我有一张表，它根据用户名存储员工通过 Time Machine 的签到时间。如果一个员工多次打卡，那么他的签到记录会有多条记录，它们之间只有几秒钟的时间差。显然只有第一条记录是有效的。所有其他条目均无效，必须从表中删除。如果我可以选择员工当前日期的所有签到记录，我该怎么做？
db中的数据如下。
Username Checktime CheckType
HRA001 7/29/2012 8:16:44 AM Check-In
HRA001 7/29/2012 8:16:46 AM Check-In
HRA001 7/29/2012 8:16:50 AM Check-In
HRA001 7/29/2012 8:16:53 AM Check-In
尝试这个：
;WITH users_CTE as (
select rank() over (partition by Username order by Checktime) as rnk from users
)
DELETE FROM users_CTE where rnk <> 1
--对于您的第二个要求，请尝试此查询
;WITH users_CTE as (
select *,rank() over (partition by Username order by Checktime) as rnk from users
)
,CTE2 as (select Username,MIN(CheckTime) as minTime,DATEADD(mi,1,MIN(CheckTime)) as maxTime from users_CTE
group by Username)
delete from users where Checktime in(
select c1.Checktime from users_CTE c1 left join CTE2 c2
on c1.Checktime > c2.minTime and c1.Checktime <= c2.maxTime
where c2.Username is not null and c1.Username in(
select c1.Username from users_CTE c1 left join CTE2 c2
on c1.Checktime > c2.minTime and c1.Checktime <= c2.maxTime
group by c1.Username,c2.Username
having COUNT(*) > 1))
--对于您更改的要求，请在下面检查此查询
alter table users add flag varchar(2)
;WITH users_CTE as (
select *,rank() over (partition by Username order by Checktime) as rnk from users
)
,CTE2 as (select Username,MIN(CheckTime) as minTime,DATEADD(mi,1,MIN(CheckTime)) as maxTime from users_CTE
group by Username)
update u SET u.flag = 'd' from users_CTE u inner join (
select c1.Checktime from users_CTE c1 left join CTE2 c2
on c1.Checktime > c2.minTime and c1.Checktime <= c2.maxTime
where c2.Username is not null and c1.Username in(
select c1.Username from users_CTE c1 left join CTE2 c2
on c1.Checktime > c2.minTime and c1.Checktime <= c2.maxTime
group by c1.Username,c2.Username
having COUNT(*) > 1)) a
on u.Checktime=a.Checktime
--用DeletFlag检查最新查询
;WITH users_CTE as
(
select *,row_number() over (partition by Username order by Checktime) as row from users
)
,CTE as(
select row,Username,Checktime,CheckType,0 as totalSeconds,'N' as Delflag from users_CTE where row=1
union all
select t.row,t.Username,t.Checktime,t.CheckType,CASE WHEN (c.totalSeconds + DATEDIFF(SECOND,c.Checktime,t.Checktime)) >= 60 then 0 else (c.totalSeconds + DATEDIFF(SECOND,c.Checktime,t.Checktime)) end as totalSeconds,
CASE WHEN (c.totalSeconds + DATEDIFF(SECOND,c.Checktime,t.Checktime)) >= 60 then 'N' else 'Y' end as Delflag
--CASE WHEN c.totalSeconds <= 60 then 'Y' else 'N' end as Delflag
from users_CTE t inner join CTE c
on t.row=c.row+1
)
select Username,Checktime,CheckType,Delflag from CTE
DELETE FROM timesheet
WHERE timesheetRecordId <>(
SELECT TOP 1 timesheetRecordId from timesheet
WHERE checkInDate=todaysDate AND employeeId=empId ORDER BY checkInTime ASC
)
AND checkInDate=today's date AND empolyeeId=empId;
您可以尝试按签到时间对记录进行排名，然后删除每个员工每天所有排名大于 1 的记录。
为什么不在将签入插入到数据库之前验证签入。如果该用户存在任何签到，则在此日期和该日期之间，然后不执行任何其他操作插入它
您应该能够按时间对所有记录进行排序，从每个员工上一次的时间中减去最近的时间，如果结果小于某个阈值，则删除具有最近时间的行。
试试这个查询： Delete from employee where employee.checkin (select checkin from employee where count(checkin)>1);
我认为您不能在同一语句的子查询中从 Delete 语句中指定目标表。所以你不能用一个单独的 Delete 语句来做到这一点。
您可以做的是编写一个存储过程。在您的存储过程中，您应该创建一个包含此查询返回的 PK 的临时表：
select cht.pkey
from CheckTimeTable as cht
where exists ( select pkey
from CheckTimeTable
where username = cht.userName
and checkType = 'check-IN'
and Checktime >= subtime(cht.Checktime, '0 0:0:15.000000')
and Checktime < cht.Checktime);
然后编写另一条语句从原始表 CheckTimeTable 中删除这些 PK。
请注意，上面的查询是针对 MySQL 的，因此您需要找到从 DBMS 的时间戳中减去 15 秒的方法。在 MySQL 中它是这样完成的：
subtime(cht.Checktime, '0 0:0:15.000000')
此查询将返回具有来自同一用户的另一个 CheckTime 记录的任何 CheckTime 记录，类型为 Check-In，并且在 15 秒内早于其自己的检查时间。