({
name: String,
athletes: [{ type: Schema.Types.ObjectId, ref: 'Athletes' }]
})

One athlete have a region but a region have multiple athletes. I don't understand how is that possible, when i create an athlete the model region knows if it's on this region or this region.

i had this to my server.js :

Athletes.find({name : datas.name}, function(err, name){
        if(name == ""){
            Athletes.create(datas, function(err, ath){
                   if(err || ath.name){
                       console.log(err)
                   }
                   Regions.create(regionName, function(err){
                    if(err){
                        console.log(err)
                    }
                })
           })

        }
    }) 

I check before create a new user if is not in the database, is not i create the athlete, but i don't know how to had his region in the Region model.

Answer 1

Maybe it would be easier for you to reverse this. Have regions only having the name. And then add regionId on the Athlete model:

region: {
  type: mongoose.Schema.Types.ObjectId,
  ref: 'Regions',
  index: true,
}

Then any time you update an Athlete, you're only updating this, athlete, object. Alternatively, dealing with arrays is not as simple.

With this you can query (and populate), eg all athletes from given regions:

Athletes.find({ region: [region1, region2]})
  .populate('region')

Now, updating region name is simplified, you do it in one place, not on all athlete instances.

There are more complex scenarios where you want to be querying things by various criteria, but most of the times, this would be the simplest way.