以下是我的表格和样本数据
DECLARE @Employee_Log table(ID int,eid int, ecode varchar(100), emp_startdate date)
INSERT INTO @Employee_Log
SELECT 1, 1, 'aaa','2019-01-01'
UNION ALL
SELECT 2, 1, 'aaa','2019-01-05'
UNION ALL
SELECT 3, 1, 'bbb','2019-01-03'
UNION ALL
SELECT 4, 2, 'aaa','2019-01-03'
UNION ALL
SELECT 5, 1, 'aaa','2019-02-01'
UNION ALL
SELECT 6, 1, 'aaa','2019-02-15'
UNION ALL
SELECT 7, 1, 'aaa','2019-02-19'
UNION ALL
SELECT 8, 1, 'aaa','2019-02-28'
在上面的数据中,我想删除基于eid
和ecode
的重复。如果emp_startdate
在7天内,则获取最新数据并忽略其余数据。
我尝试了以下代码,但如何添加周范围的条件检查
SELECT
ROW_NUMBER() OVER(PARTITION BY eid,ecode ORDER BY emp_startdate desc) as rownum,
ID,eid,ecode,emp_startdate
FROM @Employee_Log
我想要的结果如下所示
ID eid ecode emp_startdate
2 1 aaa 2019-01-05
5 1 aaa 2019-02-01
4 2 aaa 2019-01-03
7 1 aaa 2019-02-19
8 1 aaa 2019-02-28
3 1 bbb 2019-01-03
如果在同样的7天内发生超过2个事件,我仍然不确定你想要发生什么。但是这个解决方案将获得所有系列日期的最新日期,其中日期之间的差异为7天或更短。
select ID,eid,ecode,emp_startdate
from
(
select ID,
eid,
ecode,
emp_startdate,
datediff(day
,emp_startdate
,lead(emp_startdate)
over
(partition by eid,ecode order by emp_startdate)) l
from @Employee_Log
) a
where l is null or l>7
ID eid ecode emp_startdate
-- --- ----- -------------
3 1 bbb 2019-01-03
2 1 aaa 2019-01-05
5 1 aaa 2019-02-01
7 1 aaa 2019-02-19
8 1 aaa 2019-02-28
4 2 aaa 2019-01-03
以下查询将为您提供您在问题中要求使用简明英语的内容,但您的示例数据和所需输出与您自己的问题相矛盾:
SELECT *
FROM
(
SELECT
ROW_NUMBER() OVER (PARTITION BY eid , ecode , YEAR(emp_startdate)
, DATEPART(WEEK, emp_startdate)
ORDER BY emp_startdate DESC
) AS rownum
, ID
, eid
, ecode
, emp_startdate
FROM @Employee_Log
) x
WHERE x.rownum = 1;