c# open file, path starting with %userprofile%

I have a simple problem. I have a path to a file in user directory that looks like this:

%USERPROFILE%\AppData\Local\MyProg\settings.file

When I try to open it as a file

ostream = new FileStream(fileName, FileMode.Open);

It spits error because it tries to add %userprofile% to the current directory, so it becomes:

C:\Program Files\MyProg\%USERPROFILE%\AppData\Local\MyProg\settings.file

How do I make it recognise that a path starting with %USERPROFILE% is an absolute, not a relative path?

PS: I cannot use

Environment.GetFolderPath(Environment.SpecialFolder.ApplicationData)

Because I need to just open the file by its name. User specifies the name. If user specifies "settings.file", I need to open a file relative to program dir, if user specifies a path starting with %USERPROFILE% or some other thing that converts to C:\something, I need to open it as well!

Solution 1:

Use Environment.ExpandEnvironmentVariables on the path before using it.

var pathWithEnv = @"%USERPROFILE%\AppData\Local\MyProg\settings.file";
var filePath = Environment.ExpandEnvironmentVariables(pathWithEnv);

using(ostream = new FileStream(filePath, FileMode.Open))
{
   //...
}

Solution 2:

Try using ExpandEnvironmentVariables on the path.

Solution 3:

Use the Environment.ExpandEnvironmentVariables static method:

string fileName= Environment.ExpandEnvironmentVariables(fileName);
ostream = new FileStream(fileName, FileMode.Open);