C# Making the second and third OpenFileDialog appear above the console?

Question

I'm writing a console program that can accept 1 to 3 files. I'm using OpenFileDialog three times to accept the files, but the second time and third the file dialog is behind the console window, making it hard to notice. Any way to get it to appear above?

An image of the problem: http://img205.imageshack.us/img205/5312/problemr.png

The relevant code is:

static bool loadFile(ref List<string> ls)
{
 OpenFileDialog f = new OpenFileDialog();
 if (f.ShowDialog() == DialogResult.OK)
 {
  Console.WriteLine("Loaded file {0}", f.FileName);
  ls.Add(f.FileName);
  return true;
 }
 else
 {
  return false;
 }
}

[STAThread]
static void Main(string[] args)
{
 //sanity check
 if (args.Length > 3)
 {
  Console.WriteLine("Sorry, this program currently supports a maximum of three different reports to analyze at a time.");
  return;
 }
 else if (args.Length == 0) 
 {
   List<string> fL = new List<string>();
   for (int k = 0; k < 3; k++)
   {
     if (!loadFile(ref fL)) break;
   }
   if (fL.Count == 0)
   {
   InfoDisplay.HelpMessage();
   return;
   }
   else
   {
   args = fL.ToArray();
   }
 }

        //main program

Answer 1

Found a similar post here. Not tested, but give it a shot and let me know.

foreach(Process p in Process.GetProcesses)
{
    if(p.MainWindowTitle = <the main UI form caption>)
    {
     if(IsIconic(p.MainWindowHandle.ToInt32) != 0)
     {
      ShowWindow(p.MainWindowHandle.ToInt32, &H1);
      ShowWindow(f.Handle.ToInt32, &H1);
     }
     else
     {
      SetForegroundWindow(p.MainWindowHandle.ToInt32);
      SetForegroundWindow(f.Handle.ToInt32);
     }
    }
}

Answer 2

I know this doesn't directly answer the question, but the OpenFileDialog has a property called "MultiSelect" which indicates whether or not the user can select more than one file. Once the user does select the file(s), the property FileNames (string[]) gets populated with all the file names. You can then just do a check like this:

if(dialog.FileNames.Length > 3)
{
   //only 3 are allowed
}

Answer 3

I encountered the same problem in a console application.

OpenFileDialog.ShowDialog can be called with a "parent window" handle argument. I create a dummy Form and use that as parent window argument (the dummy form is invisible, since I don't call Show() on it).

The following code works for me:

Form dummyForm = new System.Windows.Forms.Form();

OpenFileDialog myOpenFileDialog1 = new OpenFileDialog();  
if (myOpenFileDialog1.ShowDialog(dummyForm) == DialogResult.OK) {  
    fileName1 = myOpenFileDialog1.FileName;  
}

OpenFileDialog myOpenFileDialog2 = new OpenFileDialog();
if (myOpenFileDialog2.ShowDialog(dummyForm) == DialogResult.OK) {
    fileName2 = myOpenFileDialog2.FileName;
}

Answer 4

OpenFileDialog.ShowDialog can be called with a "parent window" handle argument. I create a dummy Form and use that as parent window argument (the dummy form is invisible, since I don't call Show() on it).

The Form is a bit heavy-weight, perhaps a label instead, since it just takes a IWin32Window?

OpenFileDialog.ShowDialog(new Label());

It seems to work.

If someone could come up with a reasonable solution to this (now-) resurrected question, that would be great.

Aaron