I am writing a program with a listbox filled with items, and when you right click the item, context menu will pop up and have an option to choose call \"Open with\", and my
You can use this way to show "Open With" dialog directly:
string FilePath = "C:\\Text.txt";//Your File Path
System.Diagnostics.Process proc = new System.Diagnostics.Process();
proc.EnableRaisingEvents = false;
proc.StartInfo.FileName = "rundll32.exe";
proc.StartInfo.Arguments = "shell32,OpenAs_RunDLL " + FilePath;
proc.Start();