Android房间持久性库 - 多对多关系

时间:2018-04-24 15:57:10

标签: database sqlite android-room android-architecture-components

我正试图在Room Persistence Library中实现多对多关系。我正在尝试创建Notes App with Tags

这个想法是:

  • 笔记会有多个标签。

  • 标签会有多个音符。

  • 在RecyclerView中显示所有注释以及标记。

为实现这一目标,我创建了两个模型 Note.java Tag.java TagJoin 模型来存储关系b / w笔记和标签。使用@Relation注释很容易实现一对一

以下是我的模特

@Entity(tableName = "notes")
public class Note {
    @PrimaryKey
    @NonNull
    public final String id;

    @ColumnInfo(name = "note")
    public String note;

    @Ignore
    public List<Tag> tags;

    @Ignore
    public Note(String note) {
        this(UUID.randomUUID().toString(), note);
    }

    public Note(String id, String note) {
        this.id = id;
        this.note = note;
    }

    public String getId() {
        return id;
    }

    public String getNote() {
        return note;
    }

    public void setNote(String note) {
        this.note = note;
    }

    public List<Tag> getTags() {
        return tags;
    }

    public void setTags(List<Tag> tags) {
        this.tags = tags;
    }

    @Entity(tableName = "note_tag_join",
            primaryKeys = {"noteId", "tagId"},
            foreignKeys = {
                    @ForeignKey(
                            entity = Note.class,
                            parentColumns = "id",
                            childColumns = "noteId",
                            onDelete = CASCADE),
                    @ForeignKey(
                            entity = Tag.class,
                            parentColumns = "id",
                            childColumns = "tagId",
                            onDelete = CASCADE)},
            indices = {
                    @Index(value = "noteId"),
                    @Index(value = "tagId")
            }
    )

    public static class TagJoin {
        @NonNull
        public final String noteId;
        @NonNull
        public final String tagId;

        public TagJoin(String noteId, String tagId) {
            this.noteId = noteId;
            this.tagId = tagId;
        }
    }
}

标签型号:

@Entity(tableName = "tags", indices = {@Index(value = "name", unique = true)})
public class Tag {
    @PrimaryKey
    @NonNull
    public String id;

    @ColumnInfo(name = "name")
    public String name;

    @Ignore
    public Tag(String name) {
        this(UUID.randomUUID().toString(), name);
    }

    public Tag(String id, String name) {
        this.id = id;
        this.name = name;
    }


    public String getId() {
        return id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

Notes Dao:

@Dao
public interface NoteDao {

    @Query("SELECT * FROM notes ORDER BY id DESC")
    LiveData<List<Note>> getAllNotes();

    @Insert
    long insert(Note note);

    @Update
    void update(Note note);

    @Delete
    void delete(Note note);

    @Query("DELETE FROM notes")
    void deleteAll();

    @Query("SELECT COUNT(*) FROM notes")
    int getNotesCount();

    @Query("SELECT notes.* FROM notes\n" +
            "INNER JOIN note_tag_join ON notes.id=note_tag_join.noteId\n" +
            "WHERE note_tag_join.tagId=:tagId")
    List<Note> getAllNotesOfTag(String tagId);

    @Insert
    void insert(Note.TagJoin... joins);

    @Delete
    void delete(Note.TagJoin... joins);
}

到目前为止一切都很好。现在我想在RecyclerView中显示Notes,但我找不到一种方法来同时获取所有Notes以及Tags。 一种方法是,在onBindViewHolder方法中获取每个音符的标签,我认为这是错误的,因为我们必须在每次显示行时查询数据库

请提供建议。

PS:我已经遵​​循了本文提供的代码 https://commonsware.com/AndroidArch/previews/mn-relations-in-room

0 个答案:

没有答案