从路径中获取文件名

Get a file name from a path(从路径中获取文件名)

问题描述

从路径中获取文件名的最简单方法是什么?

What is the simplest way to get the file name that from a path?

string filename = "C:\MyDirectory\MyFile.bat"

在这个例子中，我应该得到MyFile".无需扩展.

In this example, I should get "MyFile". without extension.

推荐答案

_splitpath 应该做你需要的.您当然可以手动完成，但 _splitpath 也可以处理所有特殊情况.

_splitpath should do what you need. You could of course do it manually but _splitpath handles all special cases as well.

正如 BillHoag 提到的，建议使用更安全的 _splitpath 版本，称为 _splitpath_s 可用时.

As BillHoag mentioned it is recommended to use the more safe version of _splitpath called _splitpath_s when available.

或者如果你想要一些便携的东西，你可以做这样的事情

Or if you want something portable you could just do something like this

std::vector<std::string> splitpath(
  const std::string& str
  , const std::set<char> delimiters)
{
  std::vector<std::string> result;

  char const* pch = str.c_str();
  char const* start = pch;
  for(; *pch; ++pch)
  {
    if (delimiters.find(*pch) != delimiters.end())
    {
      if (start != pch)
      {
        std::string str(start, pch);
        result.push_back(str);
      }
      else
      {
        result.push_back("");
      }
      start = pch + 1;
    }
  }
  result.push_back(start);

  return result;
}

...
std::set<char> delims{'\'};

std::vector<std::string> path = splitpath("C:\MyDirectory\MyFile.bat", delims);
cout << path.back() << endl;

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