Question

我有这个查询，我从三个表中选择

select 
    min(t.ReminderDt) as 'rem dt',  
    m.Group_Id, m.AccountNumber 
from 
    ACE_AccsLevelTran t, ACE_AccsLevelMaster m 
where 
    t.MasterAccNumber = m.AccountNumber 
group by  
    m.Group_Id, m.AccountNumber;

这导致：

rem dt | Group_Id| AccountNumber
--------------------------------    
2/8/2013 | 3 | 4216985
2/22/2013 | 4 | 4274863
2/7/2013 | 3 | 4366383
2/28/2013 | 4 | 7151712

如何获得3和4的最小日期行，如结果 -

2/7/2013 | 3 | 4366383
2/22/2013 | 4 | 4274863

Answer 1

只需从群组中移除account_number，然后在min()行上使用max()或select将其包围：

select min(t.ReminderDt) as 'rem dt',  m.Group_Id, min(m.AccountNumber)
from ACE_AccsLevelTran t,     ACE_AccsLevelMaster m 
where t.MasterAccNumber=m.AccountNumber
group by  m.Group_Id

返回任意帐号。要获得具有最小值的行，最好的方法是使用row_number()：

select *
from (select t.ReminderDt) as 'rem dt',  m.Group_Id, m.AccountNumber,
            row_number() over (partition by group_id order by reminderdt desc) as seqnum
      from ACE_AccsLevelTran t join ACE_AccsLevelMaster m 
           on t.MasterAccNumber=m.AccountNumber
     ) t
where seqnum = 1

此外，您应该学习此标准中使用的ANSI标准JOIN语法。

Answer 2

如果您的accountNumber也是唯一的，您也可以这样做：

Select m.Group_Id ,X.MinReminderDT,m.AccountNumber 
from ACE_AccsLevelMaster m join(
select min(t.ReminderDt) as MinReminderDT,t.MasterAccNumber
from ACE_AccsLevelTran t
Group By t.MasterAccNumber) X on X.MasterAccNumber=m.AccountNumber

sql最小值来自一列，另一列唯一的第三列

2 个答案: