Question

如何在MySQL中使用SQL查找特定用户每天登录的信息？

我有一张这样的桌子。

+----------+---------------------+---------+
| login_id | login_date          | user_id |
+----------+---------------------+---------+
|     8494 | 2016-03-01 00:00:00 |   20703 |
|    22403 | 2016-03-01 00:00:00 |   53473 |
|    23965 | 2016-03-01 00:00:00 |   79722 |
|    30173 | 2016-03-01 00:00:00 |   36396 |
|    34928 | 2016-03-02 00:00:00 |   20703 |
|    38740 | 2016-03-02 00:00:00 |   15758 |
|    42769 | 2016-03-02 00:00:00 |   79722 |
|    44364 | 2016-03-02 00:00:00 |   79722 |
|    45440 | 2016-03-03 00:00:00 |   20703 |
|    49050 | 2016-03-03 00:00:00 |   36396 |
|    50273 | 2016-03-03 00:00:00 |   79722 |
|    50344 | 2016-03-04 00:00:00 |   20703 |
|    51360 | 2016-03-04 00:00:00 |   20703 |
|    54404 | 2016-03-04 00:00:00 |   53473 |
|    61533 | 2016-03-04 00:00:00 |   79722 |
|    72852 | 2016-03-05 00:00:00 |   20703 |
|    74546 | 2016-03-05 00:00:00 |   38289 |
|    76487 | 2016-03-05 00:00:00 |   62529 |
|    82439 | 2016-03-05 00:00:00 |   36396 |
|    90006 | 2016-03-05 00:00:00 |   36396 |
|    90404 | 2016-03-06 00:00:00 |   20703 |
+----------+---------------------+---------+

答案就像是

user_id, consecutive
20703, 'YES'
53473, 'NO'

每天从2016-03-01 - 2016-03-06登录20703登录，但53473只登录了两次。范围将从整个表格的开始到结束。

Answer 1

嗯，你可以这样做：

select user_id,
       ( count(distinct login_date) = cnt_ld ) as everyday_flag
from t cross join
     (select count(distinct login_date) as cnt_ld
      from t
     ) tt
group by user_id, tt.cnt_ld;

请注意，这会将用户与表中的所有日期进行比较。如果有一天没有登录而您仍想要包含那一天，那么请使用最大值减去最小值进行比较。

这也会返回一个“布尔”标志（实际为0或1）。您可以使用case将其转换为'YES' / 'NO'，但我发现该号码更易于使用。

Answer 2

我在tbl中获取最大和最小日期然后如果每个用户ID的计数等于最大和最小日期之间的日期（+1天），那么它是连续登录否则否。

select t1.user_id, 
    case when count(*)=datediff(t2.maxdt, t2.mindt)+1
    then 'YES' else 'NO'  end as consecutive
from tbl t1
join (select max(login_date) as maxdt,
             min(login_date) as mindt
          from tbl) t2
group by t1.user_id
order by t1.user_id;

Answer 3

我只需使用带有group by表达式的case来查找连续登录

select user_id, 
       case when count(distinct login_date) <> datediff(max(login_date), min(login_date))+1
            then 'No' else 'Yes' end consecutive
from table t
group by user_id

如果您要查找具有开始日期到结束日期的连续登录的用户，那么您也可以通过cross joins进行此操作。

select user_id, case when counts <> total_counts then 'No' else 'Yes' end consecutive 
from (select t.user_id, count(distinct login_date) counts
      from table t 
      group by t.user_id )t cross join (
select count(distinct login_date) total_counts from table) tt

了解特定用户是否每天登录

3 个答案: