MySql SUM ALIAS

Question

我的mysql别名有问题。

我有这个问题：

SELECT (`number_of_rooms`) AS total, id_room_type, 
     COUNT( fk_room_type ) AS reservation , 
     SUM(number_of_rooms - reservation) AS result
FROM room_type
    LEFT JOIN room_type_reservation 
         ON id_room_type = fk_room_type
WHERE  result > 10
GROUP BY id_room_type

我的问题从SUM，cannot recognize reservation开始，然后我想将结果用于where条件。喜欢（where result > 10）

Answer 1

要对聚合函数的结果应用谓词（过滤条件），请使用Having子句。 Where子句表达式仅适用于在任何聚合之前创建的中间结果集。

 SELECT (`number_of_rooms`) AS total, id_room_type,
     COUNT( fk_room_type ) AS reservation , 
     SUM(number_of_rooms - reservation) AS result
 FROM room_type
     LEFT JOIN room_type_reservation 
         ON id_room_type = fk_room_type
 GROUP BY id_room_type
 Having SUM(number_of_rooms - reservation) > 10

Answer 2

不是100％但据我所知，您不能在声明中使用别名，这就是您遇到列问题的原因。试试这个：

 SELECT (`number_of_rooms`) AS total, id_room_type,
     COUNT( fk_room_type ) AS reservation , 
     SUM(number_of_rooms - COUNT( fk_room_type ) ) AS result
 FROM room_type
     LEFT JOIN room_type_reservation 
         ON id_room_type = fk_room_type
 GROUP BY id_room_type
 Having SUM(number_of_rooms - COUNT( fk_room_type ) ) > 10

Answer 3

一种方法是将其包装成一个多 SELECT

SELECT t.*, t.number_of_rooms - t.reservation AS result FROM
  (
    SELECT (`number_of_rooms`) AS total, id_room_type, 
     COUNT( fk_room_type ) AS reservation , 
     SUM(number_of_rooms - reservation) AS result
     FROM room_type
     LEFT JOIN room_type_reservation 
         ON id_room_type = fk_room_type
     WHERE  result > 10
     GROUP BY id_room_type
   ) t