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Uninstaller comes second

Posted on 2006-11-29
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Last Modified: 2010-04-16
I have followed this article http://www.codeproject.com/useritems/DeployUninstall.asp and created uninstaller for my setup project. However, I have translated it into C# from VB.net

my code is like this:

            [STAThread]
            static void Main()
            {

                  string[] arguments = Environment.GetCommandLineArgs();

                  foreach(string argument in arguments)
                  {
                        string[] parameters = argument.Split('=');
                        if (parameters[0].ToLower() == "/u")
                        {
                              string productCode = parameters[1];
                              string path = Environment.GetFolderPath(Environment.SpecialFolder.System);
                              Process proc = new Process();
                              proc.StartInfo.FileName = string.Concat(path,"\\msiexec.exe");
                              proc.StartInfo.Arguments = string.Concat(" /i ", productCode);
                              proc.Start();
                              Application.Exit();
                        }
                  }

                  Application.Run(new Form1());
            }

The uninstaller works ok and it unistalls my project. Only once issue here is that when I click uninstaller, it starts the application first and the uninstallation window comes second.  I have put "Application.Run(new Form1());" code at last line (after uninstallation code), why it is still be executed? What can I do to avoid start the project and only uninstall it?

Thanks a lots!
0
Comment
Question by:boy8964
3 Comments
 
LVL 2

Expert Comment

by:BogdyPtr
ID: 18043530
Remove Application.Run(new Form1());
In this way the uninstaller will be started and no form will be shown.
0
 
LVL 4

Author Comment

by:boy8964
ID: 18048779
if I remove Application.Run(new Form1()); my application wot start!
0
 
LVL 3

Accepted Solution

by:
bvwang earned 500 total points
ID: 18081834
if (arguments == 1) {
Application.Run(new Form1());
}
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