Question

使用的表如下：

CREATE TABLE IF NOT EXISTS planta(
codigo int PRIMARY KEY NOT NULL AUTO_INCREMENT,
especialidad varchar(25) NOT NULL
)ENGINE=InnoDB;

CREATE TABLE IF NOT EXISTS habitacion(
id int PRIMARY KEY NOT NULL AUTO_INCREMENT,
numero_camas int NOT NULL,
planta_id int NOT NULL,
FOREIGN KEY (planta_id) REFERENCES planta(codigo)
)ENGINE=InnoDB;

CREATE TABLE IF NOT EXISTS paciente(
dni varchar(9) PRIMARY KEY NOT NULL,
num_ss varchar(10) NOT NULL,
nombre varchar(20) NOT NULL,
direccion varchar(50) NOT NULL,
tratamiento mediumtext NOT NULL,
diagnostico mediumtext NOT NULL,
habitacion_id int NOT NULL,
medico_id int NOT NULL,
FOREIGN KEY (habitacion_id) REFERENCES habitacion(id),
FOREIGN KEY (medico_id) REFERENCES medico(num_colegiado)
)ENGINE=InnoDB;

查询是这样的：

SELECT planta.codigo AS Floor_id, habitacion.id AS Room_id, numero_camas - count(dni) AS Free_beds 
FROM habitacion, paciente, planta 
WHERE planta_id = planta.codigo AND habitacion_id = habitacion.id 
GROUP BY planta.codigo, habitacion.id;

它返回以下结果：

Floor id | Room id | Free beds
    1         1          1    
    1         2          1    
    2         3          3

但是我想要这个：

Floor id | Rooms | Free beds
    1        2         2    
    2        1         3

我认为我有很多问题，因为首先我必须计算每个房间有多少张床（我通过减去房间中的床数减去分配给该房间的人数来做到这一点。） / p>

我想知道是否可以通过当前查询将结果分组。

Answer 1

从不在FROM子句中使用逗号。始终使用正确的，明确的，标准 JOIN语法。

您只需要正确的GROUP BY逻辑即可。我想就是这样

SELECT pl.codigo AS Floor_id, count(count distinct h.id) as num_rooms,
       MAX(numero_camas) - count(distinct h.id) AS Free_beds 
FROM habitacion h join
     paciente p
     on p.habitacion_id = h.id   -- just a guess that this is the right join condition
     planta pl
     on h.planta_id = pl.codigo
GROUP BY pl.codigo

Answer 2

根据戈登·利诺夫（Gordon Linoff）的回答和您的陈述，我想通过这一微小的更改，您应该会得到正确的结果（我想需要对不同的房间进行计数，并且床位数量减去分配的人的总和）上）：

SELECT 
     pl.codigo AS Floor_id,  
     count(distinct habitacion_id) as num_rooms, 
     SUM(numero_camas - count(dni)) AS Free_beds 
FROM habitacion h 
join paciente p on p.habitacion_id = h.id 
join condition planta pl on h.planta_id = pl.codigo 
GROUP BY pl.codigo

Answer 3

您需要两个汇总：每个居住区的床位数，然后是每层可用床位数的总和。一种方法是：

select planta_id, count(*) as rooms, sum(available - occupied) as free_beds
from
(
  select
    planta_id,
    h.id as habitacion_id,
    numero_camas as available,
    (select count(*) from paciente p where p.habitacion_id = h.id) as occupied
  from habitacion h
) habitation_aggregated
group by planta_id
order by planta_id;

您看到我在FROM子句中有一个子查询，它代表一个聚合。这是处理多种聚合的一种非常好的节省方法。

GROUP BY分组查询的结果

3 个答案: