2016-10-05 7 views
0

Ich habe versucht, auf Login-System mit MySQL zu erstellen. aber ich weiß nicht, dass es nicht funktioniert. Es gibt mein Versuch jedes Mal, wenn das Problem ist?vb.net kann keine Verbindung zu mysql

Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click 
    mysqlconn = New MySqlConnection 
    mysqlconn.ConnectionString = "server=localhost;userid=root;password=;database=tes" 
    Dim reader As MySqlDataReader 

    Try 
     mysqlconn.Open() 
     Dim query As String = "select * from kullanicilar where 'username=" & TextBox1.Text & "' and 'password=" & TextBox2.Text & "'" 
     komut = New MySqlCommand(query, mysqlconn) 
     reader = komut.ExecuteReader 
     Dim girdimi As Integer = 0 

     If reader.Read Then 
      girdimi = girdimi + 1 
     End If 

     If girdimi = 1 Then 
      Form2.Show() 
     ElseIf girdimi = 0 Then 
      MessageBox.Show("try again") 
     End If 
     mysqlconn.Close() 
    Catch ex As MySqlException 
     MessageBox.Show(ex.Message) 
    Finally 
     mysqlconn.Dispose() 

    End Try 
End Sub 
+0

select * from kullanicilar wo 'username = foo' und 'password = bar' wählen sollte * aus kullanicilar wo username =‘ foo 'und password =' ​​bar ' – Paul

+0

MessageBox.Show (Abfrage) – Paul

+0

Oh danke, es funktioniert – MuhammedM2

Antwort

0

es auf diese Weise versuchen (basierend auf Datareader MSDN article):

Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click 
    mysqlconn = New MySqlConnection 
    mysqlconn.ConnectionString = "server=localhost;userid=root;password=;database=tes" 
    Dim reader As MySqlDataReader 

    Try 
     mysqlconn.Open() 
     Dim query As String = "select * from kullanicilar where username='" & TextBox1.Text & "' and password='" & TextBox2.Text & "'" 
     komut = New MySqlCommand(query, mysqlconn) 
     reader = komut.ExecuteReader() 
     Dim girdimi As Integer = 0 

     If reader.HasRows Then 
      Do While reader.Read() 
       girdimi = girdimi + 1 
       If girdimi = 1 Then 
        Form2.Show() 
       ElseIf girdimi = 0 Then 
        MessageBox.Show("try again") 
       End If 
     Loop 
     End If 

     mysqlconn.Close() 
    Catch ex As MySqlException 
     MessageBox.Show(ex.Message) 
    Finally 
     mysqlconn.Dispose() 

    End Try 
End Sub 
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