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 th

Answer 1

Currently there is no native solution to this problem. I have created this https://issuetracker.google.com/issues/62848977 on Google's issue tracker and the Architecture Components Team said they will adding a native solution in or after v1.0 of Room library.

Temporary Workaround:

Meanwhile you can use the solution mentioned by tknell.

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

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

    petDao.insertAll(pets);
}

Answer 2

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.

Answer 3

As Room does not manage the Relations of the entities, you have to set the userId on each pet yourself and save them. As long as there are not too many pets at once, I'd use an insertAll method to keep it short.

@Dao
public interface PetDao {
    @Insert
    void insertAll(List<Pet> pets);
}

I don't think there's any better way at the moment.

To make the handling easier, I'd use an abstraction in the layer above the DAOs:

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

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

    petDao.insertAll(pets);
}

Answer 4

Now at v2.1.0 Room seems to be not suitable for models with nested relations. It needed lots of boilerplate code to maintain them. E.g. manual insert of lists, creating and mapping local IDs.

This relations-mapping operations are done out of box by Requery https://github.com/requery/requery Additionaly it does not have issues with inserting Enums and have some converters for other complex types like URI.

Answer 5

There is no native solution till any update in Room Library but you can do this by a trick. Find below mentioned.

1. Just Create a User with Pets (Ignore pets). Add getter and setter. Notice that we have to set our Id's manually later and can't use autogenerate.

   @Entity
   public class User {
         @PrimaryKey
         public int id; 
   
         @Ignore
         private List<Pet> petList;
   }
   

2. Create a Pet.

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

3. The UserDao should be an abstract class instead of an Interface. Then finally in your UserDao.

   @Insert
   public abstract void insertUser(User user);
   
   @Insert
   public abstract void insertPetList(List<Pet> pets);
   
   @Query("SELECT * FROM User WHERE id =:id")
   public abstract User getUser(int id);
   
   @Query("SELECT * FROM Pet WHERE userId =:userId")
   public abstract List<Pet> getPetList(int userId);
   
   public void insertUserWithPet(User user) {
       List<Pet> pets = user.getPetList();
       for (int i = 0; i < pets.size(); i++) {
           pets.get(i).setUserId(user.getId());
       }
       insertPetList(pets);
       insertUser(user);
   }
   
   public User getUserWithPets(int id) {
       User user = getUser(id);
       List<Pet> pets = getPetList(id);
       user.setPetList(pets);
       return user;
   }
   

Your problem can be solved by this without creating UserWithPets POJO.

Answer 6

I managed to insert it properly with a relatively simple workaround. Here are my entities:

   @Entity
public class Recipe {
    @PrimaryKey(autoGenerate = true)
    public long id;
    public String name;
    public String description;
    public String imageUrl;
    public int addedOn;
   }


  @Entity
public class Ingredient {
   @PrimaryKey(autoGenerate = true)
   public long id;
   public long recipeId; 
   public String name;
   public String quantity;
  }

public class RecipeWithIngredients {
   @Embedded
   public  Recipe recipe;
   @Relation(parentColumn = "id",entityColumn = "recipeId",entity = Ingredient.class)
   public List<Ingredient> ingredients;

I am using autoGenerate for auto-increment value(long is used with a purpoes). Here is my solution:

  @Dao
public abstract class RecipeDao {

  public  void insert(RecipeWithIngredients recipeWithIngredients){
    long id=insertRecipe(recipeWithIngredients.getRecipe());
    recipeWithIngredients.getIngredients().forEach(i->i.setRecipeId(id));
    insertAll(recipeWithIngredients.getIngredients());
  }

public void delete(RecipeWithIngredients recipeWithIngredients){
    delete(recipeWithIngredients.getRecipe(),recipeWithIngredients.getIngredients());
  }

 @Insert
 abstract  void insertAll(List<Ingredient> ingredients);
 @Insert
 abstract long insertRecipe(Recipe recipe); //return type is the key here.

 @Transaction
 @Delete
 abstract void delete(Recipe recipe,List<Ingredient> ingredients);

 @Transaction
 @Query("SELECT * FROM Recipe")
 public abstract List<RecipeWithIngredients> loadAll();
 }

I had problem linking the entities, auto generate produced "recipeId=0" all the time. Inserting the recipe entity firstly fixed it for me.