Process.start 以应用程序和文件名作为变量

Question

Have been searching, but surprisingly could not find this specific question:

With C# I want (by clicking a button in a form) to run a certain file, with an certain application.

When using "Process.start(variable)" I can only pick one of the two.

And by using "Process.startinfo.filename" (like: https://docs.microsoft.com/en-us/dotnet/api/system.diagnostics.processstartinfo.filename?view=net-5.0 ) this also seems to be the case.

Isn't is possible to just combine both in some "easy" way?

Thanks.

Answer 1

Typically you would run a file with an application using a command argument (ie 'notepad.exe file.txt').

If that is possible with the application(s) you are attempting to launch, then you would simply need to set the Filename property of StartInfo to the name, if in the PATH, or the full path of the application and the Arguments property to the path of the file.

var process = new Process();
process.StartInfo.FileName = "notepad.exe";
process.StartInfo.Arguments = "C:\\{pathToFile}\\file.txt";
process.Start();

The above code would launch notepad opening file.txt. You can simply replace the FileName and Arguments with variables containing the paths to the application and file.