Mysql 多查询'；需要什么？

我有三张桌子： -房间：关于房间的信息 -床：房间中每个床的信息（类型，…）链接到带有roomID的房间 -分配：关于分配给一个组的房间的床位数的信息（一个房间可以有两个组）

我想要一份有床位数和房间内分配床位数的清单

样本：

SELECT 
COUNT(beds.id) AS beds
FROM rooms 
JOIN beds ON rooms.id=beds.roomID 
WHERE rooms.id='124'

给出房间中的床数（8）

给我房间里分配的床位数：7（1+6）

给我：

beds | allocated
----------------
16   | 56

• 16张床位（由于分配表中有2条记录，所以计数两次）
• 56张分配床位（7*8=56）

如何进行查询以获得正确的结果

beds | allocated
----------------
8    | 7

---

谢谢

要获得正确的床位数量，您只需在您的

计数中添加一个distinct

COUNT(distinct beds.id) AS beds

为了避免分配床位的倍增，您可以使用以下解决方法：

SUM(requestsAllocations.bedsAllocated) / COUNT(distinct beds.id) as allocated

最后一个问题是

SELECT  COUNT(distinct t3.id) AS beds,
        SUM(t2.bedsAllocated) / COUNT(distinct t3.id) AS allocated
FROM    rooms t1
JOIN    requestsAllocations t2
ON      t2.roomID = t1.id
JOIN    beds t3
ON      t1.id = t3.roomID
WHERE   t1.id='124'

我还为您的表提供了别名以提高可读性
您必须在子查询中至少进行一次聚合（无论是
计数
还是
总和
），以便重复记录不会影响聚合结果。或者，您可以简单地在选择列表中聚合两个相关子查询：

select rooms.id,
       (select count(*) from beds where beds.roomID=rooms.id) as bed_count,
       (select sum(bedsAllocated) from requestsAllocations where requestsAllocations.roomID=rooms.id) as allocated_count
from rooms

SELECT  COUNT(distinct t3.id) AS beds,
        SUM(t2.bedsAllocated) / COUNT(distinct t3.id) AS allocated
FROM    rooms t1
JOIN    requestsAllocations t2
ON      t2.roomID = t1.id
JOIN    beds t3
ON      t1.id = t3.roomID
WHERE   t1.id='124'

select rooms.id,
       (select count(*) from beds where beds.roomID=rooms.id) as bed_count,
       (select sum(bedsAllocated) from requestsAllocations where requestsAllocations.roomID=rooms.id) as allocated_count
from rooms