Problem mit ausgabe von table

MacCoremag · 26. Juli 2012, 12:10

Sub MYSQL_Login(ByVal IPAdress As String, ByVal benutzer As String, ByVal passwort As String, ByVal datebasename As String)
Dim conn As MySqlConnection
conn = New MySqlConnection
conn.ConnectionString = "server=" & IPAdress & ";" & "user id=" & benutzer & ";" & "password=" & passwort & ";" & "database=" & datebasename & ";"
Try
conn.Open()
Catch ex As MySqlException
MsgBox(ex.Message, MsgBoxStyle.Critical, "Error")
End Try
If conn.Ping() = True Then
Label13.Text = "online"
NotifyIcon1.BalloonTipText = "sccessfully logged in!"
NotifyIcon1.ShowBalloonTip(2000)
End If
End Sub
Sub MYSQL_Command(ByVal command As String, ByVal output As Object)
Dim myAdapter As New MySqlDataAdapter
Dim myCommand As New MySqlCommand
myCommand.CommandText = command
myAdapter.SelectCommand = myCommand
Dim myData As MySqlDataReader
myData = myCommand.ExecuteReader()
output = myData.ToString
End Sub
Private Sub Button14_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button14.Click
MYSQL_Login(IP, TextBox35.Text, TextBox34.Text, "world")
MYSQL_Command("SELECT player FROM gm_ticket", 'Hier kommt der "Outputort" rein!)
End Sub

Der "Outputort" soll die listbox1 sein, wenn ich

eingebe klappt es nicht. Folgender Fehler:

Fehler 1 Für den Parameter "item" von "Public Function Add(item As Object) As Integer" wurde kein Argument angegeben.

Annoxoli · 26. Juli 2012, 12:21

Probier mal:

MacCoremag · 26. Juli 2012, 12:28

EDIT: Geht nicht ganz

JEtzt habe ich ein anderes Problem:

Sub MYSQL(ByVal command As String, ByVal output As Object, ByVal IPAdress As String, ByVal benutzer As String, ByVal passwort As String, ByVal datebasename As String)
Dim conn As MySqlConnection
conn = New MySqlConnection
conn.ConnectionString = "server=" & IPAdress & ";" & "user id=" & benutzer & ";" & "password=" & passwort & ";" & "database=" & datebasename & ";"
Try
conn.Open()
Catch ex As MySqlException
MsgBox(ex.Message, MsgBoxStyle.Critical, "Error")
End Try
If conn.Ping() = True Then
Label13.Text = "online"
NotifyIcon1.BalloonTipText = "sccessfully logged in!"
NotifyIcon1.ShowBalloonTip(2000)
End If
Wait(3000)
Dim myAdapter As New MySqlDataAdapter
Dim myCommand As New MySqlCommand
myCommand.CommandText = command
myAdapter.SelectCommand = myCommand
Dim myData As MySqlDataReader
myData = myCommand.ExecuteReader() 'Jetzt kommt hier der Fehler!
output = myData.ToString
End Sub

Fehler: Connection must be valid and open.