Android 无此类列错误：如果数据库中没有项，如何防止此错误？

我想我的头衔说明了这一点。。。使用以下代码：

public int getCorrectAnswersPrev(int level, String player) {
        int count = 0;
        level = level - 1;
        Cursor c = myDataBase.rawQuery(
                "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL =" + level
                        + " AND ANSWERCORRECT = 1 AND PLAYER=" + player, null);
        c.moveToFirst();
        count = c.getInt(0);
        c.close();
        myDataBase.close();

        return count;

    }

Cursor c = myDataBase.rawQuery(
    "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL = ? " +
    " AND ANSWERCORRECT = 1 AND PLAYER = ?",
    new String[] { String.valueOf(level), player }
);                      

有时候一个玩家根本没有回答某些问题，所以这意味着在一个关卡开始时，那里没有特定的物品。。。虽然我将count初始化为0，但我发现它仍然执行查询并返回错误。但是，如果数据库中没有项目，有什么方法可以防止这种情况发生吗

Cursor c = myDataBase.rawQuery(
    "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL = ? " +
    " AND ANSWERCORRECT = 1 AND PLAYER = ?",
    new String[] { String.valueOf(level), player }
);                      

编辑：

Cursor c = myDataBase.rawQuery(
    "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL = ? " +
    " AND ANSWERCORRECT = 1 AND PLAYER = ?",
    new String[] { String.valueOf(level), player }
);                      

例如：

  Cursor c = myDataBase.rawQuery(select ifnull(max(tm.no),0) as serial_no "
                        + "from table_bun as tb inner join table_mar"
                        + " as tm on tb.id = tm.id "
                        + "where tb.bid = " + bId
                        + " and tb.sid =  " + SId
                        + " and tb.eby  = '" + UId + "'", null);

Cursor c = myDataBase.rawQuery(
    "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL = ? " +
    " AND ANSWERCORRECT = 1 AND PLAYER = ?",
    new String[] { String.valueOf(level), player }
);                      

---

连接后，player中的值可能会被视为列名，因此不是：

Cursor c = myDataBase.rawQuery(
    "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL =" + level
    + " AND ANSWERCORRECT = 1 AND PLAYER=" + player, null);

Cursor c = myDataBase.rawQuery(
    "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL = ? " +
    " AND ANSWERCORRECT = 1 AND PLAYER = ?",
    new String[] { String.valueOf(level), player }
);                      

尝试使用参数执行此操作：

Cursor c = myDataBase.rawQuery(
    "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL = ? " +
    " AND ANSWERCORRECT = 1 AND PLAYER = ?",
    new String[] { String.valueOf(level), player }
);                      

如果没有记录，游标将返回null

Cursor c = myDataBase.rawQuery(
    "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL = ? " +
    " AND ANSWERCORRECT = 1 AND PLAYER = ?",
    new String[] { String.valueOf(level), player }
);                      

在使用moveToFirst之前，您应该测试c==null，如下所示：

Cursor c = myDataBase.rawQuery(
    "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL = ? " +
    " AND ANSWERCORRECT = 1 AND PLAYER = ?",
    new String[] { String.valueOf(level), player }
);                      

if(c != null){
    c.moveToFirst();


}

抛出一个错误

Cursor c = myDataBase.rawQuery(
    "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL = ? " +
    " AND ANSWERCORRECT = 1 AND PLAYER = ?",
    new String[] { String.valueOf(level), player }
);                      

应该是

Cursor c = myDataBase.rawQuery(
    "SELECT COUNT(*) FROM ANSWERS WHERE LEVEL = ? " +
    " AND ANSWERCORRECT = 1 AND PLAYER = ?",
    new String[] { String.valueOf(level), player }
);                      

SELECT COUNT(*) FROM ANSWERS WHERE LEVEL =1 AND ANSWERCORRECT = 1 AND PLAYER='Bog'

---

问题在查询中。。请在这里发布完整的错误跟踪，这样我们可以帮助您更好地做到这一点！添加Kalai Selvan和@Luis的解决方案是否明智，或者这不是必需的？如果需要，您也可以添加。对于字符串，您必须使用单引号@MatthiasVanbWhat来表示SQL注入的风险？嗨，michal，谢谢，但我在变量前后使用了“”：@MatthiasVanb是的，你可以这样做，但它很容易受到攻击。即使你已经学会了这一点，也要试着用正确的方法来做：-。好的，但我认为这主要是在我处理edittext字段时的情况？还是我错了？@MatthiasVanb一般说来，是的。在我看来，最好从一开始就把事情做好。因为有可能你们不会在真正重要的项目中做这样的事情，我不是说这不是，即使在不同的编程语言中。