Question

我有一张桌子

Create table t (
  employee varchar2(10), 
  job varchar2(5), 
  start_date date, 
  end_date date, 
  FTE number
);

包含此数据：

insert into t values (1111,1,19-May-2008,18-May-2010,100);
insert into t values (1111,1,19-May-2010,30-Jun-2010,100);
insert into t values (1111,1,01-Jul-2010,13-Feb-2011,100);
insert into t values (1111,1,14-Feb-2011,13-Feb-2012,100);
insert into t values (1111,1,14-Feb-2012,13-Feb-2013,100);
insert into t values (1111,1,14-Feb-2013,27-Nov-2013,100);
insert into t values (1111,1,15-Nov-2013,27-Nov-2013,100);
insert into t values (1111,1,28-Nov-2013,24-Aug-2014,60);
insert into t values (1111,1,25-Aug-2014,24-Aug-2015,60);
insert into t values (1111,1,15-Nov-2014,24-Aug-2015,60);
insert into t values (1111,1,25-Aug-2015,14-Nov-2015,68.571);
insert into t values (1111,1,15-Nov-2015,31-Mar-2016,68.571);
insert into t values (1111,1,01-Apr-2016,11-Sep-2016,68.571);
insert into t values (1111,1,12-Sep-2016,29-Sep-2016,72.857);
insert into t values (1111,1,30-Sep-2016,10-Nov-2016,91.429);
insert into t values (1111,1,11-Nov-2016,14-Nov-2016,100);
insert into t values (1111,1,15-Nov-2016,23-Apr-2017,100);
insert into t values (1111,1,24-Apr-2017,14-Jan-2018,60);
insert into t values (1111,1,15-Jan-2018,31-May-2019,71.429);
insert into t values (1111,1,01-Jun-2019,31-Dec-2049,100);

我希望按照正确的开始日期和结束日期对FTE数据进行分组显示，以便它显示为

1111    1   19/05/2008  27/11/2013  100 
1111    1   28/11/2013  24/08/2015  60
1111    1   25/08/2015  11/09/2016  68.571
1111    1   12/09/2016  29/09/2016  72.857
1111    1   30/09/2016  10/11/2016  91.429 
1111    1   11/11/2016  23/04/2017  100
1111    1   24/04/2017  14/01/2018  60 
1111    1   15/01/2018  31/05/2019  71.429
1111    1   1/06/2019   31/12/2049  100

我该怎么做？我试图做的是列出FTE发生变化的时间，以及发生变化时的相应开始日期和结束日期。

Answer 1

这是Tabibitosan方法的直接应用，用于在序列中查找“连续”行的集合。在应用标准聚合之前，两个分析row_number()函数的差异会创建所需的附加分组标记。

select   employee, job, min(start_date) as start_date, max(end_date) as end_date, FTE
from     ( select employee, job, start_date, end_date, FTE,
                  row_number() over (partition by employee, job      order by start_date) -
                  row_number() over (partition by employee, job, FTE order by start_date) 
                                                                                    as grp
           from t
         )
group by employee, job, FTE, grp
order by employee, start_date
;


EMPLOYEE   JOB   START_DATE  END_DATE           FTE
---------- ----- ----------- ----------- ----------
1111       1     19-May-2008 27-Nov-2013        100
1111       1     28-Nov-2013 24-Aug-2015         60
1111       1     25-Aug-2015 11-Sep-2016     68.571
1111       1     12-Sep-2016 29-Sep-2016     72.857
1111       1     30-Sep-2016 10-Nov-2016     91.429
1111       1     11-Nov-2016 23-Apr-2017        100
1111       1     24-Apr-2017 14-Jan-2018         60
1111       1     15-Jan-2018 31-May-2019     71.429
1111       1     01-Jun-2019 31-Dec-2049        100

9 rows selected.

在oracle sql中按日期分组的数据

1 个答案: