Question

我是被动的英语.. 我有一个数据库连接

           String sql = "INSERT INTO " + TABLE_TRANSAKSI
                    + " (kdCabang, kdRoti, noNota ) " + " VALUES ( '"
                    + kdCabang + "' , '" + cursor.getString(0) + "' , '"
                    + noNota + "'";

            dbHelper.getWritableDatabase().execSQL(sql);
            Log.d("INSERT noNota", sql);

我认为我的代码已经是真的..但是在我的日志中......

09-08 15:23:53.811: ERROR/Database(21975): Failure 1 (near "'1'": syntax error) on 0x343fc0 when preparing 'INSERT INTO TRANSAKSI (kdCabang, kdRoti, noNota )  VALUES ( 'TKRS' , 'KRKJ' , '1''.

谁能说出我的错？谢谢be4

Answer 1

你错过了一个右括号。

应该是：

String sql = "INSERT INTO " + TABLE_TRANSAKSI
                    + " (kdCabang, kdRoti, noNota ) " + " VALUES ( '"
                    + kdCabang + "' , '" + cursor.getString(0) + "' , '"
                    + noNota + "')";

生产：

'INSERT INTO TRANSAKSI (kdCabang, kdRoti, noNota )  VALUES ( 'TKRS' , 'KRKJ' , '1')'.

所有这些都在错误信息中 - 你需要学会阅读和理解它们。

Answer 2

您最后没有关闭支架请检查第4行

String sql = "INSERT INTO " + TABLE_TRANSAKSI
                    + " (kdCabang, kdRoti, noNota ) " + " VALUES ( '"
                    + kdCabang + "' , '" + cursor.getString(0) + "' , '"
                    + noNota + "')";

Answer 3

关闭值的括号，如下所示

 String sql = "INSERT INTO " + TABLE_TRANSAKSI
                    + " (kdCabang, kdRoti, noNota ) " + " VALUES ( '"
                    + kdCabang + "' , '" + cursor.getString(0) + "' , '"
                    + noNota + "')";

为什么查询无法执行？

3 个答案: