C# WPF StartupEvent get path

Question

My objective is that when I open a file with a specific extension it will open my program in C#. That part is done but now I want to get the path of the file that I clicked on previously.

This is just the main code of App.xaml.cs :

/// <summary>
/// Logique d'interaction pour App.xaml
/// </summary>
public partial class App : Application
{
    private void Application_Startup(object sender, StartupEventArgs e)
    {

    }
}

Here are my two commands to associate my extension to my program.

    ftype GDevDB=N:\Divers\GDevDB\bin\Debug\GDevDB.exe "%1"
    assoc .GDevDB=GDevDB

Solution :

I've put the command in a batch file and the %1 was the one of the batch file and it was returning an empty string value.

Answer 1

The file you clicked on will be in the first command line argument. The arguments are in the array of strings e.Args , in the StartupEventArgs . For the file you clicked on, you thus need to get e.Args[0] .

Answer 2

You can also use Environment.GetCommandLineArgs() If you need those somewhere else. Index 0 is the path to your program. Index 1 is the file you are looking for

Answer 3

Try to do this:

 public partial class App : Application
        {
            private void Application_Startup(object sender, StartupEventArgs e)
            {
                foreach (string arg in e.Args)
                {
                    //Do something
                }
            }
        }