Hi! I'm new to the VB.NET programming, well programming in general actually. I hope I am posting this in the right section.

I'm writing an application that pops up an OpenFile dialog, then writes the path of the selected file in a .txt file, so that an outside application can read it.

Here is the code that is supposed to do that:

Private Sub btnIzberi_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnIzberi.Click

        Dim strFileName As String
        With OpenFileDialog1
            .Filter = "Text files (*.txt)|*.txt"
            .FilterIndex = 1
            .InitialDirectory = "C:\"
            .Title = "Choose a text file"
        End With

        If OpenFileDialog1.ShowDialog() = Windows.Forms.DialogResult.OK Then
            strFileName = OpenFileDialog1.FileName
            Dim pisalnik As New StreamWriter(MyFile)
            pisalnik.WriteLine(strFileName)
            pisalnik.Close()
        Else
            Exit Sub
        End If

        txtAnalizaCrkInParov.Text = "You selected the file: " & strFileName

    End Sub


There is also a button that launches the external application that is supposed to read the file.
My problem is that the application CANNOT read the file I created (MyFile), therefore can't do it's job right.

Why won't StreamWriter let go of the file, so that an external application can use it? My best guess is that it has something to do with the OpenFileDialog, but I have no idea what, or how to fix it.

Please help me, I'm dying over here! :\


Greetz, cheh

Slovenian Hardstyle Community

OK, never mind, I've solved the problem. You change the working directory with open file dialog. Helps if you use a special folder instead of an absolute path (ex. .InitialDirectory = Environment.SpecialFolder.MyDocuments).

Also you have to set the working directory to the outsite process you are launching. :)

http://www.hardstyle.si/