在Windows XP虚拟机上使用process.start的C＃Windows Forms应用程序

Question

Here is the situation. We have a very old COBOL exe that will only run on Windows XP. In order to run this exe we have set up a virtual XP machine. Each week the user runs a series of BAT files.

I'm trying to write a C# Windows app that will run on the VM and then execute the COBOL exe. The plan is to use Process.Start to call the EXE. Everything works fine IF MY WINDOWS C# EXE runs in the same directory as the COBOL exe.

When I move the C# exe out of the folder, I get a bad return code and the cobol does not get executed.

Without going into a lot of detail, I want to be able to run the COBOL from outside of the folder. The reason is that we have several regional offices and each one has its own folder and its own copy of the cobol exe inside of each folder. The goal here is to let the user pick each regional office and then the C# program will execute each program in each directory.

In summary. If I run C# inside the folder where the COBOL exe lives, IT WORKS. BUT if I move C# outside of that folder it DOES NOT

I hope all this makes sense. Here are some additional details:

1. Created in Visual Studio 2012 as a C# Windows Application
2. Platform target x86 (I've also used Any CPU)
3. Target framework .NET Framework 4
4. COBOL is Microfocus from 1997 (yes I know. We need to get rid of it)
5. VM is Windows XP
6. If I change the exe to Notepad it runs fine

7. If I do start run from windows and paste this: K:\\AMSapp\\MYTESTFOLDER\\AMSBLD.exe it runs fine .

8. C# code for button click event (note: if I change the filename to this K:\\AMSapp\\MYTESTFOLDER\\AMSBLD.exe i get the same results. It does not run ) Also the message box is showing 255 for a return code when it does not work. If the C# program is moved inside of the cobol folder it returns a zero.

    private void button1_Click(object sender, EventArgs e) { try { int exitCode; Process process = new Process(); process.StartInfo.FileName = @"\\\\SERVER23021\\PRODDATA\\AMSapp\\MYTESTFOLDER\\AMSBLD.exe"; process.StartInfo.ErrorDialog = true; process.Start(); process.WaitForExit(1000 * 60 * 5); // Wait up to five minutes. exitCode = process.ExitCode; MessageBox.Show("My exit code = " + exitCode.ToString()); MessageBox.Show("my path and file name: " + process.StartInfo.FileName); } catch (Exception ex) { MessageBox.Show(ex.Message); } } 

Answer 1

尝试将进程工作目录设置为cobol exe文件的相同路径。

process.StartInfo.WorkingDirectory = System.IO.Path.GetFullPath(process.StartInfo.FileName);

Answer 2

I think, you should use ProcessStartInfo.Domain to set the domain and after that path to exe within ProcessStartInfo.FileName . Instance of ProcessStartInfo pass into Process.Start method

ProcessStartInfo psi = new ProcessStartInfo();
psi.Domain = @"SERVER23021";
psi.FileName = @"\PRODDATA\AMSapp\MYTESTFOLDER\AMSBLD.exe";

Process.Start(psi);

If you need to personate your calling you can set Username and Password . But you must remember that you need to provide them as a pair. UseShellExecute must be set to false and WorkingDirectory must be also provided. More: https://msdn.microsoft.com/en-us/library/system.diagnostics.processstartinfo.password(v=vs.110).aspx