Thank you for the explanation. I have rearranged my query and it works now
(surprisingly fast too) -
On Mon, Apr 23, 2018 at 9:58 PM, David G. Johnston <
david.g.johns...@gmail.com> wrote:
> On Mon, Apr 23, 2018 at 12:47 PM, Alexander Farber <
> alexander.far...@gmail.com> wrote:
>
>>
On Mon, Apr 23, 2018 at 12:47 PM, Alexander Farber <
alexander.far...@gmail.com> wrote:
> SELECT
> u.elo,
> AVG(c.played - c.prev_played) AS avg_time_per_move,
> (SELECT ROUND(AVG(score), 1) FROM words_moves
>
Good evening,
On Mon, Apr 23, 2018 at 12:56 PM, Alexander Farber <
alexander.far...@gmail.com> wrote:
>
> On Mon, Apr 23, 2018 at 12:15 PM, Daniel Verite
> wrote:
>
>>
>> You may use a correlated subquery in the SELECT clause,
>>
>
>SELECT
> u.elo,
>
Thank you, Daniel -
On Mon, Apr 23, 2018 at 12:15 PM, Daniel Verite
wrote:
>
> You may use a correlated subquery in the SELECT clause,
> like this:
>
> SELECT
>u.elo,
>u.uid,
>(SELECT AVG(score) FROM words_moves WHERE uid=u.uid),
>s.given,
>s.photo
>
thi
Alexander Farber wrote:
> Here is such a query for the best player
>
> # SELECT AVG(score) FROM words_moves WHERE uid = 1201;
> avg
> -
> 18.4803525523319868
>
> However I am not sure, how to "marry" the 2 queries?
>
> I have tried to add words_moves through
Hello,
in PostgreSQL 10.3 I run the following query to find top 10 players with
the best ELO rating:
# SELECT
u.elo,
u.uid,
s.given,
s.photo
FROM words_users u
JOIN words_social s USING (uid)
WHERE u.elo > 1500
-- take the most recent record from words_social (sto
