Question

我有以下查询给我一个ROW_NUMBER，作为正在进行的案例中的“阶段”。此查询在INSERT INTO下运行以填充表。在那一刻，我们不得不每个月完全重新构建表，以使用连续的行号捕获阶段中的每个更改（当进程继续时，每个阶段的一行）。在某些时候，表格会变得太大而无法做到这一点，我希望能够追加，但问题是，我想不出一种方法可以让行号从中断处继续。我在测试表中尝试了它，当我运行查询两次，分别在2015年11月进行两半时，每个案例中从11月上半月到下半年的任何阶段的行数都从1开始，如果是在第二轮中，而不是继续。

ModifiedRecordID列标识了案例，您可以看到PARTITION BY被分组的内容。

有没有办法做我正在考虑的事情？

SELECT  agc.ObjectTypeCode
   , ev.Name                           AS EntityName
   , agc.AttributeId
   , ea.AttributeName
   , ind.createdbyname                 AS CaseCreatedBy
   , agc.CallingUserName
   , agc.CallingUserId
   , LEFT(CAST(agc.OldFieldValue AS NVARCHAR(MAX)), ISNULL(NULLIF(CHARINDEX(',', CAST(agc.OldFieldValue AS NVARCHAR(MAX))) -1, -1), '')) AS ChangedField
   , UPPER(RIGHT(CAST(agc.OldFieldValue AS NVARCHAR(MAX)), LEN(CAST(agc.OldFieldValue AS NVARCHAR(MAX))) - CHARINDEX(',', CAST(agc.OldFieldValue AS NVARCHAR(MAX))))) AS PreviousGuidValue
   , wkt.ptl_name                      AS WorkType
   , ModifiedRecordId
   , ind.ticketnumber                  AS CaseRef
   , agc.ActionId
   , ind.createdon                     AS MatterCreated    
   , agc.LogDateTime                   AS AuditedDate
   , agc.AuditId
   , ROW_NUMBER() OVER(PARTITION BY ModifiedRecordId ORDER BY agc.LogDateTime ASC) Stage

FROM AuditGuidChange agc LEFT JOIN EntityView ev
                   ON agc.ObjectTypeCode = ev.ObjectTypeCode
                 LEFT JOIN EntityAttribute ea
                   ON agc.AttributeId = ea.ColumnNumber
                     AND agc.ObjectTypeCode = ea.MappedObjectCode
                 LEFT JOIN Peppermint_Data.dbo.incident ind
                   ON agc.modifiedrecordid = ind.incidentid
                 LEFT JOIN Filteredptl_worktype wkt
                   ON agc.PreviousGuidValue = wkt.ptl_worktypeid
WHERE LEFT(CAST(agc.OldFieldValue AS NVARCHAR(MAX)), ISNULL(NULLIF(CHARINDEX(',', CAST(agc.OldFieldValue AS NVARCHAR(MAX))) -1, -1), '')) = 'ptl_worktype'
AND CAST(agc.LogDateTime AS DATE) BETWEEN @sDate AND @eDate;

Answer 1

declare @offset int = 12;
select id, ROW_NUMBER() over (order by id) + @offset  as rn 
from table;

我可以获取row_number以在稍后的查询中继续序列

1 个答案: