gpt4 book ai didi

mySQL 子查询出现限制错误

转载 作者:行者123 更新时间:2023-11-29 19:44:43 25 4
gpt4 key购买 nike

我有两个游戏部分,但遇到了问题。第一部分的比赛什么时候开始,我不想在第二部分再次展示她。对于第一部分中的表演游戏,我正在使用:

SELECT *,platform.platformName FROM games 
JOIN platform ON(platform.PlatformID=games.GamePlatform)
WHERE GameDate <= NOW() AND GameSlide = 0
GROUP BY GameName
HAVING 1
ORDER BY `games`.`GameDate` DESC LIMIT 8

对于第二部分的表演游戏:

SELECT *,platform.platformName 
FROM games
JOIN platform ON(platform.PlatformID=games.GamePlatform)
WHERE platform.PlatformID = 2 AND GameSlide = 0
GROUP BY GameName
HAVING 1
order by rand(dayofyear(CURRENT_DATE))
LIMIT 8

我尝试创建这样的子查询:

SELECT *,platform.platformName 
FROM games
JOIN platform ON(platform.PlatformID=games.GamePlatform)
WHERE games.GameID
NOT IN(
SELECT GameID
FROM games
WHERE GameDate <= NOW() AND GameSlide = 0
GROUP BY GameName
HAVING 1
ORDER BY `games`.`GameDate` DESC LIMIT 8
) AND platform.PlatformID = 2 AND GameSlide = 0
GROUP BY GameName
HAVING 1
order by rand(dayofyear(CURRENT_DATE)) LIMIT 8

但 MySQL 调用错误:#1235 - 此版本的 MariaDB 尚不支持“LIMIT & IN/ALL/ANY/SOME 子查询”

你能帮我一下吗?我为我的英语感到抱歉。我尽力了。

最佳答案

我已经做到了。我用选择创建了一个新 View 。

CREATE VIEW Novinky AS SELECT games.GameID FROM games WHERE games.GameDate <= NOW() AND games.GameSlide = 0 GROUP BY games.GameName HAVING 1 ORDER BY `games`.`GameDate` DESC LIMIT 8

然后将 NOT IN 与 SELECT from View 结合使用:

SELECT *,platform.platformName FROM games JOIN platform ON(platform.PlatformID=games.GamePlatform) WHERE GameID NOT IN (SELECT * FROM novinky) AND platform.PlatformID = 2 AND GameSlide = 0 GROUP BY GameName HAVING 1 order by rand(dayofyear(CURRENT_DATE)) LIMIT 8

关于mySQL 子查询出现限制错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41100473/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com