How to improve my Room database architecture?

Question

I am new to databases. I am not a professional developer. I would like your advice. I want to create a database that manages the students in a class. Students belong to only one class. I present to you my model. Can you let me know if this is correct please?

My data class: Gru

@Entity(tableName = "groupe_table")
data class Gru (
 @PrimaryKey(autoGenerate = true)  @ColumnInfo(name = "idGroup") var idGroup: Int=0,
 @ColumnInfo(name = "nameGroupG") var nameGroupG : String
):Parcelable


@Entity(tableName = "user_table", foreignKeys = arrayOf(
 ForeignKey(entity = Gru::class,
     parentColumns = arrayOf("idGroup"),
     childColumns = arrayOf("id"),
     onDelete = ForeignKey.CASCADE)
))

@Parcelize
data class User(@PrimaryKey(autoGenerate = true)  @ColumnInfo(name = "id") var id: Int=0,
             @ColumnInfo(name = "nameGroup") var nameGroup: String,
             @ColumnInfo(name = "firstName") var firstName: String,
             @ColumnInfo(name = "lastName") var lastName: String,
             @ColumnInfo(name = "nbTeam") var nbTeam: String
):Parcelable

@Entity(tableName = "eval_table", foreignKeys = arrayOf(
 ForeignKey(entity = User::class,
     parentColumns = arrayOf("id"),
     childColumns = arrayOf("idEval"),
     onDelete = ForeignKey.CASCADE)
))


data class Eval(@PrimaryKey(autoGenerate = true) @ColumnInfo(name = "idEval") var idEval: Int=0,
             @ColumnInfo(name = "note_classement")    var note_classement: String,
             @ColumnInfo(name = "note_attaque")    var note_attaque: String,
             @ColumnInfo(name = "note_passe")    var note_passe: String,
             @ColumnInfo(name = "note_afl2")    var note_afl2: String,
             @ColumnInfo(name = "note_afl3")    var note_afl3: String,
             @ColumnInfo(name = "note_sur_vingt")    var note_sur_vingt: String)

Here my dataclass to create relations

data class GruWithUser(
    var idGroup: Int,
    var nameGroupG: String,
    var id: Int,
    var nameGroup: String,
    var firstName: String,
    var lastName: String,
    var nbTeam: String
):Parcelable

and the last dataclass relation: User With Eval

@Parcelize
data class UserWithEval(
    var id: Int,
    var nameGroup: String,
    var firstName: String,
    var lastName: String,
    var nbTeam: String,
    var note_attaque: String,
    var note_passe: String,
    var note_classement: String,
    var note_afl2: String,
    var note_afl3: String,
    var note_sur_vingt: String
): Parcelable

Thanks you so much for your help

Answer 1

Issue 1

You appear to have an issue that will likely cause some frustration if not addressed.

That is a User, has it's primary key as the reference to the parent group (Gru). As such a Group could only have a single User (Student) as the primary key, for the User must be unique.

Likewise for Eval's.

So you could consider the following:-

@Entity(tableName = "groupe_table")
data class Gru (
    @PrimaryKey(autoGenerate = true)  @ColumnInfo(name = "idGroup") var idGroup: Int=0,
    @ColumnInfo(name = "nameGroupG") var nameGroupG : String
): Parcelable


@Entity(tableName = "user_table", foreignKeys = arrayOf(
    ForeignKey(entity = Gru::class,
        parentColumns = arrayOf("idGroup"),
        //childColumns = arrayOf("id"), //<<<<< REMOVED
        childColumns = ["gru_id_reference"], //<<<<< REPLACED WITH

        onDelete = ForeignKey.CASCADE)
))

@Parcelize
data class User(@PrimaryKey(autoGenerate = true)  @ColumnInfo(name = "id") var id: Int=0,
                @ColumnInfo(name = "nameGroup") var nameGroup: String,
                @ColumnInfo(name = "firstName") var firstName: String,
                @ColumnInfo(name = "lastName") var lastName: String,
                @ColumnInfo(name = "nbTeam") var nbTeam: String,
                @ColumnInfo(index = true)  //<<<<< ADDED (may be more efficient)
                var gru_id_reference: Int //<<<<< ADDED
):Parcelable

