如何将具有不同值但名称相同且使用datediff的主键分组

Question

所以我有三个表：

来宾可以进行预订，并且您可以在预订表中根据主键“ pID”查看来宾进行了哪种预订。 我只想显示豪华客房的预订以及客人在那儿度过的天数。 我什至不知道我在做什么是正确的方法。 我希望有一个人可以帮助我。

保留：

pID              |begindate    | enddate     |   
------------------------------------------------------
COD12            | 2014-07-15  | 2014-07-18  |
COD400           | 2014-07-20  | 2014-07-21  |
KOD12            | 2014-07-01  | 2014-07-07  |

豪华房桌：

pID              |city         |    
---------------------------------
COD12            | Corona      | 
COD400           | Corona      | 
KOD12            | Kentucky    | 

Small room table:

pID              |city         |    
---------------------------------
COD600           | Corona      |
MOD10            | Madrid      |
KOD20            | Kentucky    | 

我想要的是：在豪华客房中待客的天数，请注意，Corona有两个房间：

city             |amountofdays |    
---------------------------------
Corona           | 4           |
Kentucky         | 6           |

我试图做的是：

SELECT datediff(reservation.enddate,reservation.begindate) as amountofdays, luxeroom.city FROM reservation
INNER JOIN luxeroom ON reservation.pID = luxeroom.pID
GROUP BY luxeroom.pID, luxeroom.city; 

this resulted in:

city             |amountofdays |    
---------------------------------
Corona           | 3           |
Corona           | 1           |
Kentucky         | 6           |

Answer 1

group by以下方式固定group by ：

SELECT sum(datediff(r.enddate, r.begindate)) as amountofdays, l.city
FROM reservation r JOIN
     luxeroom l
     ON r.pID = l.pID
GROUP BY l.city; 

您想要每个城市一个房间，因此GROUP BY应该只包括city 。

Answer 2

您需要分组和求和

SELECT sum(datediff(a.enddate, a.begindate)) as amountofdays, b.city
FROM reservation a 
JOIN luxeroom b  ON a.pID = b.pID
GROUP BY b.city;