是否需要多个子查询才能从同一个表中的两个字段中选择最大值

Question

我对SQL相对较新。我相信在我的Oracle查询中需要多个子选择来实现我感兴趣的结果，但是我一直无法使查询正常工作。

我需要从“客户”表中提取基础数据，并从第二个表“Customer_Issue”中提取附加数据。还有其他表中包含我需要的其他字段，但我已通过加入“客户”表成功地加入/选择了该数据

我需要从“Customer_Issue”获得的数据位于“Service”，“Service_Issue “和”服务版本“。每个服务有多个服务问题，每个服务问题有多个服务版本。

对于每个服务，我只需要选择最大的服务问题，对于那个选定的服务问题，我只需要选择最大的服务版本。

通过一个子选择，我已经能够选择最大服务问题如下：

Select c.customer_id, ci.service, ci.service_issue 
from customer c 
left outer join CUSTOMER_issue ci on c.CUSTOMER_ID = ci.CUSTOMER_ID 
join (select CUSTOMER_ID, service, max(service_ISSUE) as service_ISSUE 
from CUSTOMER_issue 
group by CUSTOMER_ID, service) ci1 
on ci1.CUSTOMER_ID = ci.CUSTOMER_ID and ci1.service = ci.service 
and ci1.service_issue = ci.service_issue 

但是，我不能得到后续的分选工作，以获得最大的服务版本。以下是我已经尝试（在查询中，它是直接低于上面的代码）

join (select CUSTOMER_ID, service, service_ISSUE, max (service_VERSION) as service_VERSION 
from CUSTOMER_issue 
group by CUSTOMER_ID, service, service_ISSUE) ci2 
on ci1.CUSTOMER_ID = ci2.CUSTOMER_ID and ci1.service = ci2.service and 
ci1.service_ISSUE = ci2.service_ISSUE and ci1.service_VERSION = ci2.service_VERSION 

Answer 1

我怀疑你想上的最高发行的最高版本。我并不十分确定，但这就是我解释这个问题的方式。

如果是这样，最好的解决办法是使用row_number()：

select c.customer_id, ci.service, ci.service_issue  
from customer c left outer join 
    (select ci.*, 
      ROW_NUMBER() over (partition by ci.customer_id, ci.service 
           order by service_issue desc, service_version desc 
           ) as seqnum 
     from CUSTOMER_issue ci 
    ) ci 
    on c.CUSTOMER_ID = ci.CUSTOMER_ID and 
     ci.seqnum = 1 

对于每个客户ID /服务组合，其列举的问题的基础上，order by条款。最高的问题将得到“1”的最高service_version。我的猜测是，这是你想要的。

Answer 2

像这样的东西应该努力解决您的子查询的问题，如果我理解你的要求：

SELECT 
    c.customer_id, 
    ci1.service, 
    ci1.service_name, 
    ci1.service_issue, 
    ci2.service_version 
FROM customer c 
    JOIN (
     SELECT CUSTOMER_ID, service, service_name, max(service_ISSUE) as service_ISSUE 
     FROM CUSTOMER_issue 
     GROUP BY CUSTOMER_ID, service, service_name 
     ) ci1 ON c.CUSTOMER_ID = ci1.CUSTOMER_ID 
    JOIN (
     SELECT CUSTOMER_ID, service, max(service_VERSION) as service_VERSION 
     FROM CUSTOMER_issue 
     GROUP BY CUSTOMER_ID, service 
     ) ci2 ON c.CUSTOMER_ID = ci2.CUSTOMER_ID AND ci1.service = ci2.service 

有可能是根据您的需要一个更简单的方法。