How can I show the “Open with” file dialog?

Question

Is there any simple way to open the \"Open with\" file dialog?

Answer 1

Using

System.Diagnostics.Process.Start(path);

The file will be openened with the default program, if no default program is defined the open with dialog will be shown.

You can use the the function:

[DllImport("shell32.dll", SetLastError = true)]
extern public static bool 
       ShellExecuteEx(ref ShellExecuteInfo lpExecInfo);

You have an example to use this function on: this link