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我遇到了这个问题:

错误:已经有一个打开的 DataReader 与此 Connection 关联,必须先关闭它。

请看一下我的代码:

 Dim sqlQuery As String = "SELECT * FROM users"
    Dim myAdapter As New MySqlDataAdapter

    If txtUsername.Text = String.Empty And txtPassword.Text = String.Empty Then
        MsgBox("Enter username and password", MsgBoxStyle.Exclamation, "Tea Sparkle POS")
    Else
        Dim sqlquerry = "Select * From users where username = '" + txtUsername.Text + "' And password= '" + txtPassword.Text + "'"
        Dim myCommand As New MySqlCommand()
        myCommand.Connection = SQLConnection
        myCommand.CommandText = sqlquerry
        'Starting The Query
        myAdapter.SelectCommand = myCommand
        Dim mydata As MySqlDataReader
        mydata = myCommand.ExecuteReader()
        'To check the Username and password and to validate the login a
        If mydata.HasRows = 0 Then
            MsgBox("Invalid Login")
            txtPassword.Clear()
            txtUsername.Clear()
        Else
            Dim authorityid = 0
            While mydata.Read()
                authorityid = mydata.GetInt32("authorityid")
            End While
            MsgBox("Welcome " + txtUsername.Text + "!")
            If authorityid = 1 Then
                MainForm.Show()
            Else
                MainForm.Show()
            End If
            Me.Hide()
        End If
    End If


    Private Sub Login_Load(sender As System.Object, e As System.EventArgs) Handles MyBase.Load

    SQLConnection.ConnectionString = ServerString

    Try
        If SQLConnection.State = ConnectionState.Closed Then
            SQLConnection.Open()
        Else
            SQLConnection.Close()
        End If
    Catch ex As Exception
        MsgBox(ex.ToString)
    End Try

End Sub

此错误在此行中:

 mydata = myCommand.ExecuteReader()

这有什么问题?任何帮助都非常感谢。

4

1 回答 1

3

这有什么问题?

好吧,看起来您正在重用现有连接:

myCommand.Connection = SQLConnection

不要那样做。每次需要与数据库通信时创建一个新连接,并在完成后关闭它,使用Using语句确保即使抛出异常也将其关闭。

此外,Using为您的命令使用一个语句,为您的读者使用另一个语句 - 这些都是您应该关闭的资源。

哦,而且看起来您正在 UI 线程中执行此操作,这是一个坏主意,因为在进行数据库访问时您的 UI 将无响应。

于 2013-02-05T19:28:17.750 回答