SQL Server-行号()->是否再次重置?
我们在当前编写的查询中面临一个特定的问题。 以下是一个例子:SQL Server-行号()->是否再次重置?,sql,sql-server,window-functions,Sql,Sql Server,Window Functions,我们在当前编写的查询中面临一个特定的问题。 以下是一个例子: Doc. ID | Timestamp | Employee 01 | 01 | A 01 | 02 | B 01 | 03 | B 01 | 04 | C 01 | 05 | A 01 | 06 | A
Doc. ID | Timestamp | Employee
01 | 01 | A
01 | 02 | B
01 | 03 | B
01 | 04 | C
01 | 05 | A
01 | 06 | A
我们希望实现的是:
Doc. ID | Timestamp | Employee
01 | 01 | A
01 | 03 | B
01 | 04 | C
01 | 06 | A
这是我们的方法,但不起作用:
SELECT [Doc. ID], [Timestamp], [Employee]
,ROW_NUMBER() OVER (PARTITION BY [Doc. ID],[Employee] order by [Employee] desc) as "RN"
FROM XY
WHERE "RN" = 1
但不幸的是,这不起作用,因为在底部再次找到A时,行号不会重置。如果没有where条款,我们收到的结果是:
Doc. ID | Timestamp | Employee | RN
01 | 01 | A | 1
01 | 02 | B | 1
01 | 03 | B | 2
01 | 04 | C | 1
01 | 05 | A | 2
01 | 06 | A | 3
我认为实现正确的解决方案只需要多一点 在下一行的员工价值中使用lead to peak:
select xy.*
from (select xy.*,
lead(employee) over (partition by docid order by timestamp) as next_employee
from xy
) xy
where next_employee is null or next_employee <> employee;
我想你想要:
SELECT [doc. ID], MAX([Timestamp]) AS [Timestamp], employee
FROM (SELECT t.*,
row_number() over (order by [Timestamp]) as seq1,
row_number() over (partition by [doc. ID], employee order by [Timestamp]) as seq2
FROM XY t
) t
GROUP BY [doc. ID], employee, (seq1 - seq2)
ORDER BY [Timestamp];
谢谢,成功了!我们也考虑了一些超前/滞后的问题,但不知何故,我们找不到合适的解决方案-@阿肯。您更喜欢具有两个窗口函数和聚合的更复杂的解决方案?我可以补充一点,在这一次之后大约10分钟。