2 个查询的单个结果 ORACLE答案

【问题标题】：single result for 2 queries ORACLE2 个查询的单个结果 ORACLE
【发布时间】：2022-11-30 18:18:37
【问题描述】：

我有一个 POST 表、一个 CATEGORY 表、一个 ACTION 表和 ACTION_TYPE 表，我解释说 ACTION 表包含所有已执行的操作，而表 ACTION_TYPE 包含操作详细信息，例如 ID = 4 的 ACTION 具有 ACTION_TYPE_ID = 1对于 POST_ID 6，这意味着对 50 号帖子进行了操作，我们可以对一个 post_id 执行多个操作

POST 表

 id         title       content     category_id
----------  ----------  ----------  ------------  
 1          title1      Text...     1
 2          title2      Text...     1
 3          title3      Text...     1
 4          title4      Text...     3
 5          title5      Text...     2
 6          title6      Text...     1

类别表

 id         name      
----------  ----------  
 1          category_1      
 2          category_2
 3          category_3

ACTION_TYPE 表

 id         name      
----------  ----------  
 1          updated
 2          deleted
 3          restored
 4          hided

行动表

id         post_id         action_type_id    date
----------  ----------      --------------    -----
 1          1               1                 2017-01-01
 2          1               1                 2017-02-15
 3          1               3                 2018-06-10
 4          6               1                 2019-08-01
 5          5               2                 2019-12-09
 6          2               3                 2020-04-27
 7          2               1                 2020-07-29
 8          3               2                 2021-03-13

现在我解释一下这个案例，我实际上有两个查询，一个查询计算每个类别的帖子，另一个查询按类别计算对每个帖子执行的操作，这非常有效。

这是我的第一个查询

select categories,  count(*) as cnt_posts_per_cat
   from(
        select 
        case
        when p.category_id is not null then c.name
        end as categories
        from post p  
        left join category c on p.category _id = c.id
        )
   group by categories
;

这带来了这个结果

categories  cnt_posts_per_cat
----------  -------------------   
category_1  4
category_2  1
category_3  1

Ans 这是我的第二个查询

select categories,  count(*) as cnt_actions_per_cat
   from(
        select  distinct  ac.post_id  AS action_post_id, max(ac.date) over (partition by  ac.post_id) as max_date, 
        case
        when ac.action_type_id is not null then act.name
        end as actions,
        case
        when p.category_id is not null then c.name
        else 'na'
        end as categories
        from action ac
        left join post p on ac.post_id = p.id  
        left join category c on p.category _id = c.id
        left join action_type act on ac.action_type_id = act.id
        where act.name in ('restored','deleted','updated')
        )
   group by categories
;

这带来了这个正确的结果，因为它选择了每个 action_type 的最后一个动作

categories  cnt_actions_per_cat
----------  -------------------   
category_1  3
category_2  1
category_3  na

但我想同时为两个查询创建一个结果表，如下所示：这是预期的结果

categories  cnt_posts_per_cat   cnt_actions_per_cat
----------  -----------------   -------------------  
category_1  4                   3
category_2  1                   1
category_3  1                   na

我正在尝试 union 和 union all 但它不正确它返回以下结果

categories  cnt_posts_per_cat   
----------  -----------------   
category_1  7                  
category_2  2                  
category_3  1

此致

【问题讨论】：

标签： oracle oracle11g oracle-sqldeveloper

【解决方案1】：

相关子查询怎么样？

样本数据：

SQL> with
  2  post (id, category_id) as
  3    (select 1, 1 from dual union all
  4     select 2, 1 from dual union all
  5     select 3, 1 from dual union all
  6     select 4, 3 from dual union all
  7     select 5, 2 from dual union all
  8     select 6, 1 from dual
  9    ),
 10  category (id, name) as
 11    (select 1, 'category_1' from dual union all
 12     select 2, 'category_2' from dual union all
 13     select 3, 'category_3' from dual
 14    ),
 15  action_type (id, name) as
 16    (select 1, 'updated' from dual union all
 17     select 2, 'deleted' from dual union all
 18     select 3, 'restored' from dual union all
 19     select 4, 'hided' from dual
 20    ),
 21  action (id, post_id, action_type_id) as
 22    (select 1, 1, 1 from dual union all
 23     select 2, 1, 1 from dual union all
 24     select 3, 1, 3 from dual union all
 25     select 4, 6, 1 from dual union all
 26     select 5, 5, 2 from dual union all
 27     select 6, 2, 3 from dual union all
 28     select 7, 2, 1 from dual union all
 29     select 8, 3, 2 from dual
 30    )

查询从这里开始：

 31  select c.name,
 32    --
 33    (select count(*)
 34     from post p
 35     where p.category_id = c.id
 36    ) cnt_posts_per_cat,
 37    --
 38    (select count(*)
 39     from action a join post p        on p.id = a.post_id
 40                   join action_type t on t.id = a.id
 41     where p.category_id = c.id
 42       and t.name in ('restored', 'deleted', 'updated')
 43    ) cnt_actions_per_cat
 44  from category c
 45  order by c.name;

NAME       CNT_POSTS_PER_CAT CNT_ACTIONS_PER_CAT
---------- ----------------- -------------------
category_1                 4                   3
category_2                 1                   0
category_3                 1                   0

SQL>

【讨论】：