如何正确格式化 OledbCommand 上的更新语法？

Question

更新语法错误。 笔记： 已更新代码，现在我没有收到任何错误；但是，数据不会在数据库上更新... 编辑：希望按照建议重命名命令 非常感谢任何帮助...

   OleDbDataAdapter dataAdapter = new OleDbDataAdapter();
                OleDbCommand Command;
               // OleDbParameter parameter;

                Command = new OleDbCommand("UPDATE Log SET Serial_Number = @Serial_Number , Material_Number = @Material_Number , Submitted = @Submitted , prt_Count1 = @prt_Count1 , prt_Count2 = @prt_Count2 , prt_Count3 = @prt_Count3 , Main_Relief = @Main_Relief , Crtg1 = @Crtg1 , Crtg2 = @Crtg2 , Crtg3 = @Crtg3 , Crtg4 = @Crtg4 , Crtg5 = @Crtg5 , Crtg6 = @Crtg6 , Crtg7 = @Crtg7 , Crtg8 = @Crtg8 , Crtg9 = @Crtg9 , Crtg10 = @Crtg10 , Crtg11 = @Crtg11 , Crtg12 = @Crtg12 , Rework = @Rework , aflrp_Repair = @aflrp_Repair , Operator_Initials = @Operator_Initials , Remarks = @Remarks , Pass_FailCrtg1 = @Pass_FailCrtg1 , Pass_FailCrtg2 = @Pass_FailCrtg2 , Pass_FailCrtg3 = @Pass_FailCrtg3 , Pass_FailCrtg4 = @Pass_FailCrtg4 , Pass_FailCrtg5 = @Pass_FailCrtg5 , Pass_FailCrtg6 = @Pass_FailCrtg6 , Pass_FailCrtg7 = @Pass_FailCrtg7 , Pass_FailCrtg8 = @Pass_FailCrtg8 , Pass_FailCrtg9 = @Pass_FailCrtg9 , Pass_FailCrtg10 = @Pass_FailCrtg10 , Pass_FailCrtg11 = @Pass_FailCrtg11 , Pass_FailCrtg12 = @Pass_FailCrtg12 " +
        "WHERE ID = @ID", con);

                Command.Parameters.AddWithValue("@ID", OleDbType.Numeric).Value = int.Parse(txtBoxID.Text);
                Command.Parameters.AddWithValue("@Serial_Number", OleDbType.VarChar).Value = textBox1.Text;
                Command.Parameters.AddWithValue("@Material_Number", OleDbType.VarChar).Value = textBox2.Text;
                Command.Parameters.AddWithValue("@Submitted", OleDbType.DBTime).Value = thisDay1;
                Command.Parameters.AddWithValue("@prt_Count1", OleDbType.Numeric).Value = int.Parse(textBox3.Text);
                Command.Parameters.AddWithValue("@prt_Count2", OleDbType.Numeric).Value = int.Parse(textBox4.Text);
                Command.Parameters.AddWithValue("@prt_Count3", OleDbType.Numeric).Value = int.Parse(textBox5.Text);
                Command.Parameters.AddWithValue("@Main_Relief", OleDbType.Numeric).Value = int.Parse(textBox6.Text);
                Command.Parameters.AddWithValue("@Crtg1", OleDbType.Numeric).Value = int.Parse(textBox7.Text);
                Command.Parameters.AddWithValue("@Crtg2", OleDbType.Numeric).Value = int.Parse(textBox8.Text);
                Command.Parameters.AddWithValue("@Crtg3", OleDbType.Numeric).Value = int.Parse(textBox10.Text);
                Command.Parameters.AddWithValue("@Crtg4", OleDbType.Numeric).Value = int.Parse(textBox11.Text);
                Command.Parameters.AddWithValue("@Crtg5", OleDbType.Numeric).Value = int.Parse(textBox12.Text);
                Command.Parameters.AddWithValue("@Crtg6", OleDbType.Numeric).Value = int.Parse(textBox13.Text);
                Command.Parameters.AddWithValue("@Crtg7", OleDbType.Numeric).Value = int.Parse(textBox14.Text);
                Command.Parameters.AddWithValue("@Crtg8", OleDbType.Numeric).Value = int.Parse(textBox15.Text);
                Command.Parameters.AddWithValue("@Crtg9", OleDbType.Numeric).Value = int.Parse(textBox16.Text);
                Command.Parameters.AddWithValue("@Crtg10", OleDbType.Numeric).Value = int.Parse(textBox17.Text);
                Command.Parameters.AddWithValue("@Crtg11", OleDbType.Numeric).Value = int.Parse(textBox18.Text);
                Command.Parameters.AddWithValue("@Crtg12", OleDbType.Numeric).Value = int.Parse(textBox19.Text);
                Command.Parameters.AddWithValue("@Rework", OleDbType.VarChar).Value = textBox20.Text;
                Command.Parameters.AddWithValue("@aflrp_Repair", OleDbType.VarChar).Value = textBox21.Text;
                Command.Parameters.AddWithValue("@Operator_Initials", OleDbType.VarChar).Value = textBox22.Text;
                Command.Parameters.AddWithValue("@Remarks", OleDbType.VarChar).Value = textBox23.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg1", OleDbType.VarChar).Value = textBox24.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg2", OleDbType.VarChar).Value = textBox25.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg3", OleDbType.VarChar).Value = textBox26.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg4", OleDbType.VarChar).Value = textBox27.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg5", OleDbType.VarChar).Value = textBox28.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg6", OleDbType.VarChar).Value = textBox29.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg7", OleDbType.VarChar).Value = textBox30.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg8", OleDbType.VarChar).Value = textBox31.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg9", OleDbType.VarChar).Value = textBox32.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg10", OleDbType.VarChar).Value = textBox33.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg11", OleDbType.VarChar).Value = textBox34.Text;
                Command.Parameters.AddWithValue("@Pass_FailCrtg12", OleDbType.VarChar).Value = textBox35.Text;

                dataAdapter.UpdateCommand = Command;

Answer 1

简单的 UPDATE 查询通常如下所示：

UPDATE table 
SET column1 = @parameter1, column2 = @parameter2 ...
WHERE columnX = @parameterX ...

您似乎修改了 INSERT VALUES 语句，但语法不正确

无论您使用? 还是@xyz for your parameter placeholders，OLEDB 都不挑剔，但您必须使用 AddWithValue（或等效的）与您使用的一样多的占位符，并且以相同的顺序。与其他提供者不同，您不能重用命名参数。这意味着对于这样的语句：

UPDATE person
SET name = @n, age = @a
WHERE id = @i

您的 C# 将需要以下内容：

command.Parameters.AddWithValue("thename", nameTextbox.Text);      //will be used as the value for @n
command.Parameters.AddWithValue("theage", ageNumericUpDown.Value); //will be used as the value for @a
command.Parameters.AddWithValue("theid", idVariable);              //will be used as the value for @i

增加了3个参数；我故意使名称不同，以强调它们无关紧要。添加的数量和顺序至关重要。你不能这样做：

UPDATE person
SET birthname = @n, currentname = @n
WHERE id = @i

//not enough parameters AND the wrong order. You cannot re-use @n unlike on other DBs
command.Parameters.AddWithValue("@i", ageNumericUpDown.Value);
command.Parameters.AddWithValue("@n", nameTextbox.Text);

_{“如果你愿意，可以使用@xyz，但假设它们都只是?s”}

---

如果您想更改值并再次运行查询，则给参数起一个好听的名称很有用：

command.Parameters["theid"].Value = idVariable+1;