java - SQLite数据库更新功能找不到行ID？

Question

当用户点击使用其游戏货币时，我正在尝试更新该用户的“CashStack”。到目前为止，我发现用户的 ID 应该是 6，但是当我尝试在数据库更新函数中使用这个硬编码值时，我收到了 null 对象引用错误。谁能帮我理解为什么这个值不正确？我正在 Android Studio 中工作并使用 SQLite。

正在使用的函数:public void updateCashStack(int newCashStack){..}(见下文)、 FiveDollar.setOnClickListener(..)

如何查找 ID 号:

int lastID = 0;
for (int i = 0; i < 10; i++) {
    if(userIDs[i] != 0) {
        lastID = userIDs[i];
        Toast.makeText(ChoosePokerStyle.this, "User ID = " + lastID, Toast.LENGTH_LONG).show( );
    }
}

运行应用程序时显示的内容:

数据库代码:

public static final String DATABASE_NAME = "rockfinn.db";
private SQLiteDatabase db;
public static final String TABLE_NAME = "user_data";
public static final String COL1 = "ID";
public static final String COL2 = "NICKNAME";
public static final String COL3 = "CASHSTACK";

public DatabaseHelper(Context context) {super(context, DATABASE_NAME, null, 1);}

@Override
public void onCreate(SQLiteDatabase db) {
    String createTable = "CREATE TABLE " + TABLE_NAME + " (ID INTEGER PRIMARY KEY AUTOINCREMENT, " +
            "NICKNAME TEXT, " +
            "CASHSTACK INT)";
    db.execSQL(createTable);
}

@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    db.execSQL("DROP TABLE IF EXISTS " + TABLE_NAME);
}

public boolean addData(String nickname) {
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues contentValues = new ContentValues();
    contentValues.put(COL2, nickname);
    contentValues.put(COL3, 1000);

    long result = db.insert(TABLE_NAME, null, contentValues);

    if(result == -1) {
        return false;
    }
    else {
        return true;
    }
}

public void updateCashStack(int newCashStack)
{
    try
    {
        ContentValues values = new ContentValues();
        values.put(COL3, newCashStack);
        String whereClause = "id = ?";
        db.update(TABLE_NAME, values, whereClause, new String[] {"6"});
        //db.setTransactionSuccessful();
    }
    catch (Exception e)
    {
        Log.e("Error in transaction", "Error", e);
    }
    finally
    {
      // db.endTransaction();
    }
}

public Cursor getListContents() {
    SQLiteDatabase db = this.getWritableDatabase();
    Cursor data = db.rawQuery("SELECT * FROM " + TABLE_NAME, null);
    return data;
}

}

FiveDollar.setOnClickListener 代码:

fiveDollar.setOnClickListener(new View.OnClickListener( ) {
    @Override
    public void onClick(View v)
    {
       int five = 5;
       int cashStackAsInt = Integer.parseInt(cashStack);
       cashStackAsInt = cashStackAsInt - five;
       userCashStackPrompt.setText(Integer.toString(cashStackAsInt));
       myDB.updateCashStack(cashStackAsInt);
       Toast.makeText(ChoosePokerStyle.this, "You have bet $5", Toast.LENGTH_LONG).show( );
    }
}

Answer 1

您的表格中没有标题为 id 的列。它应该是ID。尝试如下:

whereClause = "ID = ?";