download_count_view_page() and files with same filename

Great catch. And great fix. I looked at your code and I am impressed. Your knowledge of mysql surpassses mine.

The filename that is recorded in {file_downloads} is indeed the actual file name. Not the filename (a.k.a file description) that appears to users in the node.

$result = db_query("SELECT fd.filename, fd.count, fd.timestamp, f.nid, n.type FROM {file_downloads} fd JOIN {files} f ON BINARY fd.filename LIKE SUBSTRING(f.filepath FROM 7) JOIN {node} n ON n.nid = f.nid" . tablesort_sql($header));

I will test this quickly and apply the patch.

Note to you & everyone : if you look at the download_count.install, I am just not doing anything for a pgsql database. Do you know what the pgsql syntax would be for LIKE SUBSTRING(f.filepath FROM 7)... would it be the same ? Is the database creation the same, also, for a pgsql database ?

Thanks a million!

I also had to modify theme_download_count_body($node), darko.

I also fixed the Drupal 5.x version.

I used a slightly different approach because the SUBSTRING(f.filepath FROM 7) did not work in all scenarios. I used CONCAT with the File System Path.

Here's the solution I used:

$fileDirectoryPath = file_directory_path() . '/';

... JOIN {files} f ON f.filepath = CONCAT('%s', fd.filename) ... $fileDirectoryPath);

Etc.

Thanks again! Let me know if we'll run into case-sensitive glitches with that fix.

The fix is now in CVS and new versions will be built within the next 24 hours.

I surfed the MySQL doc, but didn't find what I was looking for, i.e. the equivalent of the php function strstr. I wanted to use something like: WHERE the filename is a substring of filepath... Maybe I had to use something like SUBSTRING(f.filepath FROM -COUNT(fd.filename)).