如何从路径中获取文件名的词干？

Question

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

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")

Answer 1

Just use 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 :

// 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();

Answer 2

What about using std::string? eg

  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;

Answer 3

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

#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:

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::filesystem instead until standard libraries are fully conforming.

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

Answer 4

You can easily extract the file:

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

Answer 5

// 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 );

Answer 6

You could use the _splitpath_s function 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:

#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::string but you could modify this to use TCHAR* ( wchar , char )...

For your specific case:

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;
}

Answer 7

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

You can get a filename in the correct format by calling SHGetFileInfo with the SHGFI_DISPLAYNAME flag.

Answer 8

Here you 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).