1. add new Form<br>2. add Button on Form1<br>3.add event onClick for the button, with code below<br><br> Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click<br> Dim frm2 As Form2<br> frm2 = New Form2<br> frm2.Show()<br> \'frm2.ShowDialog()<br><br> End Sub
จากคุณ
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maddog [2012-04-08 10:52:17]
ความคิดเห็น #29265 (จาก IP: 103.247.60.98)
Private Sub LinkL4_LinkClicked(ByVal sender As System.Object, ByVal e As System.Windows.Forms.LinkLabelLinkClickedEventArgs) Handles LinkL4.LinkClicked Dim myFileDialog As OpenFileDialog = New OpenFileDialog() With myFileDialog .Filter = "Text files (*.txt)|*.txt|All files (*.*)|*.*" .FilterIndex = 1 .InitialDirectory = "C:\My Documents" .Title = "Open File" .CheckFileExists = False End With
**ลิ้งข้อมูลเข้าไปแล้วแต่ข้อมูลไม่มาต้องทำยังไงต่อคะ If myFileDialog.ShowDialog() = DialogResult.OK Then Console.WriteLine(myFileDialog.FileName) End If myFileDialog = Nothing End Sub