Room Android，在两个表上选择

Question

我是新来的房间，我正在努力。我想做一个简单的选择在两个名为项目和任务表。这是两个类和第三个连接它们的类:

@Entity
public class Project {
@PrimaryKey(autoGenerate = true)
private final long id;
@NonNull
private final String name;
@ColorInt
private final Integer color;
}

@Entity(foreignKeys = @ForeignKey(entity = Project.class, parentColumns = "id", childColumns = "projectId"))
public class Task {

@PrimaryKey(autoGenerate = true)
private long idTask;

// clé commune
private long projectId;

@NonNull
private String nameTask;

public long creationTimestamp;
// constructor, getters, setters
}

public class TaskWithProject {
@Embedded public Project project;
@Relation(parentColumn = "id", entityColumn = "projectId")
public Task task;
}

我在我的刀课上尝试了很多东西。以下是其中两个:

@Dao
public interface TaskDao {
(...)

// Solution 1
@Query("SELECT * FROM Task t JOIN Project p ON t.projectId = p.id")
LiveData<List<TaskWithProject>> getTaskWithProject();

// Solution 2 from https://developer.android.com/training/data-storage/room/relationships
@Transaction
@Query("SELECT * FROM Task")
LiveData<List<TaskWithProject>> getTaskWithProject();

我的数据库中填充了这些数据(在数据库检查器中没有问题):

projectDao.insertProject(new Project(0, "Projet Tartampion", 0xFFEADAD1));
projectDao.insertProject(new Project(0, "Projet Lucidia", 0xFFB4CDBA));
projectDao.insertProject(new Project(0, "Projet " + "Circus", 0xFFA3CED2));
taskDao.insertTask(new Task(0, 1, "Task 1 - Tartampion"));
taskDao.insertTask(new Task(0, 1, "Task 2 - Tartampion"));
taskDao.insertTask(new Task(0, 2, "Task 1 - Lucidia"));

我用这个函数获取数据:

public LiveData<List<TaskViewStateItem>> getAllTasks() {
return Transformations.map(repository.getTaskWithProject(), tasks -> {
List<TaskViewStateItem> liststateitems = new ArrayList<>();
for (TaskWithProject t : tasks) {
Log.i(TAG, "getAllTasks: "+ t.toString());
liststateitems.add(new TaskViewStateItem(t.task.getIdTask(), t.task.getNameTask(), t.project.getName(), t.project.getColor(), t.task.getCreationTimestamp()));
}
}
return liststateitems;
});
}

根据上面的解决方案1，我有3个条目，但有一个重复的。

I/Log ViewModel: getAllTasks: TaskWithProject{project=Project{id=1, name='Projet Tartampion', color=-1385775}, task=Task{idTask=2, projectId=1, nameTask='Task 2 - Tartampion', creationTimestamp=1634723235}}
I/Log ViewModel: getAllTasks: TaskWithProject{project=Project{id=1, name='Projet Tartampion', color=-1385775}, task=Task{idTask=2, projectId=1, nameTask='Task 2 - Tartampion', creationTimestamp=1634723235}}
getAllTasks: TaskWithProject{project=Project{id=2, name='Projet Lucidia', color=-4928070}, task=Task{idTask=3, projectId=2, nameTask='Task 1 - Lucidia', creationTimestamp=1634723235}}

解决方案2不编译

error: Not sure how to convert a Cursor to this method's return type (com.cleanup.todoc.model.TaskWithProject).
LiveData<List<TaskWithProject>> getTaskWithProject();
^

我不知道我做错了什么。所以非常感谢你给我的任何帮助

Answer 1

由于您的TaskWithProject类嵌入了项目，因此它期望获得项目的列，然后从中构建任务(s)，因此当您使用SELECT * FROM Task时，它没有相关的列(与连接它确实但可能不像预期的那样工作)。

• TaskWithProject是ProjectWithTask.

所以使用SELECT * FROM project或使用POJO与任务嵌入和项目作为@关系(与SELECT * FROM task)。