How to Open OpenFileDiologe Box from Business One

Question

Hi....
        I want to Select .txt File From Required Location. on click event of Button.
Please Help

Answer 1

Hi venkat,
use the following code on item pressed event..
Case SAPbouiCOM.BoEventTypes.et_ITEM_PRESSED
Select Case pVal.ItemUID
Case "Btnbrowse"                   
Me.OpenFile()
end select
Private Sub OpenFile(ByRef BubbleEvent As Boolean)
Try
ShowFolderBrowserThread = New Threading.Thread(AddressOf ShowFolderBrowser)
If ShowFolderBrowserThread.ThreadState = ThreadState.Unstarted Then ShowFolderBrowserThread.SetApartmentState(ApartmentState.STA)
ShowFolderBrowserThread.Start()
ElseIf ShowFolderBrowserThread.ThreadState = ThreadState.Stopped Then
ShowFolderBrowserThread.Start()
ShowFolderBrowserThread.Join()
End If
Catch ex As Exception
sbo_application.MessageBox("OpenFile" & ex.Message)
End Try
Private Sub ShowFolderBrowser()
Dim MyTest As New OpenFileDialog
Dim MyProcs() As System.Diagnostics.Process
Dim filename As String
MyProcs = process.GetProcessesByName("SAP Business One")
If MyProcs.Length = 1 Then
For i As Integer = 0 To MyProcs.Length - 1
'// WindowWraper concepts are used to access open dialogue box of windows
Dim MyWindow As New WindowWrapper(MyProcs(i).MainWindowHandle)
MyTest.Filter = "Excel files (.xls)|.xls|Document files (.doc)|.doc|Presentation (.ppt)|.ppt|Adobe PDF Files (.pdf)|.pdf|Text Documents (.txt)|.txt" '|All Files (.)|."
'MyTest.InitialDirectory = "C:\Program Files\SAP\SAP Business One\Attachments"
'// Initial path for open dialogue box
'// We can change as per the requirements
MyTest.InitialDirectory = "C:\Program Files\SAP\SAP Business One\Attachments"
If MyTest.ShowDialog(MyWindow) = DialogResult.OK Then
                        filename = MyTest.FileName
.your validations........
Else
           System.Windows.Forms.Application.ExitThread()
   End If
end sub
Public Class WindowWrapper
Implements System.Windows.Forms.IWin32Window
Private _hwnd As IntPtr
Public Sub New(ByVal handle As IntPtr)
_hwnd = handle
End Sub
Public ReadOnly Property Handle() As System.IntPtr Implements System.Windows.Forms.IWin32Window.Handle
Get
Return _hwnd
End Get
End Property
End Class
to open the file you can use the follwing code on display button pressed.
Private Sub fileopen(ByVal filename As String)
        '// Opening specified document with reference to the filename distributed 
        Try
            System.Diagnostics.Process.Start(filename)
        Catch ex As Exception
            sbo_application.StatusBar.SetText("File Not Found in Specified Path:" & filename, SAPbouiCOM.BoMessageTime.bmt_Short, SAPbouiCOM.BoStatusBarMessageType.smt_Error)
        End Try
    End Sub
Regards,
Ganesh K