Oracle 如何对一个表中的单个查询执行多个子查询？

我有两张桌子：

大楼

和

公寓

建筑有

ID\u-BUILDING

，

BUILDING\u-NAME

（这是这个问题唯一重要的部分）

公寓有

ID\u楼

，

N\u公寓

，

总房间

我需要执行如下查询：

| BUILDING NAME | TOTAL APARTMENTS | TOTAL APARTMENTS WITH 1 BEDROOM | TOTAL APARTMENTS WITH 2 BEDROOMS |
|---------------|------------------|---------------------------------|----------------------------------|
| BUILDING A    | 31               | 8                               | 0                                 
|_______________________________________________________________________________________________________|
| BUILDING B    | 20               | 14                              | 11                               
|________________________________________________________________________________________________________
| BUILDING C    | 41               | 90                              | 5                                
|________________________________________________________________________________________________________

卧室的数量可以在1到5之间

为此，我提出了以下问题：

SELECT E.BUILDING_NAME as "name", COUNT(D.N_APARTMENTS) "TOTAL APARTMENTS", (SELECT COUNT(D1.TOTAL_ROOMS) FROM APARTMENT D1 WHERE D1.TOTAL_ROOMS = 1)
FROM BUILDING E
JOIN APARTMENT D 
ON E.ID_BUILDING = D.ID_BUILDING    
GROUP BY E.E.BUILDING_NAME
ORDER BY E.BUILDING_NAME;

不幸的是，这包括所有有一间房的公寓，而不是按建筑名称计算的：

| BUILDING NAME | TOTAL APARTMENTS | TOTAL APARTMENTS WITH 1 BEDROOM |
|---------------|------------------|---------------------------------|
| BUILDING A    | 31               | 122                             |
| BUILDING B    | 20               | 122                             |
| BUILDING C    | 41               | 122                             |

我确实试过答案，但它们不是完全相同的问题

我认为解决方案可能是使用多个连接（或内部连接），但我找不到正确的答案

---

提前感谢。

使用条件聚合：

SELECT E.BUILDING_NAME as "name", 
 COUNT(D.N_APARTMENTS) "TOTAL APARTMENTS", 
 SUM(CASE WHEN D.TOTAL_ROOMS = 1 THEN 1 ELSE 0 END) AS "TOTAL APARTMENTS WITH 1 BEDROOM"
FROM BUILDING E
JOIN APARTMENT D 
ON E.ID_BUILDING = D.ID_BUILDING    
GROUP BY E.E.BUILDING_NAME
ORDER BY E.BUILDING_NAME;

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它的工作原理与预期一样。我被卡住了，意识到答案和你说的一样简单，我觉得有点傻。谢谢