无法更新当前锁定的vb.net错误

我想在ms access的2个不同

表中插入2个不同的数据

它显示了这个错误


我有这样一个代码：

try
 dim sql1,sql2 as string
 sql1 = "INSERT INTO table1(something)VALUES(something)"
 cmd = new oledbcommand(sql1, connection)
 cmd.executenoquery()

 sql2 = "INSERT INTO table2(something)VALUES(something)"
 cmd2 = new oledbcommand(sql2, connection)
 cmd2.executenoquery()
catch ex as exception
  msgbox(ex.tostring())

(where these cmd1,cmd2 are defined in controlModule.)

那么，我该怎么办

感谢您的帮助。谢谢
我认为关闭连接可以解决问题，最好使用
使用
-语句：

try
    Using con As OleDbConnection = GetConnection() ' or New OlebConnection(...)
        Using cmd = con.CreateCommand()
            cmd.CommandText = "INSERT INTO table1(something)VALUES(@something)"
            cmd.Parameters.AddWithValue("@something", something)
            con.Open()
            cmd.ExecuteNonQuery()
        End Using
    End Using 

    Using con As OleDbConnection = GetConnection()
        Using cmd = con.CreateCommand()
            cmd.CommandText = "INSERT INTO table2(something)VALUES(@something)"
            cmd.Parameters.AddWithValue("@something", something)
            con.Open()
            cmd.ExecuteNonQuery()
        End Using
    End Using 
Catch ex As Exception
    msgbox(ex.tostring())
End Try 

我认为关闭连接可以解决问题，最好使用
using
-语句：

try
    Using con As OleDbConnection = GetConnection() ' or New OlebConnection(...)
        Using cmd = con.CreateCommand()
            cmd.CommandText = "INSERT INTO table1(something)VALUES(@something)"
            cmd.Parameters.AddWithValue("@something", something)
            con.Open()
            cmd.ExecuteNonQuery()
        End Using
    End Using 

    Using con As OleDbConnection = GetConnection()
        Using cmd = con.CreateCommand()
            cmd.CommandText = "INSERT INTO table2(something)VALUES(@something)"
            cmd.Parameters.AddWithValue("@something", something)
            con.Open()
            cmd.ExecuteNonQuery()
        End Using
    End Using 
Catch ex As Exception
    msgbox(ex.tostring())
End Try 

这是一个并发问题。因为代码的其他部分或MS Access本身会同时访问数据库。
事实上，在连接使用后，您不会关闭它。因此，第二次调用应该会失败，并出现该异常。相反，您应该在using语句中包装一次性用品，例如OLEDB连接、命令等。这样，即使发生异常，连接也将关闭：

    Using con As New OleDbConnection, cmd1 As OleDbCommand = con.CreateCommand, cmd2 As OleDbCommand = con.CreateCommand()
        cmd1.CommandText = "INSERT INTO table1(something)VALUES(something)"
        cmd1.ExecuteNonQuery()

        cmd2.CommandText = "INSERT INTO table2(something)VALUES(something)"
        cmd2.ExecuteNonQuery()
    End Using

这是一个并发问题。因为代码的其他部分或MS Access本身会同时访问数据库。
事实上，在连接使用后，您不会关闭它。因此，第二次调用应该会失败，并出现该异常。相反，您应该在using语句中包装一次性用品，例如OLEDB连接、命令等。这样，即使发生异常，连接也将关闭：

    Using con As New OleDbConnection, cmd1 As OleDbCommand = con.CreateCommand, cmd2 As OleDbCommand = con.CreateCommand()
        cmd1.CommandText = "INSERT INTO table1(something)VALUES(something)"
        cmd1.ExecuteNonQuery()

        cmd2.CommandText = "INSERT INTO table2(something)VALUES(something)"
        cmd2.ExecuteNonQuery()
    End Using

你有没有在MS Access中以设计模式打开的表？@Steve没有，我没有。你有没有在MS Access中以设计模式打开的表？@Steve没有，我没有。谢谢，@Tim Schmelter。“这对我很有帮助。谢谢，”蒂姆·施梅尔特说。它帮助了我。