Select 前两次不同的出现和 return rowid

Question

我有一个 table 这样的：

room_id | name    | time
1       | Kate    | 2019-09-18 10:00:00.000
1       | Michael | 2019-09-18 12:00:00.000
1       | George  | 2019-09-18 14:00:00.000
2       | Tom     | 2019-09-17 09:00:00.000
2       | Ben     | 2019-09-17 15:00:00.000
3       | Joerge  | 2019-09-16 17:00:00.000

我想select前N个不同room_ids和最后一个row_id。例如我想 select 前两个不同的 rooms_ids 和 return 一个 row_id 最后一个。结果应该是：

room_id
1
2

row_id 应该用这条记录来标识：

2       | Ben     | 2019-09-17 15:00:00.000

我已经写了我的 SQL 声明，但它不起作用：

SELECT distinct room_id
  FROM (
    SELECT DISTINCT room_id, time,
    rn = ROW_NUMBER() OVER (PARTITION BY room_id ORDER BY room_id, time)
    FROM tab
  ) AS sub
  WHERE rn <= N;

'N' 是我想要 select 有多少不同 room_ids 的值。我也不知道如何return最后一条记录的row_id。

Answer 1

这是一种选择：

SQL>  set ver off
SQL>
SQL> with test (room_id, name, time) as
  2    (select 1, 'Kate'   , to_date('18.09.2019 10:00', 'dd.mm.yyyy hh24:mi') from dual union all
  3     select 1, 'Michael', to_date('18.09.2019 12:00', 'dd.mm.yyyy hh24:mi') from dual union all
  4     select 1, 'George' , to_date('18.09.2019 14:00', 'dd.mm.yyyy hh24:mi') from dual union all
  5     select 2, 'Tom'    , to_date('17.09.2019 09:00', 'dd.mm.yyyy hh24:mi') from dual union all
  6     select 2, 'Ben'    , to_date('17.09.2019 15:00', 'dd.mm.yyyy hh24:mi') from dual union all
  7     select 3, 'Joerge' , to_date('16.09.2019 17:00', 'dd.mm.yyyy hh24:mi') from dual
  8    ),
  9  rn_room as
 10    (select room_id,
 11            row_number() over (order by room_id) rnr
 12     from (select distinct room_id from test)
 13    ),
 14  rn_time as
 15    (select room_id, name, time,
 16            row_number() over (partition by room_id order by time desc) rnt
 17     from test
 18    )
 19  select r.room_id, t.name, t.time
 20  from rn_time t join rn_room r on r.room_id = t.room_id
 21  where r.rnr <= &N
 22    and t.rnt = 1;
Enter value for n: 2

   ROOM_ID NAME    TIME
---------- ------- ----------------
         1 George  2019-09-18 14:00
         2 Ben     2019-09-17 15:00

SQL>

rn_room 对房间进行排序
rn_time 按时间对每个房间的名称进行排序
最后的 select 加入了这两个数据集

Answer 2

你可以这样试试:

with t1 as
(
  select t.*,
         row_number() over (partition by room_id order by room_id, time desc) as rn
    from tab t
) 
select room_id, name, time
from t1
where rn = 1 and room_id = N

Demo

重要的兴趣点是在 row_number() 分析函数中考虑 order by time desc 并为外部查询获取 rn = 1。

Select 前两次不同的出现和 return rowid

Select first two distinct occurrences and return rowid

sql

oracle

analytic-functions