一尘不染

关于ORACLE的Tricky GROUP BY问题

sql

我目前面临的问题是我的Oracle知识无法解决,我绝对不是数据库专家,这就是为什么我问您是否有解决SQL查询问题的想法。

这是我的问题,我有两个表,让它们分别称为DEVICE_TABLE和COUNT_TABLE

COUNT_TABLE看起来像:

    DEVICE (Int) PK         |       QUANTITY (Int)
- - - - - - - - - - - - - - - - - - - - - - - - - - -
        1001                |              4
- - - - - - - - - - - - - - - - - - - - - - - - - - -
        1002                |             20
- - - - - - - - - - - - - - - - - - - - - - - - - - - 
        1003                |              1
…

DEVICE_TABLE看起来像:

      ID (Int) PK            |      WiFi (String)            |     Email (String)          |   Bluetooth(String)           |   …
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
        1001                |             Yes               |               No            |                 No            |   …
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
        1002                |             Yes               |               Yes           |                 No            |   …
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
        1003                |             Unknown           |               Unknown       |                 Yes           |   …
…

约束是:

DEVICE_TABLE.ID = COUNT_TABLE.DEVICE

WiFi,电子邮件,蓝牙’‘是只能是渊'',o’‘或``未知’‘的字符串。

最后,预期的SQL请求结果是(基于我的示例):

         Feature        |            Yes           |              No            |            Unknown          
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
        WiFi            |             24           |                 0          |                 1                  
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
       Email            |             20           |                 4          |                 1                  
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
    Bluetooth           |              1           |                24          |                 0                   
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
…

简而言之,此请求的目的是汇总与特定功能兼容的所有设备数。

如果您对如何实现此目标有任何线索,请先谢谢您!(也许不可能…)


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2021-05-23

共1个答案

一尘不染

在Oracle 11中,可以将该pivot子句与该unpivot子句一起使用:

with 
count_table as (
     select 1001 device_id,  4 quantity from dual union all
     select 1002 device_id, 20 quantity from dual union all
     select 1003 device_id,  1 quantity from dual 
),
device_table as (
     select 1001 id, 'Yes'     wifi, 'No'       email, 'No'  bluetooth from dual union all
     select 1002 id, 'Yes'     wifi, 'Yes'      email, 'No'  bluetooth from dual union all
     select 1003 id, 'Unknown' wifi, 'Unknown'  email, 'Yes' bluetooth from dual 
)
----------------------------------------
select * from (
      select
        feature,
        yes_no_unknown,
        sum(quantity)  quantity
      from 
         count_table  c join 
         device_table d on c.device_id = d.id
      unpivot  ( yes_no_unknown
                 for feature in (wifi, email, bluetooth)
      ) 
      group by 
      feature,
      yes_no_unknown
)  
pivot ( sum (quantity)
        for yes_no_unknown in ('Yes' as yes, 'No' as no, 'Unknown' as unknown)
)
;

或者,您可能希望将两个现有表连接到包含三个所需行的值的第三个表。也可能更容易阅读:

with 
count_table as (
     select 1001 device_id,  4 quantity from dual union all
     select 1002 device_id, 20 quantity from dual union all
     select 1003 device_id,  1 quantity from dual 
),
device_table as (
     select 1001 id, 'Yes'     wifi, 'No'       email, 'No'  bluetooth from dual union all
     select 1002 id, 'Yes'     wifi, 'Yes'      email, 'No'  bluetooth from dual union all
     select 1003 id, 'Unknown' wifi, 'Unknown'  email, 'Yes' bluetooth from dual 
)
----------------------------------------
select
   f.txt,
   sum(case when ( f.txt = 'wifi'      and d.wifi      = 'Yes' ) or
                 ( f.txt = 'email'     and d.email     = 'Yes' ) or
                 ( f.txt = 'bluetooth' and d.bluetooth = 'Yes' ) 
            then   c.quantity
            else   0 end
      ) yes,
   sum(case when ( f.txt = 'wifi'      and d.wifi      = 'No' ) or
                 ( f.txt = 'email'     and d.email     = 'No' ) or
                 ( f.txt = 'bluetooth' and d.bluetooth = 'No' ) 
            then   c.quantity
            else   0 end
      ) no,
   sum(case when ( f.txt = 'wifi'      and d.wifi      = 'Unknown' ) or
                 ( f.txt = 'email'     and d.email     = 'Unknown' ) or
                 ( f.txt = 'bluetooth' and d.bluetooth = 'Unknown' ) 
            then   c.quantity
            else   0 end
      ) unknown
from 
   count_table  c                                   join 
   device_table d on c.device_id = d.id     cross   join
   (
        select 'wifi'      txt from dual union all
        select 'email'     txt from dual union all
        select 'bluetooth' txt from dual
   ) f
group by 
    f.txt;
2021-05-23