房间与条件的关系

问题描述

如何为关系添加条件?

例如，我们有对象Pet

For example, we have object 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
 }

为了吸引宠物，我们制作了对象

For getting user with pets we make object

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

如何按类型吸引宠物用户?只有狗或只有猫

How is possible to get user with pets by type? Only dogs or only cats

这是道课:

@Dao
public abstract class UserDao { 

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

推荐答案

只需使用Embedded从所有者模型中创建一个包装器，然后在DAO对象中查询JOIN.

Just create a wrapper from your owner model, using Embedded and query JOIN in your DAO object.

例如:User有许多Pet.我们将找到所有Pet，并按User的ID和Pet的年龄大于9的年龄进行过滤:

For example: User have many Pets. We will find all Pet, filter by User's id and Pet's age greater equal than 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()
}

在您的UserDao

@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>>
}

突出显示的SQL语法:

SQL Syntax highlighted:

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

这篇关于房间与条件的关系的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！