Question

我是sql新手。假设我们有一个这样的表：

+-------+----------+-----------+
|userid | statusid |   date    |
+-------+----------+-----------+
| 1     |  1       | 2018-10-10| 
| 1     |  2       | 2018-10-12|
| 2     |  1       | 2018-09-25|
| 2     |  1       | 2018-10-01|
+-------+----------+-----------+

我需要获取每个用户ID的状态ID，以使其日期尽可能接近给定的日期。说我的给定日期是2018-10-01。我该怎么做？我尝试了各种groupby和partition by，但是没有任何效果。有人可以帮忙吗？

编辑：我的数据库是亚马逊redshift

Answer 1

您可以使用row_number()窗口分析函数，按日期差的绝对值排序。

（请注意，row_number()对MySQL 8-不起作用，因此不使用该功能，而使用abs()函数。）

我不知道您的DBMS

此解决方案适用于Oracle：

with tab(userid, statusid, "date") as
(
 select 1,1,date'2018-10-10' from dual union all
 select 1,2,date'2018-10-12' from dual union all
 select 2,1,date'2018-09-25' from dual union all
 select 2,1,date'2018-10-02' from dual
)
select tt.userid, tt.statusid, tt."date"
  from
(
select t.userid, t.statusid , t."date",
       row_number() over (partition by t.userid 
                          order by abs("date" - date'2018-10-01')) as rn
  from tab t
) tt
where tt.rn = 1

Demo for Oracle

此解决方案适用于SQL Server：

with tab([userid], [statusid], [date]) as
(
 select 1,1,'2018-10-10' union all
 select 1,2,'2018-10-12' union all
 select 2,1,'2018-09-25' union all
 select 2,1,'2018-10-02' 
)
select tt.[userid], tt.[statusid], tt.[date]
  from
(
select t.[userid], t.[statusid] , t.[date], 
       row_number() over (partition by t.[userid] 
                          order by abs(datediff(day,[date],'2018-10-01'))) as rn
  from tab t
) tt
where tt.rn = 1

Demo for SQL Server

该解决方案适用于My SQL：

select tt.userid, tt.statusid, tt.date
  from
  (
   select t.userid, t.statusid , t.date,
          @rn := if(@iter = t.userid, @rn + 1, 1) as rn,
          @iter := t.userid, 
          abs(date - date'2018-10-01') as df
     from tab t
     join (select @iter := 0, @rn := 0) as q_iter
    order by t.userid, abs(date - date'2018-10-01') 
  ) tt
where tt.rn = 1

Demo for My SQL

此解决方案适用于PostGRES：

with tab(userid, statusid, date) as
(
 select 1,1,'2018-10-10' union all
 select 1,2,'2018-10-12' union all
 select 2,1,'2018-09-25' union all
 select 2,1,'2018-10-02' 
)
select tt.userid, tt.statusid, tt.date
  from
(
select t.userid, t.statusid , t.date, 
       row_number() over (partition by t.userid
                          order by abs(date::date-'2018-10-01'::date)) as rn
  from tab t
) tt
where tt.rn = 1

Demo for PostGRESql

Answer 2

通常，对于这种类型的问题，您希望日期在给定日期或之前。

如果是这样：

select t.*
from t
where t.date = (select max(t2.date)
                from t t2
                where t2.userid = t.userid and t2.date <= '2018-10-01'
               );

查找每个组中最接近给定日期的记录-SQL

2 个答案: