2017-06-21 69 views

回答

0

我没有尝试,但我认为它可能工作

SELECT * from Subscription 
WHERE id=(SELECT subscription_id 
FROM Show_has_Subscription 
GROUP BY subscription_id 
ORDER BY COUNT(subscription_id) DESC 
LIMIT 1); 
+0

完美地工作,谢谢 – Westwick

0

你可以试试这个查询:

select s.id, count(ss.show_id) as showcount 
from subscription s, subscription_show ss 
where s.id = ss.subcription_id 
group by s.id, ss.show_id 
order by showcount desc 
limit 1 
0

最简单的解决方案似乎是一个不需要join或子查询:

select ss.subcription_id, count(*) as showcount 
from subscription_show ss 
group by ss.subcription_id 
order by showcount desc 
limit 1;