Question

我目前正在从事大学管理项目，我希望计算每个建筑物的容量，当每个部门的案例可以容纳不同的部分如下（comp和选择在同一建筑物内）我写了一个查询，我得到的总容量是不需要的我的疑问是：

select sum(distinct capacity) 
from classroom  
where building in (select building from department group by building)

我得到的是660！我在这个嵌套查询中使用sum和distinct在错误的位置吗？如何获得个人建筑的容量？

department
+------------+----------+-----------+
| dept_name  | building | budget    |
+------------+----------+-----------+
| Biology    | Watson   |  90000.00 |
| Comp. Sci. | Taylor   | 100000.00 |
| Elec. Eng. | Taylor   |  85000.00 |
| Finance    | Painter  | 120000.00 |
| History    | Painter  |  50000.00 |
| Music      | Packard  |  80000.00 |
| Physics    | Watson   |  70000.00 |
+------------+----------+-----------+    
classroom
+----------+-------------+----------+
| building | room_number | capacity |
+----------+-------------+----------+
| Packard  | 101         |      500 |
| Painter  | 514         |       10 |
| Taylor   | 3128        |       70 |
| Watson   | 100         |       30 |
| Watson   | 120         |       50 |
+----------+-------------+----------+

Answer 1

由于您只想检查department表格中存在的建筑物容量，因此每个building需要计算capacity之和，并添加exists子句以显示只有那些至少有一套公寓住宅的建筑物：

select
  building, sum(capacity) as capacity
from classrom c
where exists (
  select 1
  from department d
  where c.building = d.building
  )
group by building

如果您只需要department中不存在建筑物的容量，那么请删除exists子句：

select
  building, sum(capacity) as capacity
from classrom
group by building

Answer 2

使用以下查询：

SELECT building, SUM(capacity) as total_capacity
FROM classroom 
GROUP BY building;

如果您想确保该建筑物中至少有一个部门，请使用inner join

SELECT classroom.building, SUM(capacity) as total_capacity
FROM classroom 
INNER JOIN department 
  ON classroom.building == department.building 
GROUP BY classroom.building;

如何计算所选列的单个总和

2 个答案: