MySql SUM 别名答案

【问题标题】：MySql SUM ALIASMySql SUM 别名
【发布时间】：2013-06-08 17:19:46
【问题描述】：

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）

【问题讨论】：

标签： mysql sql sum alias

【解决方案1】：

要对聚合函数的结果应用谓词（过滤条件），请使用 Have 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

【讨论】：

感谢您的回答，但 SUM(number_of_rooms - reservation) AS 结果返回我 #1054 - 'field list' 中的未知列 'reservation'

【解决方案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

【讨论】：

感谢您的回答，但我已经准备好这样做并返回给我 #1111 - 无效使用组函数
让我们看看，再试一次，可能是包围问题。 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
再次感谢，但结果不正确总 id_room_type 预订结果 20 10 201 3819 200 11 10 1990 10 22 100 900 22 25 0 22
感谢您的宝贵时间，但我找到了嵌套选择 SELECT (number_of_rooms) 作为总计、id_room_type、c.reservation、SUM(number_of_rooms - c.reservation) 作为来自 room_type LEFT JOIN ( SELECT number_of_rooms AS moutsela, fk_room_type, COUNT( fk_room_type ) AS reservation FROM room_type LEFT JOIN room_type_reservation ON id_room_type = fk_room_type GROUP BY id_room_type )c ON id_room_type = fk_room_type GROUP BY id_room_type LIMIT 0 , 30 但我会尝试找到优化我的查询的方法
这就是为什么我想设置你的表结构:)

【解决方案3】：

一种方法是将其包装到另一个 SELECT 中

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

【讨论】：