Android 房间 SQLite_ERROR 没有这样的房间 table
Android Room SQLite_ERROR no such table
我正在尝试使用 Android Room and after following this tutorial 我在尝试构建应用程序时遇到以下错误:
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。进行更改后,我清理了项目并确保它已从设备上完全卸载。
在我的 Activity
中,我使用以下行初始化 onCreate
中的内容:
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;
}
}
许可项目
@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='" + width + '\'' +
", height='" + height + '\'' +
", content='" + content + '\'' +
", type='" + type + '\'' +
'}';
}
}
PermitItemDao
@Dao
public interface PermitItemDao {
@Insert(onConflict = OnConflictStrategy.REPLACE)
long addPermitItem(PermitItem permitItem);
@Query("select * from " + TABLE_ITEMS)
ArrayList<PermitItem> 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();
}
房间名称table与其关联的实体名称相同。在你的 DAO 中,TABLE_ITEMS
需要是 PermitItem
,因为你的实体是 PermitItem
。或者,将 tableName
属性 添加到 @Entity
注释中,以告诉 Room 一些其他名称用于 table.
此错误的另一个原因可能是实体未在 AppDatabase.java 文件中列出:
@Database(entities = {XEntity.class, YEntity.class, ZEntity.class},
version = 1, exportSchema = true)
确保数据库文件夹中有最新的 db 文件,如果导出架构,请确保 app\schemas 下的 .json 架构文件已正确更新。
就我而言,问题的原因是使用了错误的table名称。我为模型 class 选择的 table 名称与 class 名称不同。我在查询数据访问对象接口时错误地使用了 class 名称。
@Query("SELECT * FROM tablename")
检查您的 table 姓名,确保它们与注释到模型 class 中的 table 姓名匹配。
希望对您有所帮助。编码愉快!
如果您刚刚添加了一个新的 table,只需更新您的数据库 class(那个 class 扩展 RoomDatabase() class)并更新实体注释
@Database(entities = [User::class, NewTableHere::class], version = 1)
abstract class AppDatabase : RoomDatabase() {
希望它可以节省您寻找答案的时间,祝您编码愉快。
看文章对我帮助很大。
我有这个 errorDatabase(entities = {Folder.class}, version = 1, exportSchema = false)
只需添加我的其他 class
@Database(entities = {Folder.class, Urls.class}, version = 1, exportSchema = false)
我的问题是TABLE NAME
:|||
我把我的 table 名字写错了:(((
我正在尝试使用 Android Room and after following this tutorial 我在尝试构建应用程序时遇到以下错误:
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。进行更改后,我清理了项目并确保它已从设备上完全卸载。
在我的 Activity
中,我使用以下行初始化 onCreate
中的内容:
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;
}
}
许可项目
@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='" + width + '\'' +
", height='" + height + '\'' +
", content='" + content + '\'' +
", type='" + type + '\'' +
'}';
}
}
PermitItemDao
@Dao
public interface PermitItemDao {
@Insert(onConflict = OnConflictStrategy.REPLACE)
long addPermitItem(PermitItem permitItem);
@Query("select * from " + TABLE_ITEMS)
ArrayList<PermitItem> 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();
}
房间名称table与其关联的实体名称相同。在你的 DAO 中,TABLE_ITEMS
需要是 PermitItem
,因为你的实体是 PermitItem
。或者,将 tableName
属性 添加到 @Entity
注释中,以告诉 Room 一些其他名称用于 table.
此错误的另一个原因可能是实体未在 AppDatabase.java 文件中列出:
@Database(entities = {XEntity.class, YEntity.class, ZEntity.class},
version = 1, exportSchema = true)
确保数据库文件夹中有最新的 db 文件,如果导出架构,请确保 app\schemas 下的 .json 架构文件已正确更新。
就我而言,问题的原因是使用了错误的table名称。我为模型 class 选择的 table 名称与 class 名称不同。我在查询数据访问对象接口时错误地使用了 class 名称。
@Query("SELECT * FROM tablename")
检查您的 table 姓名,确保它们与注释到模型 class 中的 table 姓名匹配。
希望对您有所帮助。编码愉快!
如果您刚刚添加了一个新的 table,只需更新您的数据库 class(那个 class 扩展 RoomDatabase() class)并更新实体注释
@Database(entities = [User::class, NewTableHere::class], version = 1)
abstract class AppDatabase : RoomDatabase() {
希望它可以节省您寻找答案的时间,祝您编码愉快。
看文章对我帮助很大。 我有这个 errorDatabase(entities = {Folder.class}, version = 1, exportSchema = false)
只需添加我的其他 class @Database(entities = {Folder.class, Urls.class}, version = 1, exportSchema = false)
我的问题是TABLE NAME
:|||
我把我的 table 名字写错了:(((