c# Get File Name

The ShowDialog method returns whether the user pressed OK or Cancel. This is useful information, but the actual filename is stored as a property on the dialog

private void button1_Click(object sender, EventArgs e)
    {
        OpenFileDialog newOpen = new OpenFileDialog();
        DialogResult result = newOpen.ShowDialog();
        if(result == DialogResult.OK) {
              this.textBox1.Text = newOpen.FileName;
        }
    }

You need to access the filename:

 string filename = newOpen.FileName;

or filenames, if you allowed multiple file selection:

newOpen.FileNames;

Ref.: OpenFileDialog Class

private void button1_Click(object sender, System.EventArgs e) {
    Stream myStream = null;
    OpenFileDialog openFileDialog1 = new OpenFileDialog();

    openFileDialog1.InitialDirectory = "c:\\" ;
    openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*" ;
    openFileDialog1.FilterIndex = 2 ;
    openFileDialog1.RestoreDirectory = true ;

    if(openFileDialog1.ShowDialog() == DialogResult.OK)
    {
        try
        {
            if ((myStream = openFileDialog1.OpenFile()) != null)
            {
                using (myStream)
                {
                    // Insert code to read the stream here.
                }
            }
        }
        catch (Exception ex)
        {
            MessageBox.Show("Error: Could not read file. Error: " + ex.Message);
        }
    } 
}

You need to read the FileName property of the OpenFileDialog instance. This will get you the path of the selected file.

Here is an example of using an existing file as a default, and getting a new file back:

private string open(string oldFile)
    {
        OpenFileDialog newOpen = new OpenFileDialog();
        if (!string.IsNullOrEmpty(oldFile))
        {
          newOpen.InitialDirectory = Path.GetDirectoryName(oldFile);
          newOpen.FileName = Path.GetFileName(oldFile);
        }

        newOpen.Filter = "eXtensible Markup Language File  (*.xml) |*.xml"; //Optional filter
        DialogResult result = newOpen.ShowDialog();
        if(result == DialogResult.OK) {
              return newOpen.FileName;
        }

        return string.Empty;
    }

Path.GetDirectoryName(file) : Return path

Path.GetFileName(file) : Return filename