Question

我有一个包含4行特殊情况的表...并且他们有出席

like : formate(mm/dd/rrrr)

2列名称

atten date :                   |          strtdate     
1) 4/02/2016                   |          1) 4/02/2016
2) 4/03/2016                   |          2) 4/02/2016
3) 4/05/2016                   |          3) 4/02/2016
4) 4/ 07/2016                  |          4) 4/02/2016
5) 4/08/2016                   |          5) 4/02/2016

这里我需要这样的o / p栏

fromdt                            |  todate
1) 4/02/2016                      |  4/03/2016
2) 4/03/2016                      |  4/05/2016
3) 4/05/2016                      |  4/05/2016
4)  4/07/2016                     | 4/08/2016

Answer 1

您可以使用LEAD()分析函数

来完成此操作

with sample_data as (select 1 emp_id, to_date('02/04/2016', 'dd/mm/yyyy') atten_date, to_date('02/04/2016', 'dd/mm/yyyy') strtdate from dual union all
                     select 1 emp_id, to_date('03/04/2016', 'dd/mm/yyyy') atten_date, to_date('02/04/2016', 'dd/mm/yyyy') strtdate from dual union all
                     select 1 emp_id, to_date('05/04/2016', 'dd/mm/yyyy') atten_date, to_date('02/04/2016', 'dd/mm/yyyy') strtdate from dual union all
                     select 1 emp_id, to_date('07/04/2016', 'dd/mm/yyyy') atten_date, to_date('02/04/2016', 'dd/mm/yyyy') strtdate from dual union all
                     select 1 emp_id, to_date('08/04/2016', 'dd/mm/yyyy') atten_date, to_date('02/04/2016', 'dd/mm/yyyy') strtdate from dual)
-- end of subquery mimicking a table called "sample_data" with data in it. See SQL below:
select emp_id,
       fromdt,
       todate
from   (select emp_id,
               atten_date fromdt,
               lead(atten_date) over (partition by emp_id order by atten_date) todate
        from   sample_data)
where  todate is not null;

    EMP_ID FROMDT     TODATE    
---------- ---------- ----------
         1 04/02/2016 04/03/2016
         1 04/03/2016 04/05/2016
         1 04/05/2016 04/07/2016
         1 04/07/2016 04/08/2016

如果您需要显示行号，那么您可以使用ROW_NUMBER()分析函数来执行此操作，如果该信息尚未作为源数据的一部分提供。

Answer 2

如果完全理解您的要求，此解决方案可能适合您：

with sample_data as
(
select 10 emp_id, to_date('4/02/2016','mm/dd/yyyy') attendate, to_date('4/02/2016','mm/dd/yyyy') strtdate from dual union all
select 10 emp_id, to_date('4/03/2016','mm/dd/yyyy') attendate, to_date('4/02/2016','mm/dd/yyyy') strtdate from dual union all
select 10 emp_id, to_date('4/05/2016','mm/dd/yyyy') attendate, to_date('4/02/2016','mm/dd/yyyy') strtdate from dual union all
select 10 emp_id, to_date('4/07/2016','mm/dd/yyyy') attendate, to_date('4/02/2016','mm/dd/yyyy') strtdate from dual union all
select 10 emp_id, to_date('4/08/2016','mm/dd/yyyy') attendate, to_date('4/02/2016','mm/dd/yyyy') strtdate from dual 
)
, marked_data as
(
select emp_id
     , attendate
     , strtdate
     , row_number() over (partition by emp_id, strtdate order by attendate)     as is_from
     , row_number() over (partition by emp_id, strtdate order by attendate) - 1 as is_to
  from sample_data
)  
select md1.emp_id
     , md1.attendate as fromdt
     , md2.attendate as todt
     , md1.strtdate
  from marked_data md1
  join marked_data md2
    on md1.emp_id = md2.emp_id
   and md1.strtdate = md2.strtdate
   and md1.is_from = md2.is_to

你将得到结果：

    EMP_ID  FROMDT  TODT    STRTDATE
1   10  4/2/2016    4/3/2016    4/2/2016
2   10  4/3/2016    4/5/2016    4/2/2016
3   10  4/5/2016    4/7/2016    4/2/2016
4   10  4/7/2016    4/8/2016    4/2/2016

按定期间隔约会参加日期

2 个答案: