如何將具有不同值但名稱相同且使用datediff的主鍵分組

Question

所以我有三個表：

來賓可以進行預訂，並且您可以在預訂表中根據主鍵“ pID”查看來賓進行了哪種預訂。 我只想顯示豪華客房的預訂以及客人在那兒度過的天數。 我什至不知道我在做什么是正確的方法。 我希望有一個人可以幫助我。

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  |

豪華房桌：

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

Small room table:

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

我想要的是：在豪華客房中待客的天數，請注意，Corona有兩個房間：

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

我試圖做的是：

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; 

您想要每個城市一個房間，因此GROUP BY應該只包括city 。

Answer 2

您需要分組和求和

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;