ISSUE SUMMARY:
Hi there,
I can see that friends suggestions module mostly show inactive user, there are any way to avoid that, because is not cool, in my case the only one that show is inactive user, also could this module show random users and not the same users all the time as facebook do, in my case in my personal account on my site i can see only inactive user so as i can not add them as friend they continue showing the same users, what facebook do is that when you refresh the page, show some other nearly diferents, and also show an (x) to remove the user to not appear, could be great adding this feature.
Best Regards

Romulo Ruiz
STEPS TO REPLICATE:
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RESULT
Only inactive user are shown
EXPECTED RESULT
Show only active users
If i don't send friend request display other users and not all the time the same
Add a (x) in the right corner to avoid this user to not show again
make mutual friends linkable in order user can click to see mutual friends as facebook do.
BROWSER
chrome

Hi Romulo,

Thank you for reporting this is valid bug.

Regarding behavior like FB we cant promise will create it soon

cheers!

Hi Romulo,

Here is the patch
open this file modules/mod_community_friendssuggestions/helper.php, find this :

$query = "SELECT user_id FROM ".$db->quoteName('#__community_fields_values')
                         ." WHERE ".$additionalQuery." GROUP BY user_id";

$query = "SELECT v.user_id FROM ".$db->quoteName('#__community_fields_values')
                                 ." as v LEFT JOIN ".$db->quoteName('#__users')." AS u ON v.user_id=u.id"
                                 ." WHERE u.".$db->quoteName('block')."=".$db->quote(0)
                                 ." AND (".$additionalQuery.") GROUP BY v.user_id";

." WHERE (".$additionalQuery.") GROUP BY user_id HAVING COUNT(field_id) = ".count($matchField);

." as v LEFT JOIN ".$db->quoteName('#__users')." AS u ON v.user_id=u.id"
                                 ." WHERE u.".$db->quoteName('block')."=".$db->quote(0)
                                 ." AND (".$additionalQuery.") GROUP BY user_id HAVING COUNT(field_id) = ".count($matchField);

Great! thank you very much, i appreciate that.