使用环境变量启动进程

Question

I want to start a process with the following path.

"ProgramFiles(x86)\\Philips Speech\\Device Control Center PDCC.exe"

When I type this into the console the process starts as expected but when I try to do it in code I get the following exception:

the system cannot find the file specified

This I my code so far:

var startInfo = new ProcessStartInfo("%ProgramFiles(x86)%\Philips Speech\Device Control Center PDCC.exe");
Debug.Assert(startInfo.EnvironmentVariables.ContainsKey("ProgramFiles(x86)")) //Is true
new Process(startInfo).Start(); //<- exception occures here

Does anybody have an idea if I can do this directly by giving the ProcessStartInfo class the environment variable or if I have to parse it before doing so?

Answer 1

You should use this to get the path to Program Files:

Environment.GetFolderPath(Environment.SpecialFolder.ProgramFilesX86)

Special folders enum .

Answer 2

string path = Environment.ExpanEnvironmentVariables("%ProgramFiles(x86)%\Philips Speech\Device Control Center PDCC.exe");
var startInfo = new ProcessStartInfo(path);
new Process(startInfo).Start(); 

This way you can use the variables (eg "%ProgramFiles(x86)% ) and not depend on the folder being in C:\\ or something.

Answer 3

The constructor just sets the fileName property to what you pass, so yes you will need to resolve the environment variable first.

From the source code for ProcessStartInfo :

public ProcessStartInfo(string fileName) {
     this.fileName = fileName;
}

Answer 4

或者试试这个，如果你想要像开始->运行一样的行为

Environment.ExpandEnvironmentVariables("%windir%\System32")