我正在尝试创建工作活动报告。如果设置了以下条件。
如果在后续日期范围内缺席，则标记为相同的缺席组，否则标记为新缺席组。
所以，如果我在星期一，星期三，星期三不在，那么这就是一组，标记为1。
如果我当时在星期四工作，但星期五再次缺席，则星期五会以group2为标记。
如果我在下一个星期一仍然缺席那么那仍然是第2组。
提供的示例数据是我从班次调度表中获得的数据类型，除了标识符，如果特定活动是缺席或不缺席。当然，在任何一天都有不止一组的首字母，以及不同类型的活动和缺席活动。
我试图提供输入数据和期望结果的最小工作示例。希望我能对此有所了解。
USE Sandbox
DROP TABLE Data /* Clean up after ourselves. */
CREATE TABLE Data ( /* Create table */
[Date] DATE,
Initials VARCHAR(10),
Activity VARCHAR(255),
ActivityType VARCHAR(255)
);
/* Insert data */
INSERT INTO [dbo].[Data] ([Date],[Initials],[Activity],[ActivityType]) VALUES ('2016-12-05','PersonA','Work','Work')
INSERT INTO [dbo].[Data] ([Date],[Initials],[Activity],[ActivityType]) VALUES ('2016-12-06','PersonA','Work','Work')
INSERT INTO [dbo].[Data] ([Date],[Initials],[Activity],[ActivityType]) VALUES ('2016-12-07','PersonA','AbsenceActivity','Absence')
INSERT INTO [dbo].[Data] ([Date],[Initials],[Activity],[ActivityType]) VALUES ('2016-12-08','PersonA','AbsenceActivity','Absence')
INSERT INTO [dbo].[Data] ([Date],[Initials],[Activity],[ActivityType]) VALUES ('2016-12-09','PersonA','AbsenceActivity','Absence')
INSERT INTO [dbo].[Data] ([Date],[Initials],[Activity],[ActivityType]) VALUES ('2016-12-10',NULL,NULL,'NoShift')
INSERT INTO [dbo].[Data] ([Date],[Initials],[Activity],[ActivityType]) VALUES ('2016-12-11',NULL,NULL,'NoShift')
INSERT INTO [dbo].[Data] ([Date],[Initials],[Activity],[ActivityType]) VALUES ('2016-12-12','PersonA','AbsenceActivity','Absence')
INSERT INTO [dbo].[Data] ([Date],[Initials],[Activity],[ActivityType]) VALUES ('2016-12-13','PersonA','Work','Work')
INSERT INTO [dbo].[Data] ([Date],[Initials],[Activity],[ActivityType]) VALUES ('2016-12-14','PersonA','AbsenceActivity','Absence')
INSERT INTO [dbo].[Data] ([Date],[Initials],[Activity],[ActivityType]) VALUES ('2016-12-15','PersonA','AbsenceActivity','Absence')
这种方式给了我想要的东西，但是当引入新组时，我无法让rownumber，rank或dense_rank重置。最终的期望结果。
SELECT [Date]
,Initials
,activity
,ActivityType
,rank() OVER (PARTITION BY data.activitytype, Initials ORDER BY data.date,data.initials) rownumber
FROM data
GROUP BY data.date, data.initials, activity,ActivityType
ORDER BY date
期望的结果 - 完整的工作活动日期范围
SELECT '2016-12-05' as [Date] ,'PersonA' AS [Initials] ,'Work' AS [Activity] ,'Work' AS [ActivityType] ,'0' AS [identifier] union all
SELECT '2016-12-06' as [Date] ,'PersonA' AS [Initials] ,'Work' AS [Activity] ,'Work' AS [ActivityType] ,'0' AS [identifier] union all
SELECT '2016-12-07' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union all
SELECT '2016-12-08' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union all
SELECT '2016-12-09' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union ALL
SELECT '2016-12-12' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union all
SELECT '2016-12-13' as [Date] ,'PersonA' AS [Initials] ,'Work' AS [Activity] ,'Work' AS [ActivityType] ,'0' AS [identifier] union all
SELECT '2016-12-14' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'2' AS [identifier] union all
SELECT '2016-12-15' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'2' AS [identifier]
替代结果1完整日期范围包括周期或非工作日期
SELECT '2016-12-05' as [Date] ,'PersonA' AS [Initials] ,'Work' AS [Activity] ,'Work' AS [ActivityType] ,'0' AS [identifier] union all
SELECT '2016-12-06' as [Date] ,'PersonA' AS [Initials] ,'Work' AS [Activity] ,'Work' AS [ActivityType] ,'0' AS [identifier] union all
