作者:飞飞飞070801 | 来源:互联网 | 2023-01-23 19:48
我正在尝试使用Android Room,在完成本教程之后,当我尝试构建应用程序时,我收到以下错误:
Error:(23, 27) error: There is a problem with the query: [SQLITE_ERROR] SQL error or missing database (no such table: screen_items)
这个名字很好,应该存在.在进行更改后,我清理了项目并确保从设备中完全卸载了该项目.
在我用这条线Activity
初始化东西onCreate
:
db = AppDatabase.getDatabase(getApplicationContext());
这是我的代码:
AppDatabase
@Database(entities = {PermitItem.class}, version = 1, exportSchema = false)
public abstract class AppDatabase extends RoomDatabase {
public static String DATABASE_NAME = "my_database";
public final static String TABLE_ITEMS = "screen_items";
private static AppDatabase INSTANCE;
public abstract PermitItemDao permitItemModel();
public static AppDatabase getDatabase(Context context) {
if (INSTANCE == null) {
INSTANCE = Room.databaseBuilder(context, AppDatabase.class, DATABASE_NAME).allowMainThreadQueries().build();
}
return INSTANCE;
}
public static void destroyInstance() {
INSTANCE = null;
}
}
PermitItem
@Entity
public class PermitItem {
@PrimaryKey(autoGenerate = true)
public final int id;
private String posX, posY, width, height, content, type;
public PermitItem(int id, String posX, String posY, String width, String height, String content, String type) {
this.id = id;
this.posX = posX;
this.posY = posY;
this.width = width;
this.height = height;
this.cOntent= content;
this.type = type;
}
public static PermitItemBuilder builder(){
return new PermitItemBuilder();
}
public static class PermitItemBuilder{
int id;
String posX, posY, width, height, content, type;
public PermitItemBuilder setId(int id) {
this.id = id;
return this;
}
public PermitItemBuilder setPosX(String posX) {
this.posX = posX;
return this;
}
public PermitItemBuilder setPosY(String posY) {
this.posY = posY;
return this;
}
public PermitItemBuilder setWidth(String width) {
this.width = width;
return this;
}
public PermitItemBuilder setHeight(String height) {
this.height = height;
return this;
}
public PermitItemBuilder setContent(String content) {
this.cOntent= content;
return this;
}
public PermitItemBuilder setType(String type) {
this.type = type;
return this;
}
public PermitItem build() {
return new PermitItem(id, posX, posY, width, height, content, type);
}
}
public long getId() {
return id;
}
public String getPosX() {
return posX;
}
public void setPosX(String posX) {
this.posX = posX;
}
public String getPosY() {
return posY;
}
public void setPosY(String posY) {
this.posY = posY;
}
public String getWidth() {
return width;
}
public void setWidth(String width) {
this.width = width;
}
public String getHeight() {
return height;
}
public void setHeight(String height) {
this.height = height;
}
public String getContent() {
return content;
}
public void setContent(String content) {
this.cOntent= content;
}
public String getType() {
return type;
}
public void setType(String type) {
this.type = type;
}
@Override
public String toString() {
return "PermitItem{" +
"id=" + id +
", posX='" + posX + '\'' +
", posY='" + posY + '\'' +
", + width + '\'' +
", + height + '\'' +
", cOntent='" + content + '\'' +
", type='" + type + '\'' +
'}';
}
}
PermitItemDao
@Dao
public interface PermitItemDao {
@Insert(OnConflict= OnConflictStrategy.REPLACE)
long addPermitItem(PermitItem permitItem);
@Query("select * from " + TABLE_ITEMS)
ArrayList getAllPermitItems();
@Query("select * from " + TABLE_ITEMS + " where id = :id")
PermitItem getPermitItemById(int id);
@Update(OnConflict= OnConflictStrategy.REPLACE)
void updatePermitItem(PermitItem permitItem);
@Query("delete from " + TABLE_ITEMS)
void removeAllPermitItems();
}
live-love..
70
此错误的另一个原因可能是AppDatabase.java文件中未列出实体:
@Database(entities = {XEntity.class, YEntity.class, ZEntity.class},
version = 1, exportSchema = true)
确保在databases文件夹中有最新的db文件,如果导出模式,请确保正在更新app\schemas下的.json模式文件.
1> live-love..:
此错误的另一个原因可能是AppDatabase.java文件中未列出实体:
@Database(entities = {XEntity.class, YEntity.class, ZEntity.class},
version = 1, exportSchema = true)
确保在databases文件夹中有最新的db文件,如果导出模式,请确保正在更新app\schemas下的.json模式文件.
2> CommonsWare..:
房间名称表与其关联实体相同.在您的DAO中,TABLE_ITEMS
需要是PermitItem
,因为您的实体是PermitItem
.或者,将tableName
属性添加到@Entity
注释,以告诉Room用于表的其他名称.