Question

我是Oracle新手，我需要帮助解决此问题。我有数据样本/记录表，如：

name | datetime
-----------
A    | 20140414 10:00
A    | 20140414 10:30
A    | 20140414 11:00
B    | 20140414 11:30
B    | 20140414 12:00
A    | 20140414 12:30
A    | 20140414 13:00
A    | 20140414 13:30

我需要＆＃34;分组＆＃34; /获取此表单的信息：

name | datetime_from  | datetime_to
----------------------------------
A    | 20140414 10:00 | 20140414 11:00
B    | 20140414 11:30 | 20140414 12:00
A    | 20140414 12:30 | 20140414 13:30

我无法找到任何与此类似的查询解决方案。有人可以帮帮我吗？

注意：我不想使用临时表。

谢谢，帕维尔

Answer 1

SQL> with t (name, datetime) as
  2  (
  3  select 'A', to_date('20140414 10:00','YYYYMMDD HH24:MI') from dual union all
  4  select 'A', to_date('20140414 10:30','YYYYMMDD HH24:MI') from dual union all
  5  select 'A', to_date('20140414 11:00','YYYYMMDD HH24:MI') from dual union all
  6  select 'B', to_date('20140414 11:30','YYYYMMDD HH24:MI') from dual union all
  7  select 'B', to_date('20140414 12:00','YYYYMMDD HH24:MI') from dual union all
  8  select 'A', to_date('20140414 12:30','YYYYMMDD HH24:MI') from dual union all
  9  select 'A', to_date('20140414 13:00','YYYYMMDD HH24:MI') from dual union all
 10  select 'A', to_date('20140414 13:30','YYYYMMDD HH24:MI') from dual
 11  )
 12  select name, min(datetime) datetime_from, max(datetime) datetime_to
 13  from (
 14  select name, datetime,
 15  datetime-(1/48)*(row_number() over(partition by name order by datetime)) dt
 16  from t
 17  )
 18  group by name,dt
 19  order by 2,1
 20  /

N DATETIME_FROM  DATETIME_TO                                                    
- -------------- --------------                                                 
A 20140414 10:00 20140414 11:00                                                 
B 20140414 11:30 20140414 12:00                                                 
A 20140414 12:30 20140414 13:30

Answer 2

您需要找到值相同的时段。 Oracle中最简单的方法是使用lag()函数，一些逻辑和聚合：

select name, min(datetime), max(datetime)
from (select t.*,
             sum(case when name <> prevname then 1 else 0 end) over (order by datetime) as cnt
      from (select t.*, lag(name) over (order by datetime) as prevname
            from table t
           ) t
     ) t
group by name, cnt;

对于给定的datetime值，这个名称已经开启或在该日期时间之前的次数是多少。这标识了＆＃34; constancy＆＃34;的句点，然后用于聚合。

Answer 3

由于9000建议您可以进行如下查询：

select
  a.name, 
  Max(a.datetime),
  Min(b.datetime)
  from
    table a,
    table b
  group by
    a.name
 where a.name = b.name

Oracle查询 - 没有临时表

3 个答案: