zip_open fails with error 18 when file path contains umlauts about libzip HOT 6 CLOSED

On wich operating system are you using libzip?

Can you trace through zip_open to see where the error occurs?

from libzip.

Hey, thanks for your fast reply.
I am using Windows 7 Professional x64 with libzip 1.5.1.

This is the path that gets passed to the function:

C:\exchange\OfflineDatensätze\bP Rapid\NOK\EthSpy Fehler\D0001Z001\offlineDataset.zip;

The error happens in zip_source_win32utf8.c (line 62):
/* Convert fname from UTF-8 to Windows-friendly UTF-16. */ size = MultiByteToWideChar(CP_UTF8, MB_ERR_INVALID_CHARS, fname, -1, NULL, 0); if (size == 0) { zip_error_set(error, ZIP_ER_INVAL, 0); return NULL; }

Seems like the function MultiByteToWideChar returns zero, but I can´t imagine why.

from libzip.

Maybe the file name is not encoded in UTF-8, but some other encoding like ISO-8859-1. If so, you will have to convert it to UTF-8 before passing it to libzip.

from libzip.

Okay, I have not much experience with encodings, here is the part of the code, where I try to open the archive:

c_archive = zip_open(c_fileName.c_str(), ZIP_RDONLY, &error);

c_fileName is just a std::string, I get it from a file choosing dialog.
Can this be the problem?
Can you tell me how to convert it to utf8?

from libzip.

Sorry, I'm not a Windows developer. Examine c_fileName to see wether it is in UTF-8.

from libzip.

It seems like c_fileName is Latin(ISO-8859-1) encoded.
I converted it to UTF8 now, and its working with umlauts.
Thank you for your great support!

from libzip.