Question

我有一个用户数据库，我需要帮助才能形成查询。

Table = users
fields = sign_in_count , last_sign_in_at

这两个查询给出了以下结果： -

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至08/15之间至少登录过一次的72,000名用户中的用户差异。

我需要知道这些不止一次登录的72k用户的用户数。

Answer 1

如果users表格有user_id列，则下面的查询可以正常工作。

    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

有什么问题

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

或者我错过了什么？

Answer 3

您想要查询已登录多次的所需时间范围内的用户总数。

但您只是要查询“2015-05-01”和“2015-08-01”之间的用户总数以及sign_in_count＆lt;＆gt; ''。

Answer 4

考虑在where的派生表中考虑两个count()条件的剩余部分：

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 != '')

如何获取SQL中的行数？

4 个答案: