C＃打开文件，路径先从％USERPROFILE％ [英] c# open file, path starting with %userprofile%

问题描述

我有一个简单的问题。我必须在用户目录中的文件看起来像这样的路径：

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

当我尝试打开它作为一个文件

When I try to open it as a file

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

这是因为它尝试添加吐奶误差％USERPROFILE％到当前目录，所以就变成：

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

我如何使它认识到，与％USERPROFILE％开始的路径是绝对的，而不是相对路径

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

PS：我不能使用

Environment.GetFolderPath(Environment.SpecialFolder.ApplicationData)

由于我只是需要通过其名称打开该文件。用户指定的名称。如果用户指定settings.file，我需要打开一个文件相对于程序目录，如果用户指定使用％USERPROFILE％启动或转换为其他一些事情的路径C：！\something，我需要打开它，以及

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!

推荐答案

使用的 Environment.ExpandEnvironmentVariables 使用它之前的路径。

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))
{
   //...
}

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