Running a .bat file from C# doesn't work as opening the .bat file

Question

I am trying to get the output of a process. I've achieved this by using a .bat, which contains the following text:

program.exe > output.txt < input.txt

where program.exe is my executable, output.txt is the file I want to output the data to and input.txt is just an empty file since the program wants key inputs from time to time. The .bat file works perfectly when I run it myself, but when I attempt to run it using C# code, it just doesn't finish properly. I am trying to run it using the following code:

Process.Start(path);

I've tried a lot of different stuff but this is the code that I last tried but none of my attempts worked. Also, the .bat file doesn't run properly when you run it as an administrator and the C# program I am using is requiring administrator permissions. Could this be the issue and not the actual process running?

Answer 1

You need to run the process cmd.exe /c [path] (aka the process is "cmd.exe", and your arguments are $"/c {path}" .

When you're on the command prompt you get "program execution" and ".bat/.cmd interactive scripting". Since you want to execute a batch file you need to tell CreateProcess that cmd.exe is what's actually running it.

Answer 2

The issue was just what I thought. The .bat file doesn't work properly when launched from an elevated app. To fix it I used this:

Process.Start("explorer.exe", path);

Worked perfectly. Fixed it really quickly but thought I'd leave this here if anybody finds himself stuck with the same thing.

Answer 3

In ProcessStartInfo you should set UseShellExecute to false , then set WorkingDirectory in the path where you found that the bat is working correctly. Below is an example to make you understand better:

using (Process MyProcess = new Process())
{
    var MyProcessInfo = new ProcessStartInfo()
    {
        UseShellExecute = false,
        WorkingDirectory = "path",
        FileName = "file.exe",
        Arguments = "args"
    };
    MyProcess.StartInfo = MyProcessInfo;
    MyProcess.Start();
}