select T1.dbid,T1.id,T1.headline,T4.login_name,T1.severity,T9.name,T1.fldcolumn_2 from history T10, Defect T1,users T4,statedef T9 where T1.owner = T4.dbid and T1.state = T9.id and (T10.action_timestamp > dateadd(wk,datediff(wk,0,getutcdate()-1),0)) and (T1.dbid <> 0 and ((T1.fldcolumn_2 like '%策划部%'  and T9.name in ('解决-Resolved') and T1.priority in ('本周解决','立即解决')))) order by T1.id ASC



查询出来之后出现好多重复的ID
如何确决！

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本主题的最佳答案为 [ 5 楼].

T10 没有和defect建立关联。 加个条件
and T10.entity_dbid=T1.dbid

加个 distinct   试试

select T1.dbid,T1.id,T1.headline,T3.login_name,T1.priority,T1.severity,T5.name,T1.fldcolumn_2 from Defect T1,users T3,statedef T5 where T1.owner = T3.dbid and T1.state = T5.id and (T1.dbid <> 0 and ((T1.fldcolumn_2 like '%策划部%' and T5.name in ('解决-Resolved') and T1.priority in ('本周解决','立即解决') and T1.submit_date between {ts '2008-02-24 16:00:00'} and  {ts '2008-03-01 15:59:59'}))) order by T1.id ASC


T1.submit_date between {ts '2008-02-24 16:00:00'} and  {ts '2008-03-01 15:59:59'})))  如何以提交日期改成周任务查询

T1.submit_date between

select   convert(varchar(10),   dateadd(day,   2   -   datepart(weekday,   getdate()),   getdate()),   120)   and select   convert(varchar(10),   dateadd(day,   2   -   datepart(weekday,   getdate())+6,   getdate()),   120)

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