表没有创建sqlite android

我正在制作一个应用程序,我想保存用户的联系人详细信息.但每当我尝试插入或选择一些值我得到一个错误.

DataBasae代码

public class ContactsDatabase extends SQLiteOpenHelper {

    private static final int dbVersion = 1;
    private static final String dbName = "HSsuraksha";
    private static final String tableName = "contactsDetails";
    private static final String contactId = "contactId";
    private static final String groupId = "groupId";
    private static final String groupGroupId = "groupId";
    private static final String groupTable = "groupDetails";
    private static final String contactName = "contactName";
    private static final String contactNumber = "contactNumber";
    private static final String createTable = "Create Table " + tableName + "(" + contactId + " Integer Primary Key AutoIncrement," + groupId + " Text," + contactName + " Text," + contactNumber + " Text" + ");";

    public ContactsDatabase(Context context) {
        super(context,dbName,null,dbVersion);
    }

    @Override
    public void onCreate(SQLiteDatabase sqLiteDatabase) {

        sqLiteDatabase.execSQL(createTable);

    }

    @Override
    public void onUpgrade(SQLiteDatabase sqLiteDatabase,int i,int i2) {

    }


    public void insertContacts(ContactModel contactModel,String id,ArrayList<ContactModel> contactModelArrayList) {
        SQLiteDatabase database = getWritableDatabase();
        database.beginTransaction();
        ContentValues contentValues = new ContentValues();
        for (int i = 0; i < contactModelArrayList.size(); i++) {

            contentValues.put(contactName,contactModelArrayList.get(i).getContactName());
            contentValues.put(contactNumber,contactModelArrayList.get(i).getContactNumber());
            contentValues.put(groupId,id);
            if (contentValues != null) {
                Long value = database.insert(tableName,id,contentValues);
            }

        }


        database.setTransactionSuccessful();
        database.close();
    }

    public void selectContacts(String id) {
        String query = "Select * From " + tableName + " where " + groupId + "=?";
        SQLiteDatabase database = getWritableDatabase();
        Cursor cursor = database.rawQuery(query,new String[]{id});
        while (cursor.moveToNext()) {
            cursor.getString(cursor.getColumnIndexOrThrow(contactName));
        }
        cursor.close();
        database.close();

    }
}

logcat的

08-01 18:53:57.820      672-672/example.com.pocketdocs E/SQLiteLog﹕ (1) no such table: contactsDetails
08-01 18:53:57.830      672-672/example.com.pocketdocs E/SQLiteDatabase﹕ Error inserting groupId=2 contactNumber=[phone] contactName=Aaaaaaa
    android.database.sqlite.SQLiteException: no such table: contactsDetails (code 1):,while compiling: INSERT INTO contactsDetails(groupId,contactNumber,contactName) VALUES (?,?,?)
            at android.database.sqlite.SQLiteConnection.nativePrepareStatement(Native Method)
            at android.database.sqlite.SQLiteConnection.acquirePreparedStatement(SQLiteConnection.java:893)
            at android.database.sqlite.SQLiteConnection.prepare(SQLiteConnection.java:504)
            at android.database.sqlite.SQLiteSession.prepare(SQLiteSession.java:588)
            at android.database.sqlite.SQLiteProgram.<init>(SQLiteProgram.java:58)
            at android.database.sqlite.SQLiteStatement.<init>(SQLiteStatement.java:31)
            at android.database.sqlite.SQLiteDatabase.insertWithOnConflict(SQLiteDatabase.java:1475)
            at android.database.sqlite.SQLiteDatabase.insert(SQLiteDatabase.java:1347)
            at example.com.pocketdocs.DataBase.ContactsDatabase.insertContacts(ContactsDatabase.java:54)
            at example.com.pocketdocs.Group.CreateNewGroup.onClick(CreateNewGroup.java:104)
            at android.view.View.performClick(View.java:4147)
            at android.view.View$PerformClick.run(View.java:17161)
            at android.os.Handler.handleCallback(Handler.java:615)
            at android.os.Handler.dispatchMessage(Handler.java:92)
            at android.os.Looper.loop(Looper.java:213)
            at android.app.ActivityThread.main(ActivityThread.java:4787)
            at java.lang.reflect.Method.invokeNative(Native Method)
            at java.lang.reflect.Method.invoke(Method.java:511)
            at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:789)
            at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:556)
            at dalvik.system.NativeStart.main(Native Method)

我尝试解决错误但没有成功.我的代码是什么问题???

解决方法

i have another table groupInfo with same database name,so it that the problem??

这是一个问题.这是发生的事情：

>访问具有相同数据库文件的第一个sqlite打开助手.如果数据库文件不存在,则调用onCreate()回调,以便您可以设置数据库文件.
>访问具有相同数据库文件的其他sqlite打开助手.具有给定名称的数据库文件已存在且版本正确,因此不会调用onCreate()或onUpgrade().而是刚刚打开文件.

解决方案：每个数据库文件只使用一个sqlite打开助手.将两个表的创建语句放在同一帮助器onCreate()方法中.

还要卸载您的应用程序,以便删除仅包含其他表的旧数据库文件.

请参阅链接的问题When is SQLiteOpenHelper onCreate() / onUpgrade() run?以了解有关sqlite open helper生命周期回调的更多信息.