从 Oracle 表中选择上传历史

Question

我有一个包含工厂列表（工厂 ID 和工厂名称）的表

Factory ID    Factory Name
1001          Factory1
1002          Factory2
1003          Factory3

用户将每月将一些数据上传到每个工厂的历史记录表中。并且同一月份的现有工厂数据也将被不同的用户覆盖。因此历史表如下所示。

Factory ID   Month UploadedValue UploadedBy UploadedOn
1001         01    250           User1      29.11.2016 07:28 PM
1002         01    102.12        User1      29.11.2016 07:28 PM
1001         01    400           User2      30.11.2016 12.00 PM

现在，我需要为每个工厂和每个月生成一份报告，这是最新上传的（上传者和上传时间），如果没有上传，则应显示为“未上传”。如下所示。

   Factory ID    Factory Name   Month   Last Upload by   Last Upload on
    1001          Factory1       01      User2            30.11.2016 12.00 PM
    1002          Factory2       01      User1            29.11.2016 07:28 PM
    1003          Factory3       01      Not Uploaded     Not Uploaded

请帮助获得这个。我尝试了分析功能。但没有运气。

Answer 1

要获取每个工厂每个月的最新更新，您可以在“更新”表中group by factory_id 和month，然后选择max(updatedon)。对于其他列，您可以使用keep (dense_rank last...)。所有这些都是聚合函数（而不是分析函数） - 您不会做超出需要的工作。

查询的另一部分是数据密集化。在较新的 Oracle 版本中，这可以通过“分区外连接”来完成（谷歌以阅读有关此主题的更多信息）。我在输入数据中添加了第二个“月份”以充分说明这个概念（也请参阅输出）。注意 - 我没有打扰 'Not Uploaded' 而不是 null - 如果真的需要，可以将所有值包装在 coalesce() 中以用该文本替换 null，但我看不出它添加了什么。 null 传达的信息完全相同。

with
     factories ( factory_id, factory_name ) as (
       select 1001, 'Factory1' from dual union all
       select 1002, 'Factory2' from dual union all
       select 1003, 'Factory3' from dual
     ),
     updates ( factory_id, month, uploadedvalue, uploadedby, uploadedon ) as (
       select 1001, '01', 250   , 'User1', to_date('29.11.2016 07:28 PM', 'dd.mm.yyyy hh:mi AM') from dual union all
       select 1002, '01', 102.12, 'User1', to_date('29.11.2016 07:28 PM', 'dd.mm.yyyy hh:mi AM') from dual union all
       select 1001, '01', 400   , 'User2', to_date('30.11.2016 12.00 PM', 'dd.mm.yyyy hh:mi AM') from dual
     ),
     months ( month ) as (
       select '01' from dual union all
       select '02' from dual
     )
select fm.factory_id, fm.factory_name, fm.month, s.uploadedvalue, s.uploadedby, 
       s.uploadedon
from   (
select factory_id, month, 
       min(uploadedvalue) keep (dense_rank last order by uploadedon) as uploadedvalue,
       min(uploadedby)    keep (dense_rank last order by uploadedon) as uploadedby,
       max(uploadedon) as uploadedon
from   updates
group by factory_id, month
) s
partition by (month)
right outer join (select * from factories cross join months) fm  
         on fm.factory_id = s.factory_id
        and fm.month      = s.month
order by month, factory_id
;

输出：

FACTORY_ID  FACTORY_NAME  MONTH  UPLOADEDVALUE  UPLOADEDBY  UPLOADEDON
----------  ------------  -----  -------------  ----------  -------------------
1001        Factory1      01               400  User2       30.11.2016 12.00 PM
1002        Factory2      01            102.12  User1       29.11.2016 07:28 PM
1003        Factory3      01            
1001        Factory1      02            
1002        Factory2      02            
1003        Factory3      02            

Answer 2

您可以使用RANK() 函数。假设表格是FAC和FAC_HIST，代码可能如下

SELECT fid, fname, month, last_uploaded_by, last_uploaded_on
  FROM ( select fh.fid fid
              , fh.fname fname
              , COALESCE(fd.month, 1) month
              , COALESCE(RANK() FIRST OVER (PARTITION BY fd.fid, fd.month 
                                                ORDER BY fd.last_uploaded_on DESC)
                         , -1) rnk
              , COALESCE(fd.uploaded_by, 'not uploaded') last_uploaded_by
              , COALESCE(fd.last_uploaded_on, 'not uploaded') last_uploaded_on
           FROM fac fh
           LEFT OUTER JOIN fac_hist fd
             ON fh.fid = fd.fid
                )
 WHERE rnk <= 1;