Buat form dengan visual studio 2010

Buat form dengan visual studio 2010

Klik project – add reference –pilih .net - pilih mysql.data
Buat Koneksi
Aktifkan Xampp – start mysql & Apache
Masuk ke browser ketik localhost/phpmyadmin/ - buat database aplikasivbnet – buat table login dengan
2 tabel – username varchar 25 latin_swedist_1, password varchar 25 latin_swedist_1 – oke
Kemudian insert username dan password : admin - 123456
Imports MySql.Data.MySqlClient
Public Class Form1
Public str As String =
"server=localhost;user=root;password=;database=aplikasivbnet"
Public cn As MySqlConnection = New MySqlConnection(str)
Private Sub Timer1_Tick(ByVal sender As System.Object, ByVal e As
System.EventArgs) Handles Timer1.Tick
Try
ProgressBar1.Increment(10)
If ProgressBar1.Value = ProgressBar1.Maximum Then
Dim cn As MySqlConnection = New MySqlConnection(str)
cn.Open()
Dim qry As String = "SELECT * FROM login where username='" &

TextBox1.Text & "' AND password='" & TextBox2.Text & "'"
Dim cmd As MySqlCommand = New MySqlCommand(qry, cn)
Dim da As MySqlDataReader = cmd.ExecuteReader
If (da.Read() = True) Then
ProgressBar1.Value = 0

MessageBox.Show("Login berhasil",
"Login Information",
MessageBoxButtons.OK,
MessageBoxIcon.Information)
Else
Timer1.Stop()
ProgressBar1.Value = 0
MessageBox.Show("Login gagal",
"Login Information",
MessageBoxButtons.OK, MessageBoxIcon.Error)
End If
Timer1.Stop()
End If
Catch ex As Exception

MsgBox(ex.Message)
Finally
cn.Close()
End Try
End Sub
Private Sub login_Click(ByVal sender As System.Object, ByVal e As
System.EventArgs) Handles login.Click
If TextBox1.Text = "" Then
: Exit Sub
End If
If TextBox2.Text = "" Then
: Exit Sub
End If
Timer1.Start()
End Sub
End Class