Starting an .exe from the same folder of my C# program

Question

I made a small program in c# with a button that is supposed to open another .exe file.

It works fine if I use:

private void start_Click(object sender, RoutedEventArgs e)
        {
            System.Diagnostics.Process.Start(@"path to file");
        }

But not if I want it to run an .exe from the same folder, i basically wanted something like:

private void start_Click(object sender, RoutedEventArgs e)
        {
            System.Diagnostics.Process.Start(@"program.exe");
        }

What am I missing, I've tried a solution from this website:

var startIngo = new ProcessStartInfo();
startIngo.WorkingDirectory = // working directory
// set additional properties 

Process proc = Process.Start(startIngo);

But Visual c# doesn't recognize "ProcessStartInfo" at all...

Answer 1

What your looking for is:

Application.StartupPath

It will return the startup path that your executable was started in.

If you are using WPF, try this instead:

String appStartPath = System.IO.Path.GetDirectoryName(Process.GetCurrentProcess().MainModule.FileName);

Answer 2

You can do:

var startupPath = System.IO.Path.GetDirectoryName(
                       System.Reflection.Assembly.GetEntryAssembly().Location);
var programPath = System.IO.Path.Combine(startupPath, "program.exe");

System.Diagnostics.Process.Start(programPath);

Answer 3

ProcessStartInfo is in the System.Diagnostics namespace - you need to import that namespace at the top of your cs file using a using System.Diagnostics; statement for the compiler to recognise ProcessStartInfo without specifying the namespace explicitly where you use the class.

Answer 4

You could also try System.IO.Path.GetDirectoryName(System.Reflection.Assembly.GetExecutingAssembly().GetName().CodeBase);

To get your local path. For example

//in your imports/using section
using System.IO
using System.Reflection
using System.Diagnostics;

//in your code to execute
Process.start(Path.GetDirectoryName(Aseembly.GetExecutingAssembly().GetName().CodeBase) + "\\program.exe")

Answer 5

There are two cases:

1. The application was started directly - start up path can be extracted from the command-line.

2. The application was started indirectly - e.g. from a unit-test, start up path can not be extracted from the command-line, however you can read it from the current directory into a static variable during the start-up (before the user has a chance to change it (e.g. using a file open/save dialog)).