我有两个表:
Quest
- (int) id
- (text) characters
User
- (int) id
- (text) characters
条目如下所示:
任务
id | characters
1 | abcdefgh
2 | mkorti
3 | afoxi
4 | bac
用户
id | characters
1 | abcd
现在我想为用户选择最简单的任务。最简单的任务是 quest.characters 和 user.characters 交集最多的任务。所以在这个例子中,列表看起来像这样(对于 user.id = 1):
questid | easiness
4 | 100
1 | 50
3 | 40
2 | 0
easiness 简单地显示匹配的百分比。 MySQL 是否可以像这样创建列的交集?表现如何?事实上,我确实也有关系(任务 -> 角色和用户 -> 角色),但我猜它的性能不是很好。因为有几千个任务和几千个字符。
更新 #1
好吧,关系似乎仍然是要走的路,好吧。现在我的表看起来像这样:
CREATE TABLE IF NOT EXISTS `quest` (
`questid` int(10) unsigned NOT NULL AUTO_INCREMENT,
PRIMARY KEY (`questid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 ;
CREATE TABLE IF NOT EXISTS `questcharacters` (
`questid` int(10) unsigned NOT NULL,
`characterid` int(10) unsigned NOT NULL,
PRIMARY KEY (`questid`,`characterid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
CREATE TABLE IF NOT EXISTS `single_character` (
`characterid` int(10) unsigned NOT NULL AUTO_INCREMENT,
`single_char` varchar(10) NOT NULL,
PRIMARY KEY (`characterid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
CREATE TABLE IF NOT EXISTS `user` (
`userid` int(10) unsigned NOT NULL AUTO_INCREMENT,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
CREATE TABLE IF NOT EXISTS `usercharacters` (
`userid` int(10) unsigned NOT NULL,
`characterid` int(10) unsigned NOT NULL,
PRIMARY KEY (`userid`,`characterid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
PS.: 不要奇怪为什么 single_char
有 VARCHAR(10) 作为数据类型,但我将使用多字节值,我不知道 MySQL 是如何处理它们的 char(1 ).所以我在那里很慷慨。
更新#2
我现在的查询是:
SELECT usercharacters.userid, questcharacters.questid
FROM `usercharacters`
LEFT OUTER JOIN questcharacters ON usercharacters.characterid = usercharacters.characterid
GROUP BY questcharacters.questid, usercharacters.userid;
但是如何计算容易度/重叠字符?我必须在哪个字段上应用 COUNT()?
更新#3
好吧,看来我可以使用这个查询(使用子选择):
SELECT usercharacters.userid as uid, questcharacters.questid as qid, (SELECT COUNT(questcharacters.characterid) FROM questcharacters LEFT OUTER JOIN usercharacters ON questcharacters.characterid = usercharacters.characterid WHERE questcharacters.questid = qid) as questcount
FROM `usercharacters`
LEFT OUTER JOIN questcharacters ON usercharacters.characterid = usercharacters.characterid
GROUP BY questcharacters.questid, usercharacters.userid;
更新 #4
SELECT usercharacters.userid as uid, questcharacters.questid as qid, (SELECT COUNT(questcharacters.characterid) FROM questcharacters LEFT OUTER JOIN usercharacters ON questcharacters.characterid = usercharacters.characterid WHERE questcharacters.questid = qid) as user_knows, (SELECT COUNT(questcharacters.characterid) FROM questcharacters WHERE questcharacters.questid = qid) as total_characters
FROM `usercharacters`
LEFT OUTER JOIN questcharacters ON usercharacters.characterid = usercharacters.characterid
GROUP BY questcharacters.questid, usercharacters.userid
ORDER BY total_characters / user_knows DESC;
现在唯一缺少的是:选择轻松。 (如在 ORDER BY 子句中)。任何人都知道如何做到这一点?
最佳答案
所以这是我最终的可行解决方案:
SELECT usercharacters.userid AS uid,
questcharacters.questid AS qid,
(SELECT Count(questcharacters.characterid)
FROM questcharacters
LEFT OUTER JOIN usercharacters
ON questcharacters.characterid =
usercharacters.characterid
WHERE questcharacters.questid = qid) AS user_knows,
(SELECT Count(questcharacters.characterid)
FROM questcharacters
WHERE questcharacters.questid = qid) AS total_characters,
(SELECT ( Count(questcharacters.characterid) / (SELECT
Count(questcharacters.characterid)
FROM questcharacters
WHERE
questcharacters.questid = qid) )
FROM questcharacters
LEFT OUTER JOIN usercharacters
ON questcharacters.characterid =
usercharacters.characterid
WHERE questcharacters.questid = qid) AS ratio
FROM `usercharacters`
LEFT OUTER JOIN questcharacters
ON usercharacters.characterid = usercharacters.characterid
GROUP BY questcharacters.questid,
usercharacters.userid
ORDER BY ratio DESC;
我真的需要那么多子选择吗?
关于MySQL 与字符串相交?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17660172/