Mysql 如何计算所选列的单个和

我目前正在大学管理项目中工作，我想计算每栋建筑的容量，如果每个部门可以在同一栋建筑中容纳不同的部分，如下面的comp和elect 我写了一个查询，在那里我得到了不需要的总容量 我的问题是：

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

我得到的ans是660！！ 在这个嵌套查询中，我是否在错误的位置使用了sum和distinct？如何获得单个建筑的容量

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 |
+----------+-------------+----------+

使用以下查询：

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

如果要确保该建筑中至少有一个部门，请使用内部连接

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由于您只想检查部门表中存在的建筑物的容量，因此需要为每个建筑物计算容量总和，并添加exists子句，以仅显示至少一套公寓的居住建筑物：

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

如果您只需要容量，而不管部门中是否存在建筑，则删除exists子句：

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第一个答案是ty，但第二个查询显示了错误的输出：仍然显示相同的输出..不管怎样，没有连接的ans工作：考虑是否有帮助。如何做。。我是新加入的，声誉也没什么可提升的：

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

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