C++ 如何从路径中获取文件名的词干？

How to get the stem of a filename from a path?

提问by luac

I want to extract a const char* filename from a const char* filepath. I tried with regex but failed:

我想从 const char* 文件路径中提取一个 const char* 文件名。我尝试使用正则表达式但失败了：

const char* currentLoadedFile = "D:\files\file.lua";
char fileName[256];
if (sscanf(currentLoadedFile, "%*[^\]\%[^.].lua", fileName)) {
return (const char*)fileName; // WILL RETURN "D:\files\file!!
}

The issue is that "D:\files\file" will be returned and not the wanted "file"(note: without ".lua")

问题是将返回“D:\files\file”而不是想要的“文件”（注意：没有“.lua”）

回答by marchelbling

What about using std::string? e.g.

使用 std::string 怎么样？例如

  std::string path("d:\dir\subdir\file.ext");
  std::string filename;

  size_t pos = path.find_last_of("\");
  if(pos != std::string::npos)
    filename.assign(path.begin() + pos + 1, path.end());
  else
    filename = path;

回答by BatchyX

Just use boost::filesystem.

只需使用boost::filesystem.

#include <boost/filesystem.hpp>

std::string filename_noext;
filename_noext = boost::filesystem::path("D:\files\file.lua").stem().string().
const char* result_as_const_char = filename_noext.c_str();

or alternatively, if you want to introduce bugs yourself :

或者，如果您想自己引入错误：

// have fun defining that to the separator of the target OS.
#define PLATFORM_DIRECTORY_SEPARATOR '\'

// the following code is guaranteed to have bugs.
std::string input = "D:\files\file.lua";
std::string::size_type filename_begin = input.find_last_of(PLATFORM_DIRECTORY_SEPERATOR);
if (filename_begin == std::string::npos)
    filename_begin = 0;
else
    filename_begin++;
std::string::size_type filename_length = input.find_last_of('.');
if (filename_length != std::string::npos)
    filename_length = filename_length - filename_begin;

std::string result = input.substr(filename_begin, filename_length);

const char* bugy_result_as_const_char = result.c_str();

回答by tambre

You can do this portably and easily using the new filesystem library in C++17.

您可以使用 C++17 中的新文件系统库轻松轻松地完成此操作。

#include <cstdint>
#include <cstdio>
#include <filesystem>

int main()
{
    std::filesystem::path my_path("D:/files/file.lua");
    std::printf("filename: %s\n", my_path.filename().u8string().c_str());
    std::printf("stem: %s\n", my_path.stem().u8string().c_str());
    std::printf("extension: %s\n", my_path.extension().u8string().c_str());
}

Output:

输出：

filename: file.lua
stem: file
extension: .lua

Do note that for the time being you may need to use #include <experimental/fileystem>along with std::experimental::filesysteminstead until standard libraries are fully conforming.

请注意，在标准库完全符合之前，您可能需要暂时使用#include <experimental/fileystem>withstd::experimental::filesystem代替。

For more documentation on std::filesystemcheck out the filesystem library reference.

有关std::filesystem查看文件系统库参考的更多文档。

回答by kushvendra

You can easily extract the file:

您可以轻松提取文件：

int main()
{
    char pscL_Dir[]="/home/srfuser/kush/folder/kushvendra.txt";
    char pscL_FileName[50];
    char pscL_FilePath[100];
    char *pscL;
    pscL=strrchr(pscL_Dir,'/');
    if(pscL==NULL)
        printf("\n ERROR :INvalid DIr");
    else
    {
        strncpy(pscL_FilePath,pscL_Dir,(pscL-pscL_Dir));
        strcpy(pscL_FileName,pscL+1);
        printf("LENTH [%d}\n pscL_FilePath[%s]\n pscL_FileName[%s]",(pscL-pscL_Dir),pscL_FilePath,pscL_FileName);
    }
    return 0;
}


output: 
LENTH [25}
 pscL_FilePath[/home/srfuser/kush/folder]
 pscL_FileName[kushvendra.txt

回答by int21

// Set short name:
char *Filename;
Filename = strrchr(svFilename, '\');
if ( Filename == NULL )
    Filename = svFilename;

if ( Filename[0] == '\')
    ++Filename;
if ( !lstrlen(Filename) )
{
    Filename = svFilename;
}
fprintf( m_FileOutput, ";\n; %s\n;\n", Filename );

回答by lmiguelmh

You could use the _splitpath_sfunction to break a path name into its components. I don't know if this is standard C or is Windows specific. Anyway this is the function:

您可以使用该_splitpath_s函数将路径名分解为其组成部分。我不知道这是标准 C 还是 Windows 特定的。无论如何，这是功能：

#include <stdlib.h>
#include <string>

using std::string;

bool splitPath(string const &path, string &drive, string &directory, string &filename, string &extension) {
    // validate path
    drive.resize(_MAX_DRIVE);
    directory.resize(_MAX_DIR);
    filename.resize(_MAX_FNAME);
    extension.resize(_MAX_EXT);

    errno_t result;
    result = _splitpath_s(path.c_str(), &drive[0], drive.size(), &directory[0], directory.size(), &filename[0], filename.size(), &extension[0], extension.size()); 
    //_splitpath(path.c_str(), &drive[0], &directory[0], &filename[0], &extension[0]); //WindowsXp compatibility
    _get_errno(&result);
    if (result != 0) {
        return false;
    } else {
        //delete the blank spaces at the end
        drive = drive.c_str();
        directory = directory.c_str();
        filename = filename.c_str();
        extension = extension.c_str();
        return true;
    }
}

It is a lot easier and safe to use std::stringbut you could modify this to use TCHAR*(wchar, char)...

它使用起来更容易和安全，std::string但您可以修改它以使用TCHAR*( wchar, char)...

For your specific case:

对于您的具体情况：

int main(int argc, char *argv[]) {
    string path = argv[0];
    string drive, directory, filename, extension;
    splitPath(path, drive, directory, filename, extension);
    printf("FILE = %s%s", filename.c_str(), extension.c_str());
    return 0;
}

回答by Anders

If you are going to displaya filename to the user on Windows you should respect their shell settings (show/hide extension etc).

如果您要在 Windows 上向用户显示文件名，您应该尊重他们的 shell 设置（显示/隐藏扩展名等）。

You can get a filename in the correct format by calling SHGetFileInfowith the SHGFI_DISPLAYNAMEflag.

你可以通过调用得到正确格式的文件名SHGetFileInfo与SHGFI_DISPLAYNAME标志。

回答by celavek

Hereyou can find an example. I'm not saying it's the best and I'm sure you could improve on that but it uses only standard C++ (anyway at least what's now considered standard). Of course you won't have the features of the boost::filesystem (those functions in the example play along with plain strings and do not guarantee/check you'll actually working with a real filesystem path).

在这里你可以找到一个例子。我并不是说它是最好的，我相信你可以改进它，但它只使用标准的 C++（无论如何至少现在被认为是标准的）。当然，您不会拥有 boost::filesystem 的功能（示例中的那些函数与纯字符串一起播放，并且不保证/检查您实际上会使用真实的文件系统路径）。