Question

如何为关系添加条件？

例如，我们有对象Pet

    @Entity
     public class Pet {
         @ PrimaryKey
         int id;
         int userId;
         String name;
         String type;
         // other fields
     }

和对象用户

public class User {
     int id;
     // other fields
 }

为了吸引使用宠物的用户，我们制作了对象

public class UserAllPets {
   @Embedded
   public User user;
   @Relation(parentColumn = "id", entityColumn = "userId", entity = Pet.class)
   public List<PetNameAndId> pets;
 }

如何通过类型获取宠物用户？只有狗或只有猫

这是dao类：

@Dao
public abstract class UserDao { 

   @Query("SELECT * FROM `users`")
   public abstract UserAllPets getUserWithPets();
}

Answer 1

只需在您的所有者模型中创建一个包装器，使用Embedded并在DAO对象中查询JOIN。

例如：User有很多Pet个。我们会发现所有Pet，按User的ID过滤，Pet的年龄大于9：

@Entity(tableName = "USERS")
class User {
    var _ID: Long? = null
}

@Entity(tableName = "PETS")
class Pet {
    var _ID: Long? = null
    var _USER_ID: Long? = null
}

// Merged class extend from `User`
class UserPets : User {
    @Embedded(prefix = "PETS_")
    var pets: List<Pet> = emptyList()
}

在UserDao

中

@Dao
interface UserDao {
    @Query("SELECT USERS.*, PETS._ID AS PETS__ID, PETS._USER_ID AS PETS__USER_ID FROM USERS JOIN PETS ON PETS._USER_ID = USERS._ID WHERE PETS.AGE >= 9 GROUP BY USERS._ID")
    fun getUserPets(): LiveData<List<UserPets>>
}

Answer 2

在您的DAO中，您可以指定您想要的任何查询。所以你可以这样做：

@Query("select * from pet_table where userId = :userId and type = :type")
List<Pet> getPetsByUserAndType(int userId, String type)

或类似的东西，我不确定你的桌名是什么。这有意义吗？

Answer 3

作为最后一个选项，您可以自己编写所有查询并将它们组合在一个抽象 DAO 类中：

@Dao
public abstract class UserDao { 

    @Transaction
    @Query("SELECT * FROM `User`")
    abstract List<UserWithPets> getUsers();

    @Transaction
    @Query("SELECT * FROM `Pet` WHERE userId = :userId AND type = :type")
    abstract List<Pet> getPetsByUser(int userId, String type);

    public List<UserWithPets> getUsersWithDogs() {
        List<UserWithPets> users = getUsers();
        for (User user: users) {
            List<Pet> pets = getPetsByUser(user.id, "dog");
            user.pets = pets;
        }
        return users;
    }
}

public class UserWithPets {
   @Embedded
   public User user;
   
   @Ignore
   public List<Pet> pets;
   
   // Other stuff
   // @Relation(parentColumn = "id", entityColumn = "parentId")
   // public List<Child> children;
 }

房间与条件的关系

3 个答案: