Openfiledialog

[jaeden99]

When trying to use this control, how do you change the default filename, OPENFILEDIALOG1 is what is showing up in filename textbox? Any hint or clue to what I am doing incorrectly.

Here is the code that I am using:

With OpenFileDialog1
.InitialDirectory = "C:"
.Filter = "All Files (*.*)|*.*"
.CheckFileExists = True
TextBox1.Text = (DataGridView1.SelectedCells(0).Value.ToString())
hit = TextBox1.Text
'TextBox1.Text = (DataGridView1.SelectedCells(0).Value.ToString())

'open Dialiog
If .ShowDialog = Windows.Forms.DialogResult.OK Then
'm_InputStream = CType(OpenFileDialog1.OpenFile(), System.IO.StreamReader)
m_inputstream = New System.IO.StreamReader(OpenFileDialog1.OpenFile)
If m_inputstream Is Nothing Then
Throw New Exception("Error opening the file")
End If

' OpenFileDialog1.ToString = hit


End If
End With

[SammyB]

.FileName

With OpenFileDialog1
     .InitialDirectory = "C:"
     .Filter = "All Files (*.*)|*.*"
     .CheckFileExists = True
     .FileName = "WORK.LOG"
     If .ShowDialog = Windows.Forms.DialogResult.OK Then
         MsgBox(.FileName)
     End If
End With

[jaeden99]

How do you get it the file to open? I can pull up the file dialog and the filename is the textbox, but when I click on open, nothing happens...at least I don't see anything happening.
I would like to be able to view the actual document when I click on open.

Any suggestions.
It's like downloading an attachment and viewing it. But it is in a windows application instead of a web based. I know it probably would be easy in web.

[SammyB]

LOL, the dialog only gives you the filename, it does not open, read, or write. You'll need to use StreamWriter and/or StreamReader. It's explained at http://support.microsoft.com/kb/304427. --Sam

[mwalts]

LOL, the dialog only gives you the filename, it does not open, read, or write. You'll need to use StreamWriter and/or StreamReader. It's explained at http://support.microsoft.com/kb/304427. --Sam

Right, also it seems to me that you may just want to open the file with whatever is the default application for that file type. If that is true, then you should look into the Process class (in the System.Diagnostics namespace).

Specifically with the StartInfo.UseShellExecute property set to true.
Yes I could tell you exactly how to use it, but google it first, it'll be good practice

Good luck,

-mwalts

[SammyB]

Also, I keep forgetting the the OpenFileDialog has a OpenFile method that returns a Stream object, http://msdn2.microsoft.com/en-us/lib....openfile.aspx