SELECT '2016-12-07' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union all
SELECT '2016-12-08' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union all
SELECT '2016-12-09' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union ALL
SELECT '2016-12-10' as [Date] ,NULL AS [Initials] ,'Weekend' AS [Activity] ,'noShift' AS [ActivityType] ,NULL AS [identifier] union ALL
SELECT '2016-12-11' as [Date] ,NULL AS [Initials] ,'Weekend' AS [Activity] ,'noShift' AS [ActivityType] ,NULL AS [identifier] union ALL
SELECT '2016-12-12' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union all
SELECT '2016-12-13' as [Date] ,'PersonA' AS [Initials] ,'Work' AS [Activity] ,'Work' AS [ActivityType] ,'0' AS [identifier] union all
SELECT '2016-12-14' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'2' AS [identifier] union all
SELECT '2016-12-15' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'2' AS [identifier]
替代结果2仅包括一组首字母有活动的日期
SELECT '2016-12-07' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union all
SELECT '2016-12-08' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union all
SELECT '2016-12-09' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union all
SELECT '2016-12-10' as [Date] ,NULL AS [Initials] ,'Weekend' AS [Activity] ,'noShift' AS [ActivityType] ,NULL AS [identifier] union ALL
SELECT '2016-12-11' as [Date] ,NULL AS [Initials] ,'Weekend' AS [Activity] ,'noShift' AS [ActivityType] ,NULL AS [identifier] union ALL
SELECT '2016-12-12' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'1' AS [identifier] union all
SELECT '2016-12-14' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'2' AS [identifier] union all
SELECT '2016-12-15' as [Date] ,'PersonA' AS [Initials] ,'AbsenceActivity' AS [Activity] ,'Absence' AS [ActivityType] ,'2' AS [identifier]
答案 0 :(得分：1)
使用with (common table expression)和row_number()，我们可以为分区(Initials, activitytype)和(Initials)生成行号。
通过比较这些，我们可以按活动对连续的块进行分组，并使用dense_rank()对它们进行编号。
转换所有工作＆＃39;至0我在最终查询中使用了case表达式。
rextester：http://rextester.com/AYR27547
with cte as (
select
[Date]
, Initials
, activity
, ActivityType
, irn=row_number() over (
partition by Initials
order by [date]
)
, atrn=row_number() over (
partition by Initials, activitytype
order by [date]
)
from [data] d
group by [Date], Initials, Activity, ActivityType
)
select
[date]
, Initials
, activity
, ActivityType
, ActivityGroup = case
when ActivityType='Work'
then 0
else dense_rank() over (
partition by Initials, ActivityType
order by irn-atrn
)
end
from cte
order by [date]
结果：
+------------+----------+-----------------+--------------+---------------+
| date | Initials | activity | ActivityType | ActivityGroup |
+------------+----------+-----------------+--------------+---------------+
| 2016-12-05 | PersonA | Work | Work | 0 |
| 2016-12-06 | PersonA | Work | Work | 0 |
| 2016-12-07 | PersonA | AbsenceActivity | Absence | 1 |
| 2016-12-08 | PersonA | AbsenceActivity | Absence | 1 |
| 2016-12-09 | PersonA | AbsenceActivity | Absence | 1 |
| 2016-12-10 | NULL | NULL | NoShift | 1 |
| 2016-12-11 | NULL | NULL | NoShift | 1 |
| 2016-12-12 | PersonA | AbsenceActivity | Absence | 1 |
| 2016-12-13 | PersonA | Work | Work | 0 |
| 2016-12-14 | PersonA | AbsenceActivity | Absence | 2 |
| 2016-12-15 | PersonA | AbsenceActivity | Absence | 2 |
+------------+----------+-----------------+--------------+---------------+