我有兩個疑問:MySQL的:平均,並加入兩個查詢
SELECT
s.id AS id,
g.id AS group_id,
g.nazwa AS group_name,
s.nazwa AS station_name,
s.szerokosc AS szerokosc,
s.dlugosc AS dlugosc,
s.95 as p95,
s.98 as p98,
s.Diesel as diesel,
s.DieselTIR as dieseltir,
s.Super98 as s98,
s.SuperDiesel as sdiesel,
s.LPG as lpg,
s.ulica as ulica,
s.kodPocztowy as kod_pocztowy,
s.miasto as miasto,
w.id as wojewodztwo_id,
w.nazwa as wojewodzto_nazwa,
k.id as kraj_id,
k.kod as kraj_kod,
k.nazwa as kraj_nazwa,
s.data as date_mod,
s.active as active
FROM stacje_main s
JOIN stacje_grups g ON (s.grupa=g.id)
JOIN wojewodztwa w ON (s.wojewodztwo=w.id)
JOIN kraje k ON (w.kraj=k.id)
WHERE s.data > 0;
和
SELECT
AVG(rr.vote) as average,
COUNT(rr.station_id) counter
FROM stacje_ratings rr
GROUP BY rr.station_id;
在第二個查詢並不是所有的ID(station_id)都存在,有時加倍。 用id加入station_id,併爲每個id給出rate的平均值。
,當沒有速率,在平均和計數器有問題的值必須是0。
當我合併這些查詢我看到只有該ID的問題,即具有存在station_id。 但我想看到所有。