我编写了命令来在onCreate()
类的DatabaseManager
类中创建表,该类扩展了SQLiteOpenHelper
类。打开数据库时,首次调用onCreate()
方法。但是仍然存在一个错误,即表corporate_boxes已经存在。
开放方法
public DatabaseManager open() {
ourHelper = new DatabaseManager(context);
ourDatabase = ourHelper.getWritableDatabase();
return this;
}
Oncreate Method
@Override
public void onCreate(SQLiteDatabase db) {
db.execSQL(DATABASE_CREATE_TABLE2);
}
DATABASE_CREATE_TABLE2
是一个字符串,其定义如下,
private static final String DATABASE_CREATE_TABLE2 = "CREATE TABLE "
+ DATABASE_TABLE2 + " ( " + KEY_ROWID
+ " INTEGER PRIMARY KEY AUTOINCREMENT, " + KEY_PRODUCT
+ " TEXT NOT NULL, " + KEY_PRICE + " TEXT NOT NULL," + KEY_IMAGEID
+ " TEXT NOT NULL );";
我也试过了CREATE TABLE IF NOT EXISTS
,但它仍然给出了同样的错误。
这是正在显示的堆栈跟踪。
02-21 07:59:38.296: E/AndroidRuntime(1984): FATAL EXCEPTION: main
02-21 07:59:38.296: E/AndroidRuntime(1984): android.database.sqlite.SQLiteException: table corporate_boxes already exists (code 1): , while compiling: CREATE TABLE corporate_boxes ( _id INTEGER PRIMARY KEY AUTOINCREMENT, _name TEXT NOT NULL, _price TEXT NOT NULL,_imageid TEXT NOT NULL );
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.database.sqlite.SQLiteConnection.nativePrepareStatement(Native Method)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.database.sqlite.SQLiteConnection.acquirePreparedStatement(SQLiteConnection.java:882)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.database.sqlite.SQLiteConnection.prepare(SQLiteConnection.java:493)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.database.sqlite.SQLiteSession.prepare(SQLiteSession.java:588)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.database.sqlite.SQLiteProgram.<init>(SQLiteProgram.java:58)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.database.sqlite.SQLiteStatement.<init>(SQLiteStatement.java:31)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.database.sqlite.SQLiteDatabase.executeSql(SQLiteDatabase.java:1663)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.database.sqlite.SQLiteDatabase.execSQL(SQLiteDatabase.java:1594)
02-21 07:59:38.296: E/AndroidRuntime(1984): at appistic.services.tricouschocolates.categories.DatabaseManager.onCreate(DatabaseManager.java:199)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.database.sqlite.SQLiteOpenHelper.getDatabaseLocked(SQLiteOpenHelper.java:252)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.database.sqlite.SQLiteOpenHelper.getWritableDatabase(SQLiteOpenHelper.java:164)
02-21 07:59:38.296: E/AndroidRuntime(1984): at appistic.services.tricouschocolates.categories.DatabaseManager.open(DatabaseManager.java:116)
02-21 07:59:38.296: E/AndroidRuntime(1984): at appistic.services.tricouschocolates.Products.onItemSelected(Products.java:51)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.widget.AdapterView.fireOnSelected(AdapterView.java:892)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.widget.AdapterView.access$200(AdapterView.java:49)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.widget.AdapterView$SelectionNotifier.run(AdapterView.java:860)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.os.Handler.handleCallback(Handler.java:725)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.os.Handler.dispatchMessage(Handler.java:92)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.os.Looper.loop(Looper.java:137)
02-21 07:59:38.296: E/AndroidRuntime(1984): at android.app.ActivityThread.main(ActivityThread.java:5039)
02-21 07:59:38.296: E/AndroidRuntime(1984): at java.lang.reflect.Method.invokeNative(Native Method)
02-21 07:59:38.296: E/AndroidRuntime(1984): at java.lang.reflect.Method.invoke(Method.java:511)
02-21 07:59:38.296: E/AndroidRuntime(1984): at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:793)
02-21 07:59:38.296: E/AndroidRuntime(1984): at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:560)
02-21 07:59:38.296: E/AndroidRuntime(1984): at dalvik.system.NativeStart.main(Native Method)
答案 0 :(得分:0)
您的问题很可能是 ;
字符,因为其他代码看起来是正确的。这个角色可能会导致神秘的问题,所以尝试从DML语句中删除它,它应该有效。
我无法理解为什么在查询中使用if not exists
时也会抛出此错误,因为onCreate()
我只在第一次使用db的某个'action'时调用,然后db始终只是开放阅读或写作或两者兼而有之。
答案 1 :(得分:0)
更改数据库版本号,尝试使用它时可以使用。