Question

请您查看以下任务？

我有一个表DATA，其结构如下：

CREATE TABLE DATA
(
    EMPLOYEE nvarchar(50) NOT NULL,
    ABSENCE_START nvarchar(50) NOT NULL,
    ABSENCE_END nvarchar(50) NOT NULL
)

它包含以下数据：

insert into DATA (EMPLOYEE, ABSENCE_START, ABSENCE_END) 
values ('emp01','21/06/2017','22/06/2017'),
       ('emp01','24/06/2017','27/06/2017'),
       ('emp02','22/06/2017','24/06/2017'),
       ('emp02','22/06/2017','27/06/2017')

正如您所看到的那样，＆＃39; emp02＆＃39;员工我们有两个缺席期，其中一个（第4行）完全包含另一个缺席期（第3行）。

我需要将这两行合并为一行（第4行）。

预期结果将是：

EMPLOYEE   ABSENCE_START    ABSENSE
---------------------------------------
emp01      21/06/2017       22/06/2017
emp02      24/06/2017       27/06/2017
emp01      22/06/2017       27/06/2017

提前谢谢！

Answer 1

如果您要查找的唯一重叠修正是当一行完全包含另一行，并且您没有唯一的行标识符时，您可以使用not exists()，如下所示：

set dateformat dmy;
;with cte as (
  select 
      employee
    , absence_start= convert(date,absence_start)
    , absence_end  = convert(date,absence_end)
  from data
)    
select *
from cte d
where not exists (
    select 1
    from cte i
    where i.employee=d.employee 
      and i.absence_start <= d.absence_start
      and i.absence_end >= d.absence_end
      and (i.absence_start <> d.absence_start 
         or i.absence_end <> d.absence_end)
    )

rextester演示：http://rextester.com/PHF88226

返回：

+----------+---------------+-------------+
| employee | absence_start | absence_end |
+----------+---------------+-------------+
| emp01    | 2017-06-21    | 2017-06-22  |
| emp01    | 2017-06-24    | 2017-06-27  |
| emp02    | 2017-06-22    | 2017-06-27  |
+----------+---------------+-------------+

Answer 2

查询应如下所示：

select
*
from dbo.DATA t1
where not exists (
            select
            *
            from dbo.DATA t2
            where t1.ABSENCE_START >= t2.ABSENCE_START
                and t1.ABSENCE_END < t2.ABSENCE_END)

但是，如果你有重叠的缺席怎么办？像这样：

insert into DATA(EMPLOYEE,ABSENCE_START,ABSENCE_END) values
('emp01','21/06/2017','22/06/2017'),
('emp01','24/06/2017','27/06/2017'),
('emp02','22/06/2017','29/06/2017'),
('emp02','22/06/2017','27/06/2017')

SQL Server独特的缺席期

2 个答案: