使用用户的用户名作为打开文件的路径

Question

I recently had to make a button in C# that simply had to open a text file. The job was easy until I realized that I had no ideea how to open the file, why? Well simply because I can't think of a way to "define" the user's name in the path to the file.

Here is the code I tryed to use:

private void button5_Click(object sender, EventArgs e)
{
    try
    {
        System.Diagnostics.Process.Start("C:\\Users\\%USERNAME%\\AppData\\Roaming\\SchoolProject\\file.txt");
    }
    catch { }
}

And, it did not work.

So what's the solution to this problem? If you feel like knowing the answer please be very explicit about it, I'm new to programming languages and I don't quite understand the codes so well. ( If you can and if it's not too much to ask please include the code that should work in your answer.)

Answer 1

You can get to all of these "special" folder locations via the Environment.SpecialFolders enum and the GetFolderPath method.

In your case, you want SpecialFolder.ApplicationData . Something like:

Path.Combine(Environment.GetFolderPath(Environment.SpecialFolder.ApplicationData), "test.txt");

All the special folders can be found on MSDN .

Answer 2

You need to expand environment variable use: Environment.ExpandEnvironmentVariables

Replaces the name of each environment variable embedded in the specified string with the string equivalent of the value of the variable, then returns the resulting string.

Environment.ExpandEnvironmentVariables("C:\\Users\\%USERPROFILE%\\AppData\\Roaming\\SchoolProject\\file.txt");

This will give you the exact path.

So your code could be:

string filePath = Environment.ExpandEnvironmentVariables("C:\\Users\\%USERPROFILE%\\AppData\\Roaming\\SchoolProject\\file.txt");
System.Diagnostics.Process.Start(filePath);

Also, having an empty try-catch will not help you in determining the exception, catch specific exception or at least base class Exception and then you can log/look in the debugger, at the exception and its message.