我试图根据评级百分比获得排名,因此像mysql这样的查询
select c.id , sum((r.value * 20))/ count(r1.pagetypeid) as score, @curRank := @curRank + 1 AS rank from (SELECT @curRank := 0) cr, rating as r
inner join rateelement as r1 on r.elementid = r1.id
inner join ratesubscription as r2 on r.subscriptionid = r2.id
inner join consultant as c on r2.consultantid = c.id
where r1.displayorder not in (6) and r2.agencyid = 38
group by c.id order by score desc
但它返回错误的耙索引
查询出了什么问题?
最佳答案
使用变量进行排名通常会在group by
甚至在最新版本的MySQL中出现order by
问题。因此,使用子查询:
select x.*, (@curRank := @curRank + 1) AS rank
from (select c.id, sum((r.value * 20))/ count(r1.pagetypeid) as score
from rating r inner join
rateelement r1
on r.elementid = r1.id inner join
ratesubscription r2
on r.subscriptionid = r2.id inner join
consultant c
on r2.consultantid = c.id
where r1.displayorder not in (6) and r2.agencyid = 38
group by c.id
order by score desc
) x cross join
(SELECT @curRank := 0) cr;
关于mysql - 在mysql上获得排名会产生错误的排名,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/47826021/