Efficient friend suggestion sql query without using php - php

Question

I have a members table in which every site member is the unique id.
eg
id firstname secondname emailaddress country city gender
the second table is a friends table with the following structure
id meid friendid date
what query would i use to get friend suggestions of a particular user based on mutual friends and sorted accordingly. before i was using php to loop thru and collect mutual friends but as the site grew, php started misbehaving and running out of memory.
This is the function i was using
//-----------------------------------------------
function getFriendSuggestions($id)
{
$friendids=getFriendIdArray($id); //returns list of your friends
$networkids=getNetworkIdArray($id);//returns list of all members in your network(friends and their friends)
$diff=array_merge(array(),array_diff($networkids,$friendids));
$diff_mutual=array();
$diff_mutual_total=array();
for ($n=0;$n<count($diff);$n++)
{
$ff=getFriendIdArray($diff[$n]);
$mf=array_merge(array(),array_intersect($ff,$friendids));
$diff_mutual[]=$mf;
$diff_mutual_total[]=count($mf);
}
$diff=array_merge(array(),$diff);
$diff_mutual=array_merge(array(),$diff_mutual);
$diff_mutual_total=array_merge(array(),$diff_mutual_total);
$w=$diff_mutual_total;
arsort($w);
$d=array();
$dm=array();
foreach ($w as $key => $value)
{
$d[]=$diff[$key];
$dm[]=$diff_mutual[$key];
}
$cv=array($d,$dm);
return $cv;
}

Answer 1

We say that B is possibly-friend of A if there is a large number of entries of the kind
(B, someguy) (someguy, A)
in the database, and there is no (B, A) entry.
We know the ID of user A and let it be AID. Then we can do:
SELECT b.meid, COUNT(b.meid) AS incommon
FROM friends AS b
JOIN friends AS a ON (b.friendid = a.meid AND a.friendid = AID)
GROUP BY b.meid ORDER BY incommon DESC;
This will tell us all "possible friends" of A, including those that are
already friends of A. Then we have to exclude them:
SELECT maybe.meid, maybe.incommon FROM
( SELECT b.meid, COUNT(b.meid) AS incommon
FROM friends AS b
JOIN friends AS a ON (b.friendid = a.meid AND a.friendid = AID)
GROUP BY b.meid ORDER BY incommon DESC ) AS maybe
LEFT JOIN friends AS already ON (maybe.friendid = already.meid AND already.friendid = AID) WHERE already.friendid IS NULL;
Then we need to populate the rest of the fields:
SELECT members.firstname, members.secondname, maybe.incommon FROM
( SELECT b.meid, COUNT(b.meid) AS incommon
FROM friends AS b
JOIN friends AS a ON (b.friendid = a.meid AND a.friendid = AID)
GROUP BY b.meid ORDER BY incommon DESC ) AS maybe
LEFT JOIN friends AS already ON (maybe.friendid = already.meid AND already.friendid = AID)
JOIN members ON (members.id = maybe.meid)
WHERE already.friendid IS NULL;
This will return friend suggestions for AID, including how many people in common it has for every choice (e.g. "John Doe (15 friends in common)", etc.).

Answer 2

You want "friends of my friends, but who are not also my friends".
SELECT
me.id AS member_id,
their_friends.friendid AS suggested_friend_id,
COUNT(*) AS friends_in_common
FROM
members AS me
INNER JOIN
friends_map AS my_friends
ON my_friends.meid = me.id
INNER JOIN
friends_map AS their_friends
ON their_friends.meid = my_friends.friendid
LEFT JOIN
friends_map AS friends_with_me
ON friends_with_me.meid = their_friends.friendid
AND friends_with_me.friendid = me.id
WHERE
friends_with_me.meid IS NULL
GROUP BY
me.id,
their_friends.friendid