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I have a list of files stored in a .log in this syntax:
c:\foto\foto2003\shadow.gif D:\etc\mom.jpg I want to extract the name and the extension from this files. Can you give a example of a simple way to do this?
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To extract filename from the file, we use “GetFileName()” method of “Path” class. This method is used to get the file name and extension of the specified path string. The returned value is null if the file path is null. Syntax: public static string GetFileName (string path);
You can extract the file extension of a filename string using the os. path. splitext method. It splits the pathname path into a pair (root, ext) such that root + ext == path, and ext is empty or begins with a period and contains at most one period.
Paths include the root, the filename, or both. That is, paths can be formed by adding either the root, filename, or both, to a directory.
To extract a filename without extension, use boost::filesystem::path::stem instead of ugly std::string::find_last_of(".")
boost::filesystem::path p("c:/dir/dir/file.ext"); std::cout << "filename and extension : " << p.filename() << std::endl; // file.ext std::cout << "filename only : " << p.stem() << std::endl; // file If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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