Android Room: 使用 Room 插入关系实体

我在 Room 中使用 关系添加了一对多的关系。 我参考 这篇文章编写了以下房间关系代码。

这篇文章告诉我们如何从数据库中读取值，但是将实体存储到数据库中会导致 userId为空，这意味着这两个表之间没有关系。

我不确定什么是 insert a User和 List of Pet进入数据库的理想方法，同时具有 userId值。

1)用户实体:

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

2)宠物实体:

@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;
}

现在，为了从 DB 获取记录，我们使用以下 DAO:

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


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

剪辑

我已经在问题跟踪器上创建了这个问题 https://issuetracker.google.com/issues/62848977。希望他们会做一些关于它的事情。

64526 次浏览
小开

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);
}
小开

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);
}
小开

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.

小开

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.

小开

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.

小开

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.


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