Open File Dialog reuse

Question

This seems to be very simple but for some reason I cannot get it to work.

I am building a small C# utility that uses Open File Dialog to select an XML file from a directory. That part works fine.

The file is transferred to a web service and a response is received. No problem.

I simply want to open up the DestDirectory where the response is placed. However, no matter what I end up with the Open File Dialog opening up the directory from the first usage.
I have tried two instances of Open File Dialog on the form. OpenFD1.Reset(); Resetting all the value programatically,ie initialDirectory

It seems very simple to do what I want to do but I am not seeing the answer.

Here is part of the code: I have two buttons on the form that I want to use to open up the dialog.

    public void button1_Click(object sender, EventArgs e)
    {          
        OpenFD1.Title = "Open Test File";
        OpenFD1.Filter = "XML Files|*.xml";

        string SourceDir = AppSetting("SourceDir");
        OpenFD1.InitialDirectory = SourceDir;
        if (OpenFD1.ShowDialog() == DialogResult.OK)
        {
            sSelectedFile = OpenFD1.FileName.ToString();
            WFileName.Text = OpenFD1.FileName.ToString();
        }
        OpenFD1.CheckFileExists = true;
        OpenFD1.CheckPathExists = true;
        OpenFD1.InitialDirectory = null; ; 
    }

    private void button3_Click(object sender, EventArgs e)
    {
        OpenFD2.FileName = null;
        OpenFD2.Title = "Test Results";
        OpenFD2.Filter = "XML Files|*.xml";
        string DestDir = AppSetting("DestDir");
        OpenFD2.InitialDirectory = DestDir;
        if (OpenFD2.ShowDialog() == DialogResult.OK)
        {
            sSelectedFile = OpenFD2.FileName.ToString();
            WFileName.Text = OpenFD2.FileName.ToString();
        }
        OpenFD2.CheckFileExists=true;
        OpenFD2.CheckPathExists = true;
    }

Show your code. Explain what you mean by _"open up the DestDirectory"_. — CodeCaster, Apr 30 '15 at 15:26
Process.Start(detinationDirectory) should do the trick if you want to open the folder path in explorer — , Apr 30 '15 at 15:31
Have you tried setting `FileDialog.InitialDirectory` https://msdn.microsoft.com/en-us/library/system.windows.forms.filedialog.initialdirectory(v=vs.110).aspx — Praveen Paulose, Apr 30 '15 at 15:35
Please show your code. What is the value of OpenFD1.RestoreDirectory? — Sébastien Sevrin, Apr 30 '15 at 15:37
CodeCaster -- simply having the dialog box open n for that directory... make sense? — Sleepy, Apr 30 '15 at 16:01

score 2 · Answer 1 · answered Apr 30 '15 at 15:33

2

Try to set the OpenFD1.RestoreDirectory = false; when you create your OpenFileDialog. RestoreDirectory property makes sure that the value in Environment.CurrentDirectory will be reset before the OpenFileDialog closes.

answered Apr 30 '15 at 15:33

Cristina Alboni

1,014
9
15

That did not make any difference in how the application behaved. – Sleepy Apr 30 '15 at 19:56

Open File Dialog reuse

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