@Entity(tableName = "eval_table", foreignKeys = arrayOf(
    ForeignKey(entity = User::class,
        parentColumns = arrayOf("id"),
        childColumns = arrayOf("user_id_reference"), //<<<<< CHANGED
        onDelete = ForeignKey.CASCADE)
))

data class Eval(@PrimaryKey(autoGenerate = true) @ColumnInfo(name = "idEval") var idEval: Int=0,
                @ColumnInfo(name = "note_classement")    var note_classement: String,
                @ColumnInfo(name = "note_attaque")    var note_attaque: String,
                @ColumnInfo(name = "note_passe")    var note_passe: String,
                @ColumnInfo(name = "note_afl2")    var note_afl2: String,
                @ColumnInfo(name = "note_afl3")    var note_afl3: String,
                @ColumnInfo(name = "note_sur_vingt")    var note_sur_vingt: String,
                @ColumnInfo(index = true) var user_id_reference: Int //<<<<< ADDED
)

• See the comments
• note that there is no need to use @ColumnInfo to name a column the same name as the field/member (hence the added code doesn't, but instead uses the annotation to introduce an index on the additional column use to reference the parent).

Without going into all the other code, the following code:-

db = TheDatabase.getInstance(this)
dao = db.getAllDao()

val g1id = dao.insert(Gru(nameGroupG = "Group001"))
val g2id = dao.insert(Gru(nameGroupG = "Group002"))
val g3id = dao.insert(Gru(nameGroupG = "group003"))

val u1id = dao.insert(User(nameGroup = "Why have this here?", firstName = "Fred", lastName = "Bloggs", nbTeam = "TeamA", gru_id_reference = g1id.toInt()))
val u2id = dao.insert(User(nameGroup = "?????", firstName = "Jane", lastName = "Doe", nbTeam = "TeamX", gru_id_reference = g1id.toInt()))
val u3id = dao.insert(User(nameGroup = "?????", firstName = "Mary", lastName = "Smith", nbTeam = "TeamB", gru_id_reference = g2id.toInt()))
val u4id = dao.insert(User(nameGroup = "?????", firstName = "Tom", lastName = "Cobbely", nbTeam = "TeamC", gru_id_reference = g3id.toInt()))

var baseEval = Eval(note_classement = "CMENT_", note_attaque = "ATTQ_", note_afl2 = "AFL2_", note_afl3 = "AFL3_", note_passe = "PASSE_",  note_sur_vingt = "SV_",user_id_reference = -99)

for (i in 1..10) {
    dao.insert(
        Eval(
            note_classement = baseEval.note_classement + i,
            note_attaque = baseEval.note_classement + i,
            note_afl2 =  baseEval.note_afl2 + i,
            note_afl3 = baseEval.note_afl3 + i,
            note_passe = baseEval.note_passe + i,
            note_sur_vingt = baseEval.note_sur_vingt + i,
            user_id_reference = Random.nextInt(4) + 1
        )
    )
}

results in a database (i.e. tests the changed code) as per :-

1. The 3 Groups (Gru's/Classes) :-

   • 

2. The 4 users :-

   • 

   • Note how Fred and Jane are both in Group001

3. And the 10 Eval's spread across the 4 Users

   • 

4. A query that joins the data according to the relationships looks like:-

- here you can see that there are 3 evaluations for Group001, 2 of them for Fred and 1 for Jane etc


- (note Eval's reference a random User)

• The above data was obtained by running the code above and then using App Inspection (available in Android Studio).

Issue 2

You may well encounter subsequent issues due to both the user_table and the eval_table having a column named id. From an SQL point of view this can be overcome by qualifying the column with it's table name (see SQL used above where the tablename . column is used to disambiguate the ambiquity). However, as far as the resultant output there would still be 2 id columns. This ambiguity can be overcome using AS to rename the output column but you may then encounter issues. I would suggest ensuring that all column names are unique (so perhaps have column names userId and evalId instead of just id).