>>I have a table called friends with a user_id and a friend_id (both of these relate to an id in a users table).
>>For each friend relationship there are two rows. There are currently ONLY reciprocal relationships. So if user ids 1 and 2 are friends there will be two rows (1,2) and (2,1).
>>For 2 arbitrary ids, I need a query to get two pieced of data:
>> * Are the two users friends?
>> * How many friends do the two users have in common.
>>Is there a way to do this with one query? Currently I've only been able to figure out how to do it with two.
>>SELECT
>> EXISTS(
>> SELECT 1
>> FROM friends
>> WHERE user_id = 166324 AND friend_id = 166325) AS friends,
>> (SELECT COUNT(1)
>> FROM friends f1 JOIN friends f2 ON f1.friend_id = f2.friend_id
>> WHERE f1.user_id = 166324 AND f2.user_id = 166325) AS mutual;
>>I'm wondering if there is a better way to do this using only one query. I've tried a couple of GROUP BY approaches but they haven't worked.
>>
>
>Hi,
>this should do the job, but requires an aditional check constraint user_id <> friend_id to be on the safe side:
>
>
>SELECT count(case when c1=1 then true end)=1 as are_friend,
> count(*)-1 as common_friends
>FROM
> (
> SELECT count(*) as c1
> FROM friends
> WHERE user_id IN (USER1, USER2)
> GROUP BY case when user_id = USER2 then USER1 else USER1 end,
> friend_id
> HAVING COUNT (*) =2
> OR COUNT(case when friend_id =USER1 then true end)=1
> ) q1
>
fix:
SELECT count(case when c1=1 then true end)=1 as are_friend,
count(case when c1=2 then true end) as common_friends
FROM
(
SELECT count(*) as c1
FROM friends
WHERE user_id IN (USER1, USER2)
GROUP BY case when user_id = USER2 then USER1 else USER1 end,
friend_id
HAVING COUNT (*) =2
OR COUNT(case when friend_id =USER1 then true end)=1
) q1
regards,
Marc Mamin