Question

第一个选择是

select user_id, count(*) as count
from users
where referrer IS NOT NULL
group by referrer
order by count DESC

然后根据该查询返回的记录，我需要获取在上述查询中引用用户的用户的日期。

select user_id from users where token = IDS_FROM_LAST_QUERY

我知道我可以使用子查询并说出IN（子查询），但是我试图将子计数保留在子查询中。

所以最后我需要以下信息

user_id, count

Answer 1

select o.user_id user_id, count(*) count
from users o
join users i on o.token = i.user_id
where i.referrer is not null
group by referrer
order by count desc

Answer 2

我会使用CTE（公用表表达式）。 CTE非常便于查找一个人口，然后从CTE中查询相同或略有不同的人口。

WITH Referrer (user_id, count) AS
(
select user_id, count(*) as count
from users
where referrer IS NOT NULL
group by referrer
order by count DESC
)

select 

users.user_id
,Referrer.count

from users 

inner join Referrer.user_id = users.user_id

mysql结合两个选择？

2 个答案: