我正在制作一个应用程序，我想在其中保存用户联系人详细信息。但是每当我尝试插入或选择一些值时，我都会收到错误消息。

数据库代码

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();

    }
}

洛格卡特

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=+91 97 69 512114 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。我的代码有什么问题？？？

最佳答案

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

这是一个问题。这是发生了什么:

• 访问第一个具有相同数据库文件的 sqlite open helper。如果数据库文件不存在，则会调用 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 open helper。具有给定名称的数据库文件已经存在并且版本正确，因此不会调用 onCreate() 或 onUpgrade()。而只是打开文件。

解决方案:每个数据库文件只使用一个 sqlite open helper。将两个表的创建语句放在同一个帮助器 onCreate() 方法中。

同时卸载您的应用程序，以便删除仅包含另一个表的旧数据库文件。

查看链接问题 When is SQLiteOpenHelper onCreate() / onUpgrade() run?了解有关 sqlite open helper 生命周期回调的更多信息。

关于android - 表未创建 sqlite android，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25081316/