sql中的连接计数

我对SQL非常陌生，正在尝试用一种不愚蠢的方式做一些事情。为了提供一些背景信息，我比较了运营商分支机构中车辆的规划和实际使用情况，需要计算两者的发生率。我有一个子查询为的

，返回如下内容

PLANNED  | REALIZED
---------+----------
TRUCK    | BI-TREM
TRUCK    | TRUCK
TRUCK    | TRUCK
TRUCK    | TRUCK
TRUCK    | TRUCK
CARRETA  | CARRETA
CARRETA  | CARRETA
TRUCK    | KOMBI
TRUCK    | BI-TREM
CARRETA  | KOMBI
CARRETA  | KOMBI
TRUCK    | TRUCK
CARRETA  | CARRETA
CARRETA  | BI-TREM
CARRETA  | CARRETA
CARRETA  | CARRETA
TRUCK    | BI-TREM

VEHICLE | TOTAL_PLANNED | TOTAL_REALIZED
--------+---------------+---------------
CARRETA | 8             | 5
TRUCK   | 9             | 5
BI-TREM | 0             | 4
KOMBI   | 0             | 3

select PLANNED, 
       count(*) as TOTAL_PLANNED
       null as REALIZED,
       0 as TOTAL_REALIZED
from subquery
group by PLANNED

union all

select null as PLANNED,
       0 as TOTAL_PLANNED,
       REALIZED,
       count(*) as TOTAL_REALIZED
from subquery
group by REALIZED

我想返回如下内容

PLANNED  | REALIZED
---------+----------
TRUCK    | BI-TREM
TRUCK    | TRUCK
TRUCK    | TRUCK
TRUCK    | TRUCK
TRUCK    | TRUCK
CARRETA  | CARRETA
CARRETA  | CARRETA
TRUCK    | KOMBI
TRUCK    | BI-TREM
CARRETA  | KOMBI
CARRETA  | KOMBI
TRUCK    | TRUCK
CARRETA  | CARRETA
CARRETA  | BI-TREM
CARRETA  | CARRETA
CARRETA  | CARRETA
TRUCK    | BI-TREM

VEHICLE | TOTAL_PLANNED | TOTAL_REALIZED
--------+---------------+---------------
CARRETA | 8             | 5
TRUCK   | 9             | 5
BI-TREM | 0             | 4
KOMBI   | 0             | 3

select PLANNED, 
       count(*) as TOTAL_PLANNED
       null as REALIZED,
       0 as TOTAL_REALIZED
from subquery
group by PLANNED

union all

select null as PLANNED,
       0 as TOTAL_PLANNED,
       REALIZED,
       count(*) as TOTAL_REALIZED
from subquery
group by REALIZED

我试过以下方法

PLANNED  | REALIZED
---------+----------
TRUCK    | BI-TREM
TRUCK    | TRUCK
TRUCK    | TRUCK
TRUCK    | TRUCK
TRUCK    | TRUCK
CARRETA  | CARRETA
CARRETA  | CARRETA
TRUCK    | KOMBI
TRUCK    | BI-TREM
CARRETA  | KOMBI
CARRETA  | KOMBI
TRUCK    | TRUCK
CARRETA  | CARRETA
CARRETA  | BI-TREM
CARRETA  | CARRETA
CARRETA  | CARRETA
TRUCK    | BI-TREM

VEHICLE | TOTAL_PLANNED | TOTAL_REALIZED
--------+---------------+---------------
CARRETA | 8             | 5
TRUCK   | 9             | 5
BI-TREM | 0             | 4
KOMBI   | 0             | 3

select PLANNED, 
       count(*) as TOTAL_PLANNED
       null as REALIZED,
       0 as TOTAL_REALIZED
from subquery
group by PLANNED

union all

select null as PLANNED,
       0 as TOTAL_PLANNED,
       REALIZED,
       count(*) as TOTAL_REALIZED
from subquery
group by REALIZED

返回

CARRETA 8   NULL    0
TRUCK   9   NULL    0
NULL    0   BI-TREM 4
NULL    0   CARRETA 5
NULL    0   TRUCK   5
NULL    0   KOMBI   3

我还使用子查询尝试了所有可用的联接，但没有成功。事实上，
右连接
起作用，但这只是因为所有计划的车辆也在已实现侧：如果一些车辆不在，我将在
车辆
列中使用空值

感谢您的帮助，即使只是一个指向SQL命令的指针

注：此查询必须同时在SQL Server和Oracle上运行，因此我正在努力实现纯SQL。
尝试：

SELECT PLANNED_REALIZED AS VEHICLE, 
       SUM(TOTAL_PLANNED) AS TOTAL_PLANNED,
       SUM(TOTAL_REALIZED) AS TOTAL_REALIZED
FROM
(select PLANNED AS PLANNED_REALIZED, 
        1 as TOTAL_PLANNED
        0 as TOTAL_REALIZED
 from subquery
 union all
 select REALIZED as PLANNED_REALIZED,
        0 as TOTAL_PLANNED,
        1 as TOTAL_REALIZED
 from subquery
) SQ
GROUP BY PLANNED_REALIZED

我认为这应该奏效：

select T.VEHICLE, 
       (SELECT count(*) FROM Table AS T1 WHERE T1.PLANNED = T.VEHICLE) AS TOTAL_PLANNED
       (SELECT count(*) FROM Table AS T2 WHERE T2.REALIZED = T.VEHICLE) AS TOTAL_REALIZED,
from (SELECT DISTINCT PLANNED AS VEHICLE FROM Table
        UNION SELECT DISTINCT REALIZED AS VEHICLE FROM Table) AS T

谢谢，那也行！尽管如此，我还是要花一些时间来了解它是如何工作的，而不必使用
groupby
：）！这似乎有点复杂，我选择这个答案只是因为更多的人喜欢它。理解这两个答案需要一些时间…它可能看起来更复杂，但它只需要两次读取表格，而Simon的解决方案需要两次（n+1）读取。