Question

我刚刚在Oracle SQL Developer中为我的研究所启动一个数据库项目，并且在第一个查询中遇到了问题，在该查询中，我必须返回“建筑物名称”，“部门总数”，“拥有1个房间的部门”，“拥有2个房间的部门数”，依此类推直到5个房间。

到目前为止，我有这个：

    select  E.EDI_NOMBRE_EDIFICIO as "NOMBRE EDIFICIO",
            count(D.NRO_DEPARTAMENTO) as "TOTAL DEPTOS",
            (
             select count(D.NRO_DEPARTAMENTO) as "TOTAL DEPTOS 1 DORMITORIO"
             from (edificio E join departamento D 
             on E.id_edificio = D.ID_EDIFICIO)
             where D.TOTAL_DORMITORIOS = 1
             group by E.EDI_NOMBRE_EDIFICIO
            ) AS "DEPTOS 1 DORM"        
    from (edificio E join departamento D on E.id_edificio = D.ID_EDIFICIO)
    group by E.EDI_NOMBRE_EDIFICIO
    order BY E.EDI_NOMBRE_EDIFICIO;

问题是，我很好地获得了前2列，但是我用来获取具有1个房间的部门总数的子查询却抛出了01427错误，因为（我认为）我正在放置一个SELECT子句中的多行子查询。

如果不使用Select子句中的子查询，我不知道如何处理正确的查询。如果您能帮助我，我将不胜感激。预先感谢。

这是我编程课程的第二个学期，所以我什么都不知道。实际上，我只是阅读了“合并不同count（*）条sql查询的结果”的问题和答案，我听不懂。

Answer 1

如何将子查询替换为：

Sum(Case when D.TOTAL_DORMITORIOS = 1 then 1 else 0 end) as "DEPTOS 1 DORM"

...等等

Sum(Case when D.TOTAL_DORMITORIOS = 2 then 1 else 0 end) as "DEPTOS 2 DORM"
Sum(Case when D.TOTAL_DORMITORIOS = 3 then 1 else 0 end) as "DEPTOS 3 DORM"

Answer 2

以这种方式解决：

SELECT  E.EDI_NOMBRE_EDIFICIO as "NOMBRE EDIFICIO",
         (
         SELECT COUNT(D.NRO_DEPARTAMENTO)         
         FROM departamento D 
         WHERE E.id_edificio = D.ID_EDIFICIO
         ) AS "TOTAL DEPTOS",
         (
          SELECT COUNT(D.NRO_DEPARTAMENTO) AS "TOTAL DEPTOS 1 DORMITORIO"
          FROM departamento D
          WHERE D.TOTAL_DORMITORIOS = 1
          AND D.id_edificio = E.id_edificio
         ) AS "DEPTOS 1 DORM",
         (
          SELECT COUNT(D.NRO_DEPARTAMENTO)
          AS "TOTAL DEPTOS 1 DORMITORIO"
          FROM departamento D
          WHERE D.TOTAL_DORMITORIOS = 2
          AND D.id_edificio = E.id_edificio
         ) AS "DEPTOS 2 DORM",
         (
          SELECT COUNT(D.NRO_DEPARTAMENTO)
          AS "TOTAL DEPTOS 1 DORMITORIO"
          FROM departamento D
          WHERE D.TOTAL_DORMITORIOS = 3
          AND D.id_edificio = E.id_edificio
         ) AS "DEPTOS 3 DORM",
         (
          SELECT COUNT(D.NRO_DEPARTAMENTO)
          AS "TOTAL DEPTOS 1 DORMITORIO"
          FROM departamento D
          WHERE D.TOTAL_DORMITORIOS = 4
          AND D.id_edificio = E.id_edificio
         ) AS "DEPTOS 4 DORM",
         (
          SELECT COUNT(D.NRO_DEPARTAMENTO)
          AS "TOTAL DEPTOS 1 DORMITORIO"
          FROM departamento D
          WHERE D.TOTAL_DORMITORIOS = 5
          AND D.id_edificio = E.id_edificio
         ) AS "DEPTOS 5 DORM"
FROM EDIFICIO E
ORDER BY E.EDI_NOMBRE_EDIFICIO ASC;

Answer 3

如果您首先将departamento表中的结果归为一组，您的答案可能会更准确，因为那样您将只对表查询一次，而不是对每个id_edificio查询6次，就像这样：

SELECT e.edi_nombre_edificio AS "NOMBRE EDIFICIO",
       d.total_deptos AS "TOTAL DEPTOS",
       d.deptos_1_dorm AS "DEPTOS 1 DORM",
       d.deptos_2_dorm AS "DEPTOS 2 DORM",
       d.deptos_3_dorm AS "DEPTOS 3 DORM",
       d.deptos_4_dorm AS "DEPTOS 4 DORM",
       d.deptos_5_dorm AS "DEPTOS 5 DORM"
FROM   edificio e
       INNER JOIN (SELECT id_edificio,
                          COUNT(nro_departamento) total_deptos,
                          COUNT(CASE WHEN total_dormitorios = 1 THEN nro_departamento END) deptos_1_dorm,
                          COUNT(CASE WHEN total_dormitorios = 1 THEN nro_departamento END) deptos_2_dorm,
                          COUNT(CASE WHEN total_dormitorios = 1 THEN nro_departamento END) deptos_3_dorm,
                          COUNT(CASE WHEN total_dormitorios = 1 THEN nro_departamento END) deptos_4_dorm,
                          COUNT(CASE WHEN total_dormitorios = 1 THEN nro_departamento END) deptos_5_dorm
                   FROM   departamento) d ON e.id_edificio = d.id_edificio;

有关在查询中组合不同计数的SQL问题

3 个答案: