Android Room: Insert relation entities using Room

Question

I've added one to many relationship in Room using Relation. I referred to this post to write the following code for relation in Room.

The post tells how to read the values from the database but storing the entities into the database resulted in userId to be empty which means there is no relation between the 2 tables.

I'm not sure what is the ideal way to insert a User and List of Pet into the database while having userId value.

1) User Entity:

@Entity
public class User {
    @PrimaryKey
    public int id; // User id
}

2) Pet Entity:

@Entity
public class Pet {
    @PrimaryKey
    public int id;     // Pet id
    public int userId; // User id
    public String name;
}

3) UserWithPets POJO:

// Note: No annotation required at this class definition.
public class UserWithPets {
   @Embedded
   public User user;

   @Relation(parentColumn = "id", entityColumn = "userId", entity = Pet.class)
   public List<Pet> pets;
}

Now to fetch the records from DB we use the following DAO:

@Dao
public interface UserDao {
    @Insert
    fun insertUser(user: User)

    @Query("SELECT * FROM User")
    public List<UserWithPets> loadUsersWithPets();
}

---

EDIT

I have created this issue https://issuetracker.google.com/issues/62848977 on the issue tracker. Hopefully they will do something regarding it.

Answer 1

You can do this by changing your Dao from an interface to an abstract class.

@Dao
public abstract class UserDao {

    public void insertPetsForUser(User user, List<Pet> pets){

        for(Pet pet : pets){
            pet.setUserId(user.getId());
        }

        _insertAll(pets);
    }

    @Insert
    abstract void _insertAll(List<Pet> pets);  //this could go in a PetDao instead...

    @Insert
    public abstract void insertUser(User user);

    @Query("SELECT * FROM User")
    abstract List<UserWithPets> loadUsersWithPets();
}

You can also go further by having a User object have an @Ignored List<Pet> pets

@Entity
public class User {
    @PrimaryKey
    public int id; // User id

    @Ignored
    public List<Pet> pets
}

and then the Dao can map UserWithPets to User:

public List<User> getUsers() {
    List<UserWithPets> usersWithPets = loadUserWithPets();
    List<User> users = new ArrayList<User>(usersWithPets.size())
    for(UserWithPets userWithPets: usersWithPets) {
        userWithPets.user.pets = userWithPets.pets;
        users.add(userWithPets.user);
    }
    return users;
}

This leaves you with the full Dao:

@Dao
public abstract class UserDao {

    public void insertAll(List<User> users) {
        for(User user:users) {
           if(user.pets != null) {
               insertPetsForUser(user, user.pets);
           }
        }
        _insertAll(users);
    }

    private void insertPetsForUser(User user, List<Pet> pets){

        for(Pet pet : pets){
            pet.setUserId(user.getId());
        }

        _insertAll(pets);
    }

    public List<User> getUsersWithPetsEagerlyLoaded() {
        List<UserWithPets> usersWithPets = _loadUsersWithPets();
        List<User> users = new ArrayList<User>(usersWithPets.size())
        for(UserWithPets userWithPets: usersWithPets) {
            userWithPets.user.pets = userWithPets.pets;
            users.add(userWithPets.user);
        }
        return users;
    }


    //package private methods so that wrapper methods are used, Room allows for this, but not private methods, hence the underscores to put people off using them :)
    @Insert
    abstract void _insertAll(List<Pet> pets);

    @Insert
    abstract void _insertAll(List<User> users);

    @Query("SELECT * FROM User")
    abstract List<UserWithPets> _loadUsersWithPets();
}

You may want to have the insertAll(List<Pet>) and insertPetsForUser(User, List<Pet>) methods in a PetDAO instead... how you partition your DAOs is up to you! :)

Anyway, it's just another option. Wrapping your DAOs in DataSource objects also works.