Question

我正在使用具有以下结构的表格，我试图在两天之间找到TAT（周转时间）。但是，重叠的日子，我无法找到实际的TAT

Appln No        Start Date  End Date
1001009     01-10-15    06-10-15
1001009     02-10-15    04-10-15
1001009     03-10-15    04-10-15
1001009     03-10-15    05-10-15
1001009     04-10-15    07-10-15
1001009     09-10-15    10-10-15
1001009     12-10-15    16-10-15
1001009     14-10-15    17-10-15

从上面的示例数据中删除重叠日期后，输出将采用以下格式 -

Appln No    Start Date  End Date
1001009     01-10-15    07-10-15
1001009     09-10-15    10-10-15
1001009     12-10-15    17-10-15

由于我是sql的初学者并使用oracle sql developer，我发现很难将上述逻辑编写到代码中。欢迎任何关于这个问题的建议：）

Answer 1

试试这个：

select t1.* from myTable t1
inner join myTable t2
on t2.StartDate > t1.StartDate and t2.StartDate < t1.EndDate

Answer 2

这是一个棘手的查询。您需要通过分配分组ID来识别重叠的组。一种方法是找到重叠组的起始位置，然后累计每条记录之间的起始次数。

以下假设您的表具有主键（由于缺少更好的名称而称为id）。

这样可以聚合以获得您想要的内容：

select ApplnNo, min(start), max(end)
from (select t.*,
             sum(IsGroupStart) over (partition by ApplnNo order by start) as grp
      from (select t.*,
                   (case when exists (select 1
                                      from t t2
                                      where t2.end >= t.start and t2.start <= t.end and
                                            t2.id <> t.id
                                     )
                         then 0 else 1
                    end) as IsGroupStart
            from t
           ) t
      ) t
group by ApplnNo, grp;

有一些细微差别。 exists的确切最内部子查询取决于您如何定义重叠。这包括在开始或结束时甚至有一天重叠。

Answer 3

更多这是一个棘手的任务，因为你不能信任任何间隔的顺序。我通过删除子区间来攻击它（在其他区间完全覆盖的区间）。在此之后，我可以按照START_DATE定义的顺序查看感知间隔是否与下一个间隔重叠并应用标准分组机制。

 with subs as (
 /* first remove all intervals that are subsets of other intervals */
 select * from tst t1
 where NOT exists (select null from tst t2 where t2.start_date < t1.start_date and t1.end_date < t2.end_date)
 ),overlap as (
 select APPLN_NO, START_DATE, END_DATE,
 case when (nvl(lag(END_DATE) over (partition by APPLN_NO order by START_DATE),START_DATE-1)  < START_DATE) then 
      row_number() over (partition by APPLN_NO order by START_DATE) end grp
 from subs),
 overlap2 as (
 select 
 APPLN_NO, START_DATE, END_DATE, GRP,
 last_value(grp ignore nulls) over (partition by APPLN_NO order by START_DATE) as grp2
 from overlap)
 select 
   APPLN_NO,   min(START_DATE) START_DATE, max(END_DATE) END_DATE
 from overlap2
 group by APPLN_NO, grp2
 order by 1,2
 ;

在此处检查查询我的设置

 drop table tst ;
 create table tst 
 (appln_no number,
 start_date date,
 end_date date);

 insert into tst values (1001009, to_date('01-10-15','dd-mm-rr'),to_date('06-10-15','dd-mm-rr'));
 insert into tst values (1001009, to_date('02-10-15','dd-mm-rr'),to_date('04-10-15','dd-mm-rr'));
 insert into tst values (1001009, to_date('03-10-15','dd-mm-rr'),to_date('04-10-15','dd-mm-rr'));
 insert into tst values (1001009, to_date('03-10-15','dd-mm-rr'),to_date('05-10-15','dd-mm-rr'));
 insert into tst values (1001009, to_date('04-10-15','dd-mm-rr'),to_date('07-10-15','dd-mm-rr'));
 insert into tst values (1001009, to_date('09-10-15','dd-mm-rr'),to_date('10-10-15','dd-mm-rr'));
 insert into tst values (1001009, to_date('12-10-15','dd-mm-rr'),to_date('16-10-15','dd-mm-rr'));
 insert into tst values (1001009, to_date('13-10-15','dd-mm-rr'),to_date('14-10-15','dd-mm-rr')); /* this is added to make it more interesting */
 insert into tst values (1001009, to_date('15-10-15','dd-mm-rr'),to_date('17-10-15','dd-mm-rr'));

给

    APPLN_NO START_DATE          END_DATE          
 ---------- ------------------- -------------------
    1001009 01.10.2015 00:00:00 07.10.2015 00:00:00 
    1001009 09.10.2015 00:00:00 10.10.2015 00:00:00 
    1001009 12.10.2015 00:00:00 17.10.2015 00:00:00

正如所料。

通过比较两个日期列来确定实际的TAT（周转时间）

3 个答案: