Java android ROOM获取具有详细信息的对象

Question

我有这个:

    @Entity(tableName = "director"/*,
            indices = {@Index(value = "full_name", unique = true)}*/)
    public class Director {
        @PrimaryKey(autoGenerate = true)
        @ColumnInfo(name = "did")
        public int id;
        @ColumnInfo(name = "full_name")
        @NonNull
        public String fullName;


@Entity(tableName = "movie",
        foreignKeys = @ForeignKey(entity = Director.class,
                parentColumns = "did",
                childColumns = "directorId",
                onDelete = ForeignKey.CASCADE),
        indices = {@Index("title"), @Index("directorId")})
public class Movie {
    @PrimaryKey(autoGenerate = true)
    @ColumnInfo(name = "mid")
    public int id;
    @ColumnInfo(name = "title")
    @NonNull
    public String title;
    @ColumnInfo(name = "directorId")
    public int directorId;

Amd 当我获取所有电影时，我会获取导演 ID。我应该怎么做才能生成这个:

public class MovieWithDirector {

    Movie movie;
    Director director;
}

我可以获得一个用户，然后再获得 directerByID，但我现在不知道这是不是最好的方法

Answer 1

我会说执行此操作的更好方法是使用 join 编写查询:

SELECT  movie.title, director.fullname 
FROM director 
INNER JOIN movie 
ON movie.directorId=director.id 
WHERE director.id= :passed_value

当然，您需要自定义类来存储查询结果。如果您检查此 link 中的示例，并结合上面的查询，然后你就可以解决你的问题。