Question

我有一个具有唯一ID但重复行信息的表。

我可以使用此查询找到具有重复项的行

SELECT
    PersonAliasId, StartDateTime, GroupId, COUNT(*) as Count
FROM
    Attendance
GROUP BY
    PersonAliasId, StartDateTime, GroupId
HAVING
    COUNT(*) > 1

我可以手动删除行，同时保留此查询所需的1

Delete
From Attendance
Where Id IN(SELECT
    Id
FROM
    Attendance
Where PersonAliasId = 15
    and StartDateTime = '9/24/2017'
and GroupId = 1429
Order By ModifiedDateTIme Desc
Offset 1 Rows)

我不熟悉SQL，足以弄清楚如何使用第一个查询中的行来删除留下最新的重复项。第一个查询返回了超过3481条记录，以便手动逐一执行此操作。

如何像第一个查询一样找到重复的行，并删除除第二个之外的所有行？

Answer 1

您可以使用Common Table Expression删除重复项：

WITH Cte AS(
    SELECT *,
        Rn = ROW_NUMBER() OVER(PARTITION BY PersonAliasId, StartDateTime, GroupId 
                                ORDER BY ModifiedDateTIme DESC)
    FROM Attendance
)
DELETE FROM Cte WHERE Rn > 1;

这将保留每个PersonAliasId - StartDateTime - GroupId组合的最新记录。

Answer 2

使用MAX聚合函数识别每个组/个人组合的最新startdatetime。然后删除没有最新时间的记录。

DELETE a
FROM attendance as a
INNER JOIN (  
   SELECT
        PersonAliasId, MAX(StartDateTime) AS LatestTime, GroupId,
    FROM
        Attendance
    GROUP BY
        PersonAliasId, GroupId
    HAVING
        COUNT(*) > 1
) as b
on a.personaliasid=b.personaliasid and a.groupid=b.groupid and a.startdatetime < b.latesttime

Answer 3

与CTE答案相同 - 给Felix支票

delete 
from ( SELECT rn = ROW_NUMBER() OVER(PARTITION BY PersonAliasId, StartDateTime, GroupId 
                                     ORDER BY ModifiedDateTIme DESC)
        FROM Attendance
     ) tt 
where tt.rn > 1

在SQL中删除重复行

3 个答案: