mysql - SQL query to order by the top averages of previous data -

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given table data these columns (other columns too, but, these important ones):

gameid awayscore homescore 1      3         8 2      4         2 3      4         9 4      15        5 5      0         8 6      4         10 ...

what looking average margin of victory on previous 3 games (assuming gameid order played), so, on game 6, should (6+8+10=24, average 8), on game 5 (8+10+5=23, average 7.666), etc. , show weeks 3-game average highest.

i tried this:

select g.gameid, (select avg(scores.ave) (select abs(awayscore-homescore) ave games gs (gs.gameid<=g.gameid) order gameid desc limit 3) scores) margin `games` g group g.gameid order margin desc  limit 10;

but "unknown column 'g.gameid' in 'where clause'", assume because subquery of subquery(?), , loses reference? @ loss of how structure otherwise though work correctly (if possible @ all).

try query

select g.gameid,    (select avg(abs(awayscore-homescore))       games gs        (gs.gameid<=g.gameid)              , (gs.gameid>g.gameid-3)     ) margin `games` g order margin desc  limit 10;

sqlfiddle

update if want use gamedate (datetime) order game, can create game order (gorder) number based on sort everytime select games below query.

select g.gameid,    (select avg(abs(awayscore-homescore))               (select awayscore,homescore,@rank:=@rank+1 gorder         games,(select @rank:=0)r         order gamedate asc         )gs        (gs.gorder <= g.gorder)              , (gs.gorder > g.gorder-3)     ) margin (select games.gameid,@rank2:=@rank2+1 gorder         games,(select @rank2:=0)r2         order gamedate asc) g order margin desc  limit 10;

sqlfiddle


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