从Windows窗体向运行控制台应用程序传递多个参数

Question

I have the console application, lets name it X.exe . It works with two arguments lets say 'a' and 'a.tmp' where a is my input file name and a.tmp is output filename . On console I usually run application like: X a a.tmp but first I have to be present in location of input file 'a' else application wont work if I tried to give its absolute path. I have created windows form to run these console application, but as I said earlier application must be started in file location. I tried using process object but application was not working. I created two processes:

1. go to the file location
2. to execute application on file location

Question: can I excute these multiple commands in one go and avoid using IPC?

Answer 1

What you can use is the ProcessStartInfo.WorkingDirectory .

Eg from MS Docs - ProcessStartInfo Class

The WorkingDirectory property behaves differently when UseShellExecute is true than when UseShellExecute is false . When UseShellExecute is true , the WorkingDirectory property specifies the location of the executable. If WorkingDirectory is an empty string, the current directory is understood to contain the executable.

Note - When UseShellExecute is true , the working directory of the application that starts the executable is also the working directory of the executable.

When UseShellExecute is false , the WorkingDirectory property is not used to find the executable. Instead, its value applies to the process that is started and only has meaning within the context of the new process.

Eg

    public static void Main()
    {
        Process myProcess = new Process();

        try
        {                
            myProcess.StartInfo.UseShellExecute = true;

            // You can start any process, HelloWorld is a do-nothing example.
            myProcess.StartInfo.FileName = "X.exe"; /
            myProcess.WorkingDirectory = "C:\SomeDirectory That contains A and A.tmp"
            myProcess.Start();
        }
        catch (Exception e)
        {
            Console.WriteLine(e.Message);
        }
    }