Question

我有以下查询：

select * from [lead].[ContactFeedback] cf
where cf.LeadId in
(select LeadId from [lead].[LeadDetails]
where TeleportReference in (122096, 
122097, 
122098))
order by LeadId Desc

结果是这样的：

FeedbackDate            LeadId         
2015-01-23 16:25:13.547 95920
2015-01-23 16:25:38.960 95919
2015-01-23 16:25:19.393 95917
2015-01-23 16:25:32.837 95916
2015-01-23 16:25:59.840 95914
2015-01-23 16:26:08.840 95913
2015-01-23 16:15:01.933 95910
2015-01-23 16:22:04.820 95910
2015-01-23 16:24:40.477 95909
2015-01-23 16:24:03.523 95908
2015-01-23 16:16:44.290 95908
2015-01-23 16:17:16.047 95907
2015-01-23 16:25:11.783 95907

我想列出每个LeadId的所有前1名（最近的反馈日期）。如何在SQL Server中实现此目的？

Answer 1

select LeadId, Max(FeedbackDate) 
from [lead].[ContactFeedback] cf
where cf.LeadId in
    (select LeadId from [lead].[LeadDetails]
     where TeleportReference in (122096, 122097, 122098))
group by LeadId
order by LeadId Desc

Answer 2

使用Window Function

select leadid,feedbackdate from ( 
select row_number() over(partition by leadid order by feedbackdate desc) rn,* 
from yourtable) a
where rn=1

Answer 3

您可以执行子查询以获取每个LeadId的最大FeedBackDate。然后将详细信息表加入其中。

select cf.* 
from [lead].[ContactFeedback] cf
join (select LeadId, max(FeedBackDate) maxFeedBackDate
     from [lead].[ContactFeedback]
     group by LeadId) as maxes
on cf.LeadId = maxes.LeadId
   and cf.FeedBackDate = maxes.maxFeedBackDate
where cf.LeadId in
      (select LeadId 
       from [lead].[LeadDetails]
       where TeleportReference in (122096, 
                                   122097, 
                                   122098)
       )
order by cf.LeadId Desc

SQL - 选择具有相同ID的前1个多行

3 个答案: