01. december 2004 - 14:16
Der er
1 løsning
Openfildialog, mangler sti i textbox
Har følgende kode:
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
OpenFile()
End Sub
Private Function OpenFile() As FileStream
' Displays an OpenFileDialog and shows the read/only files.
Dim DlgOpenFile As New OpenFileDialog
DlgOpenFile.ShowReadOnly = True
Dim Fs As FileStream
If DlgOpenFile.ShowDialog() = DialogResult.OK Then
' If ReadOnlyChecked is true, uses the OpenFile method to
' open the file with read/only access.
If DlgOpenFile.ReadOnlyChecked = True Then
Return DlgOpenFile.OpenFile()
' Otherwise, opens the file with read/write access.
Else
Dim Path As String = DlgOpenFile.FileName
Return New FileStream(Path, System.IO.FileMode.Open, _
System.IO.FileAccess.ReadWrite)
End If
End If
End Function
Har en textbox, hvordan får jeg den til at vise stien til den fil jeg vil åbne via openfiledialog
01. december 2004 - 14:31
#1
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
TextBox1.Text = OpenFile()
'Dim myStream As Stream
'Dim openFileDialog1 As New OpenFileDialog
'openFileDialog1.InitialDirectory = "c:\"
'openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*"
'openFileDialog1.FilterIndex = 2
'openFileDialog1.RestoreDirectory = True
'If openFileDialog1.ShowDialog() = DialogResult.OK Then
' myStream = openFileDialog1.OpenFile()
' If Not (myStream Is Nothing) Then
' ' Insert code to read the stream here.
' myStream.Close()
' End If
'End If
End Sub
Private Function OpenFile() As String
' Displays an OpenFileDialog and shows the read/only files.
Dim path As String
Dim DlgOpenFile As New OpenFileDialog
DlgOpenFile.ShowReadOnly = True
Dim Fs As FileStream
If DlgOpenFile.ShowDialog() = DialogResult.OK Then
' If ReadOnlyChecked is true, uses the OpenFile method to
' open the file with read/only access.
'If DlgOpenFile.ReadOnlyChecked = True Then
' Return DlgOpenFile.OpenFile()
' ' Otherwise, opens the file with read/write access.
'Else
path = DlgOpenFile.FileName
'Return New FileStream(Path, System.IO.FileMode.Open, _
' System.IO.FileAccess.ReadWrite)
'End If
End If
Return path
End Function
