在字符串中引用Linux用户名以打开文件 [英] Reference Linux username in string to open file

问题描述

我有一个配置文件，其中包含另一个需要打开的文件的路径.此文件路径引用Linux用户名:

I have a config file which contains the path to another file which needs to be opened. This file path references the Linux username:

/root/${USER}/workspace/myfile.txt

其中$ USER应该转换为Linux用户名.

where $USER should be translated to the Linux username.

这不起作用，因为该字符串存储在我的配置文件中，所以我不能使用getenv().

This doesn't work and because the string is stored in my config file, I cannot use getenv().

还有另一种方法可以实现这一目标吗?

Is there another way to achieve this?

推荐答案

您可以使用UNIX路径元素，表示HOME目录.像这样:

You can use wordexp to translate "~" which is a UNIX path element meaning the HOME directory. Something like this:

#include <wordexp.h>

std::string homedir()
{
    std::string s;
    wordexp_t p;
    if(!wordexp("~", &p, 0))
    {
        if(p.we_wordc && p.we_wordv[0])
            s = p.we_wordv[0];
        wordfree(&p);
    }
    return s;
}

然后从返回的路径中提取用户名.

And then extract the username from the returned path.

但是我通常这样使用std::getenv():

auto HOME = std::getenv("HOME"); // may return nullptr
auto USER = std::getenv("USER"); // may return nullptr

这篇关于在字符串中引用Linux用户名以打开文件的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！