好像你還需要一個集合體。使用您
cnt
價值和
GROUP BY placeId, userVotedId
的
MAX()
合計:
SELECT placeId, userVotedId, max(cnt)
FROM
(
SELECT uvo.userVotedId, p.placeId, count(*) as cnt
FROM users as u
INNER JOIN users_votes as uvo
ON u.userId = uvo.userVotedId
INNER JOIN places as p
ON p.placeId = uvo.placeId
GROUP BY userVotedId, placeId
) AS RESULT
GROUP BY placeId, userVotedId
注:我改變了你的查詢中使用
JOIN
語法,而不是表之間的逗號。
編輯,根據您的評論下面應該工作:
select total.uservotedid,
total.placeid,
total.cnt
from
(
SELECT uvo.userVotedId, p.placeId, count(*) as cnt
FROM users as u
INNER JOIN users_votes as uvo
ON u.userId = uvo.userVotedId
INNER JOIN places as p
ON p.placeId = uvo.placeId
GROUP BY userVotedId, placeId
) total
inner join
(
select max(cnt) Mx, placeid
from
(
SELECT uvo.userVotedId, p.placeId, count(*) as cnt
FROM users as u
INNER JOIN users_votes as uvo
ON u.userId = uvo.userVotedId
INNER JOIN places as p
ON p.placeId = uvo.placeId
GROUP BY userVotedId, placeId
) mx
group by placeid
) src
on total.placeid = src.placeid
and total.cnt = src.mx
見SQL Fiddle with Demo
結果是:
| USERVOTEDID | PLACEID | CNT |
-------------------------------
| 65 | 11 | 1 |
| 67 | 13 | 1 |
| 67 | 25 | 1 |
| 67 | 51 | 2 |
編輯#2,如果你想有一個隨機數如果有聯繫返回,那麼你可以使用用戶變量:
select uservotedid,
placeid,
cnt
from
(
select total.uservotedid,
total.placeid,
total.cnt,
@rownum := case when @prev = total.placeid then @rownum+1 else 1 end rownum,
@prev := total.placeid pplaceid
from
(
SELECT uvo.userVotedId, p.placeId, count(*) as cnt
FROM users as u
INNER JOIN users_votes as uvo
ON u.userId = uvo.userVotedId
INNER JOIN places as p
ON p.placeId = uvo.placeId
GROUP BY userVotedId, placeId
) total
inner join
(
select max(cnt) Mx, placeid
from
(
SELECT uvo.userVotedId, p.placeId, count(*) as cnt
FROM users as u
INNER JOIN users_votes as uvo
ON u.userId = uvo.userVotedId
INNER JOIN places as p
ON p.placeId = uvo.placeId
GROUP BY userVotedId, placeId
) mx
group by placeid
) src
on total.placeid = src.placeid
and total.cnt = src.mx
order by total.placeid, total.uservotedid
) src
where rownum = 1
order by placeid, uservotedid
見SQL Fiddle with Demo
是相同的。所以,你的查詢工作! – cha
@cha我認爲是一樣的,但重點是消除雙重placeid = 51.可能有更多的可以指出的方式來指出這一點,而不是四個顛簸的紅色框;-) – KekuSemau
如果您試圖獲得頂部如果您已經根據結果的placeID進一步分組,則基於cnt的行比第一次出現時更難。在SQL Server中,我認爲你可以使用APPLY來實現這一點,但是在MySQL中不支持(?)。 – Matthew