Programmatically setting startin location when starting a process

I have an application that creates a shortcut on my desktop and allows you to drag and drop files into the shortcut to perform an action (convert a word document to PDF). Now what I am trying to do is perform this action programmatically using shellexecute (.NET Process.Start()).

The problem is that it doesnt seem to be working and I have a sneaking suspicion this has something to do with the fact that the shortcut created has the "Start in" parameter set to a specific folder.

So it looks like this:

Shortcut target: "C:\Program Files (x86)\MyPDFConvertor\MyPDFConvertor.exe"
Shortcut startin: "C:\Program Files (x86)\MyPDFConvertor\SomeSubfolder\SomeSubSubFolder"

My code was the following.

System.Diagnostics.Process.Start("C:\\Program Files (x86)\\MyPDFConvertor\\MyPDFConvertor.exe", "C:\\MyFiles\\This is a test word document.docx");

Fundamentally my question boils down to: What does "Startin" actually mean/do for shortcuts and can I replicate this functionality when startin开发者_开发技巧g an application using either shellexecute or Process.Start?

When you use Process.Start you can call it with a ProcessStartInfo which in turn happens to be able to setup a WorkingDirectory property - this way you can replicate that behaviour.

As Yahia said, set the WorkingDirectory property. You also need to quote the arguments. Here is a rough example:

//System.Diagnostics.Process.Start("C:\\Program Files (x86)\\MyPDFConvertor\\MyPDFConvertor.exe", "C:\\MyFiles\\This is a test word document.docx");
ProcessStartInfo start = new ProcessStartInfo();
//must exist, and be fully qualified:
start.FileName = Path.GetFullPath("C:\\Program Files (x86)\\MyPDFConvertor\\MyPDFConvertor.exe");
//set working directory:
start.WorkingDirectory = Path.GetFullPath("C:\Program Files (x86)\MyPDFConvertor\SomeSubfolder\SomeSubSubFolder");
//arguments must be quoted:
const char quote = '"';
start.Arguments = quote + "C:\\MyFiles\\This is a test word document.docx" + quote;
//disable the error dialog
start.ErrorDialog = false;
try
{
    Process process = Process.Start(start);
    if(process == null)
    {//started but we don't have access

    }
    else
    {
        process.WaitForExit();
        int exitCode = process.ExitCode;
    }
}
catch
{
    Console.WriteLine("failed to start the program.");
}