Question

I have two tables in my database

users table :

+-----------+---------------------+---------+---------+ 
| id        | username            | article | date    |
+-----------+---------------------+---------+---------+
|         1 | max                 |      2  |392185767| 
+-----------+---------------------+---------+---------+
|         2 | alex                |      3  |392333337| 
+-----------+---------------------+---------+---------+

user_specialtys table :

+-----------+---------------------+ 
| spc_id    | user_id             | 
+-----------+---------------------+
|         1 | 1                   |     
+-----------+---------------------+
|         2 | 1                   |     
+-----------+---------------------+
|         3 | 1                   |     
+-----------+---------------------+
|         1 | 2                   |     
+-----------+---------------------+

and there is a third table with each specialty’s id and name which is not important.

As you can see, each user can have many different specialties. Now in each user profile, I want to show a list of similar users (specialty wise!).
Something like:

$user_speciltys_in_array = $user->get_thisUser_specialtys();
$sql = "select `usrname` from users join user_specialtys " .
          "where 'they have most similarity to $user_specialtys_in_array'";

I’m not sure how to do this.

Answer 1

EDIT: Oops! I didn’t read the question carefully! You wanted the usernames! So we need to wrap the query above as a sub-query and use the user_ids to fine the usernames. (Modified below.)

Let’s look for user ID’s by counting the number of matching specialities. We’ll build our SQL query using the $user_speciltys_in_array:

$user_speciltys_in_array = $user->get_thisUser_specialtys();

$sql = 'SELECT users.user_id, score, username FROM (' .
          'SELECT user_id, COUNT(user_id) AS score FROM user_speciality';
$first = 1;
foreach ($user_speciltys_in_array as $spc_id)
{
   $sql = $sql . (($first==1) ? ' WHERE ' : ' AND ';
   $first = 0;
   $sql = $sql . "(user_speciality.spc_id=$spc_id)";
}
$sql = $sql . ' GROUP BY user_id ORDER BY score DESC LIMIT 10) userscores ';
$sql = $sql . ' LEFT JOIN users ON userscores.user_id = users.user_id';

For example, if $user_speciltys_in_array is array(1,2), you’ll get these results:

user_id  score  username
-------  -----  --------
      1      2  max
      2      1  alex
      3      1  thirdguy

I added ORDER BY score DESC and LIMIT 10 to just get the 10 best-matching users, which is often what people look for in a recommendation system (a short list of recommendations).