如何对来自 table 的单个查询执行多个子查询?
How can I do multiples sub-querys on a single query from one table?
我有两个表:BUILDING
和 APARTMENT
.
BUILDING 有 ID_BUILDING
、BUILDING_NAME
(这就是这个问题唯一重要的)
公寓有 ID_BUILDING
、N_APARTMENTS
、TOTAL_ROOMS
。
我需要做这样的查询:
| 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 whis 查询:
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,这计算了所有带 1 个房间的公寓,而不是 BUILDING_NAME
:
| BUILDING NAME | TOTAL APARTMENTS | TOTAL APARTMENTS WITH 1 BEDROOM |
|---------------|------------------|---------------------------------|
| BUILDING A | 31 | 122 |
| BUILDING B | 20 | 122 |
| BUILDING C | 41 | 122 |
我确实尝试了 here and 的答案,但它们不是同一个问题。
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。
提前致谢。
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;
我有两个表:BUILDING
和 APARTMENT
.
BUILDING 有 ID_BUILDING
、BUILDING_NAME
(这就是这个问题唯一重要的)
公寓有 ID_BUILDING
、N_APARTMENTS
、TOTAL_ROOMS
。
我需要做这样的查询:
| 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 whis 查询:
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,这计算了所有带 1 个房间的公寓,而不是 BUILDING_NAME
:
| BUILDING NAME | TOTAL APARTMENTS | TOTAL APARTMENTS WITH 1 BEDROOM |
|---------------|------------------|---------------------------------|
| BUILDING A | 31 | 122 |
| BUILDING B | 20 | 122 |
| BUILDING C | 41 | 122 |
我确实尝试了 here and
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。
提前致谢。
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;