C# process.Start filename and passing arguments

Question

I need to open the file test.mdb. The path must be fullpath built from whatever directory it is located in relative to the C# program exe

I need to pass the Arguments li

Answer 1

I don't think you need to worry about detecting the office version, just this should work for you:

        string filepath = '"' + Directory.GetCurrentDirectory() + "\\test.mdb" + '"';
        string acc_cmd_arg = "HELLO";

        using (System.Diagnostics.Process process = new System.Diagnostics.Process() )  {
            process.StartInfo.FileName = "msaccess.exe";
            process.StartInfo.Arguments = filepath + " /cmd " + acc_cmd_arg;
            process.Start(); 
        }