数据库名称：Mysql
数据库版本：MySQL 5.5.44-MariaDB - MariaDB Server
内容/问题描述：( 使用别的办法解决问题了 )
问题的SQL:
UPDATE `players`
SET `players`.`xp` = `players`.`xp` +
(SELECT count(oldplayers.`pid`) as addxp FROM (select
tmp.`pid`,tmp.`loginTime` from `players` tmp) as oldplayers
inner join `friends` tmpf on oldplayers.`pid` = tmpf.`playerB` where
tmpf.`friendship` = 2 and oldplayers.`loginTime` > (NOW() - INTERVAL 3 DAY)
and tmpf.`playerA` = 133223225)
WHERE `pid` = 133223225 ;
中间那一段(SELECT ... tmpf.`playerA` = 133223225 )
单独执行可得到资料笔数2笔, 用上述语法却是查到 0笔
大家好,
这个SQL问题的背景是我把玩家经验值xp
设计成需要好友进行登入游戏才会增加玩家经验值xp,
相关条件有:
好友登入动作需要在过去三天内进行才有效
( players.loginTime > (NOW() - INTERVAL 3 DAY) )
好友关系成立
( friends.friendship = 2 )
以上条件成立时,根据查询到的资料笔数,
用来增加玩家资料表players 的经验值xp
玩家资料表
TABLE players
pid
xp (经验值)
loginTime (登入时间)
玩家的好友资料表
TABLE friends
playerA (玩家本人pid ,资料内容可对应 players.pid)
playerB (好友的pid, 资料内容可对应 players.pid)
friendship (交往程度,2表示好友)
SQL 1 (没有问题, 可成功查询SQL 2需要的 xp)：
SELECT count(oldplayers.`pid`) as addxp
FROM (select tmp.`pid`,tmp.`loginTime` from `players` tmp) as oldplayers
inner join `friends` tmpf on oldplayers.`pid` = tmpf.`playerB`
WHERE tmpf.`friendship` = 2 and oldplayers.`loginTime` > (NOW() - INTERVAL 3
DAY) and tmpf.`playerA` = 133223225
SQL 2
(有问题的语法, 这是使用SQL 1的部分加入Update 里,
查询xp 完毕后同时更新xp, 语法执行没有出现问题, 但是结果有问题,
xp 没有增加, 经测试后得知 SQL 1的部分得到0, Why ???)
UPDATE `players`
SET `players`.`xp` = `players`.`xp` +
(SELECT count(oldplayers.`pid`) as addxp FROM (select
tmp.`pid`,tmp.`loginTime` from `players` tmp) as oldplayers
inner join `friends` tmpf on oldplayers.`pid` = tmpf.`playerB` where
tmpf.`friendship` = 2 and oldplayers.`loginTime` > (NOW() - INTERVAL 3 DAY)
and tmpf.`playerA` = 133223225)
WHERE `pid` = 133223225 ;
SQL 3 (上述的语法都只有针对单一玩家, 如果有N个玩家就要执行N次,
SQL 3改为针对所有玩家进行一次处理)
本人能力不足做不出来...
先感谢大家看完了这复杂的问题...
恳请高手帮忙
====================用别的办法解决问题了
SQL n1:
( 查询每个玩家须要加多少xp )
SELECT `playerA`,count(`playerA`) as addxp FROM `players` inner join
`friends` on `players`.`pid` = `friends`.`playerB` WHERE `friendship` = 2
and `loginTime` > (NOW() - INTERVAL 3 DAY) GROUP BY `playerA`
SQL n2:
( N个玩家更新xp 资料, 下列语法中的111 与 222 是SQL n1的查询结果addxp 与
playerA,
再用中间程式( 我的是node.js )去拼出 N 句SQL n2给MySql执行)
Update players Set xp = xp + 111 where pid = 222 ;