how to open file from a folder with same name in C#

Question

I need to run an executable file(Temp.exe) from "C:Program Files\\First Flight\\Temp Client" folder. The folder contains files with similar name as TempClientSetup.exml

I have this code

var path = @"C:\Program Files\First Flight\Temp Client";
var exepath = Path.Combine(Path.GetDirectoryName(path), "Temp.exe");
app = Application.Launch(exepath);

When I run this code, TempClientSetup.exml file is executed instead of Temp.exe

Any clues here? Pl suggest.

Answer 1

Stick a slash on the end of the path, or get rid of the Path.GetDirectoryName call.

var path = @"C:\Program Files\First Flight\Temp Client\";
var exepath = Path.Combine(Path.GetDirectoryName(path), "Temp.exe");

or

var path = @"C:\Program Files\First Flight\Temp Client";
var exepath = Path.Combine(path, "Temp.exe");

Answer 2

Here is the solution to my question for those who face this issue is, You may have to set the working directory of that exe to launch ;-) like this

var psi = new ProcessStartInfo(@"Temp.exe");
psi.WorkingDirectory = @"C:\Program Files\First Flight\Temp Client";            
app = Application.Launch(psi);