如何让这个 MySQL 查询在 MySQL 4.1.15 上工作？答案

【问题标题】：How can I get this MySQL query to work on MySQL 4.1.15?如何让这个 MySQL 查询在 MySQL 4.1.15 上工作？
【发布时间】：2012-04-17 02:08:48
【问题描述】：

以下查询适用于 MySql 5，但我的客户端使用的是 MySql 4.1.15，并且此查询因以下错误而阻塞：

#1140 - 混合 GROUP 列 (MIN(),MAX(),COUNT(),...) 如果没有 GROUP BY 子句，则 GROUP 列是非法的

升级客户端 MySql 版本不是一个选项顺便说一句。有人可以帮我重写它以在旧版本的 MySql 上工作吗？

SELECT 
   Sum( room_rev + room_rev_future ) as weekly_room_rev, 
   Sum( (`food_rev`) + (`beverage_rev`)) as weekly_catering_rev,
   (SELECT min_sales_persons 
   FROM bdp_hotel_min WHERE 
   bdp_hotel_min.hotel_num = bdp_scorecard.hotel_num 
   AND UNIX_TIMESTAMP(bdp_hotel_min.`min_override_week`) 
   < UNIX_TIMESTAMP("2011-12-02 23:59:59") 
   ORDER BY bdp_hotel_min.`min_override_week` DESC LIMIT 1 ) as override_persons,
   min_sales_persons 
FROM bdp_scorecard JOIN locations ON bdp_scorecard.hotel_num = locations.hotel_num 
WHERE bdp_scorecard.hotel_num =837 AND bdp_scorecard.hotel_num = 837 
AND UNIX_TIMESTAMP(bdp_scorecard.`created_date`) >= UNIX_TIMESTAMP("2011-11-26") 
AND UNIX_TIMESTAMP(bdp_scorecard.`created_date`) < UNIX_TIMESTAMP("2011-12-02 23:59:59")

【问题讨论】：

标签： mysql database

【解决方案1】：

无法测试，但考虑到错误似乎需要 group by ..

SELECT 
   Sum( room_rev + room_rev_future ) as weekly_room_rev, 
   Sum( (`food_rev`) + (`beverage_rev`)) as weekly_catering_rev,
   (SELECT min_sales_persons 
   FROM bdp_hotel_min WHERE 
   bdp_hotel_min.hotel_num = bdp_scorecard.hotel_num 
   AND UNIX_TIMESTAMP(bdp_hotel_min.`min_override_week`) 
   < UNIX_TIMESTAMP("2011-12-02 23:59:59") 
   ORDER BY bdp_hotel_min.`min_override_week` DESC LIMIT 1 ) as override_persons,
   min_sales_persons 
FROM bdp_scorecard JOIN locations ON bdp_scorecard.hotel_num = locations.hotel_num 
WHERE bdp_scorecard.hotel_num =837 AND bdp_scorecard.hotel_num = 837 
AND UNIX_TIMESTAMP(bdp_scorecard.`created_date`) >= UNIX_TIMESTAMP("2011-11-26") 
AND UNIX_TIMESTAMP(bdp_scorecard.`created_date`) < UNIX_TIMESTAMP("2011-12-02 23:59:59")
GROUP BY override_persons, min_sales_persons

【讨论】：