CODEWITHSUNDEEP
c++visual-studiosearchdirectoryfilesystems
Gus
Gus

Reputation: 45

C++ .find() function to search for file names in directory

I am trying to develop a small search engine application to search for my local files in my C:// directory by typing their names as input.

I received this suggestion as an implementation to the search function. However, it only allows me to search with the exact name of the file e.g "1234_0506AB.pdf"

I want my search function to take "1234" as an input and still be able to fetch "1234_0506AB.pdf".

Another suggestion was: instead of simply testing for equality via dir_entry.path().filename() == file_name, just get the filename as string dir_entry.path().filename().string() and do a .find(). For even more general searching, you could use a regex and match that with the filename.

I have very little experience in that and require some help and guidance to use .find() or regex in my code.

#include <filesystem>
#include <algorithm>

namespace fs = std::filesystem;

void search(const fs::path& directory, const fs::path& file_name)
{
    auto d = fs::directory_iterator(directory);

    auto found = std::find_if(d, end(d), [&file_name](const auto& dir_entry)
    {
        return dir_entry.path().filename() == file_name;
    });

    if (found != end(d))
    {
        // we have found what we were looking for
    }

    // ...
}

Upvotes: 1

Views: 5291

Answers (1)

Dutow
Dutow

Reputation: 5668

To use find, you should check the documentation of the method, which also contains some examples on most sites.

In your code, you accept a filename if its exactly the same as the one you are looking for:

return dir_entry.path().filename() == file_name;

To accept substring matches, you'll have to modify this check to use find instead of ==. As mentioned in the linked doc, find returns npos if it can't find a match.

return dir_entry.path().filename().string().find(file_name.string()) != std::string::npos;

If you are looking for matches only at the beginning of the string, you could use == 0instead of != npos.

But in this case, you have other options too, for example using substr to cut the filename to the desired length:

return dir_entry.path().filename().string().substr(0, file_name.size()) == file_name.string();

For a solution using regex, check the regex examples on the same site.

Upvotes: 4

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