Question

我知道这是一个愚蠢的问题，但我是php的新手，所以任何帮助都很有用。

我有两个表，product和payment：

产品表格包含Id，Room_Available，price，...
付款表格包含Id，Product_Id，Total_Room，...
产品表有很多记录

我所期望的是，如果Occupied_Room <= Room_Available，则条件会失败，状态='Y'，否则状态='N'。

cms_product.Room_Available是可用的总房间
cms_payment.Total_Room已预订客房

查询是：

SELECT SUM(cms_payment.Total_Room) as Occupied_Room,
    `cms_product`.`Room_Available` as Available_Room  
FROM 
(
    `cms_product` 
    LEFT JOIN `cms_payment` on (cms_product.`Id` = cms_payment.`Product_Id`)
)
GROUP BY `cms_product`.`Id` 
HAVING Occupied_Room <= Room_Available

Answer 1

在黑暗中有点刺，但我想你想要：

SELECT Prod.Id, 
    CASE WHEN COALESCE(SUM(Pay.Total_Room),0) <= Prod.Room_Available THEN 'Y' 
         ELSE 'N'
    END AS Status
FROM cms_product AS Prod
LEFT JOIN cms_payment AS Pay ON Prod.Id = Pay.Product_Id
GROUP BY Prod.Id, Prod.Room_Available

这将显示所有产品ID和状态（'Y' =可用或已完全预订; 'N' =已超额预订）。

Answer 2

试试这个：下表使用一些样本数据。您可以将状态更改为y或n：）

SQLFIDDLE DEMO

select product.id, 
case when coalesce(x.sm,0) <= product.total_room then 
'available' else 'unavailable' end status,
product.total_room, coalesce(x.sm,0) as rooms_required
from product 
left join (
select payment.product_id, 
sum(payment.room_required) sm
from payment 
group by payment.product_id)x
on x.product_id = product.id
group by product.id
;


| ID |      STATUS | TOTAL_ROOM | ROOMS_REQUIRED |
--------------------------------------------------
|  1 | unavailable |         10 |             12 |
|  2 |   available |         15 |             15 |
|  3 |   available |          5 |              4 |
|  4 |   available |         10 |              0 |

mysql查询查询

2 个答案: