如何获取SQL中的行数？

Question

我有一个用户数据库，需要帮助以形成查询。

Table = users
fields = sign_in_count , last_sign_in_at

这是两个查询，可为我提供以下结果：-

select *
from users
where last_sign_in_at >= '2015-08-01' and sign_in_count != ''

给了我131000行

select *
from users
where last_sign_in_at between '2015-05-01' and '2015-09-11' and
      sign_in_count <> ''

给了我203000行

因此在05/15到08/15之间至少登录过一次的72,000个用户中，用户数之间的差异。

我需要知道这72k用户中多次登录的用户数量。

Answer 1

假设users表具有user_id列，下面的查询将起作用。

    select count(user_id) from
     (
    select user_id --replace it with appropriate user identifier column
    from (select * from users
    where last_sign_in_at between '2015-05-01' and '2015-09-11' 
    and sign_in_count != ''
    and user_id not in (select user_id from users 
    where last_sign_in_at >= '2015-08-01' and sign_in_count!='')
     ) t
    group by user_id 
    having count(*) > 1
    ) t1

Answer 2

出什么问题了

Select count(*) from users
where last_sign_in_at  between '2015-05-01' and '2015-08-01'
and sign_in_count > 1

还是我错过了什么？

Answer 3

您要查询所需时间范围内已登录一次以上的用户总数。

但是您只是在查询以查找'2015-05-01'和'2015-08-01'之间的用户总数以及sign_in_count <>''。

Answer 4

考虑采取两个的其余部分where在派生表的条件为count() ：

select count(*) 
from
   (select * from users 
    where last_sign_in_at > '2015-05-01' 
    and last_sign_in_at < '215-08-01'
    and sign_in_count != '')