Question

我有两张桌子：

Quest
- (int) id 
- (text) characters

User
- (int) id
- (text) characters

参赛作品如下：

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的最交叉点。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

简单性只表示匹配了多少百分比。 MySQL有可能像这样交叉列吗？性能如何？事实上我做也有关系（任务 - ＆gt;字符和用户 - ＆gt;字符），但我想它不是很高效。因为有几千个任务，还有几千个字符。

更新＃1

好的，关系仍然似乎要走了，好的。现在我的表格看起来像这样：

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将VARCHAR（10）作为数据类型，但我将使用多字节值，我不知道MySQL如何处理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/重叠字符？我必须在哪个字段中应用COUNT（）？

更新＃3

好吧，好像我使用了这个查询（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;

现在只缺少一件事：选择简单。（如在ORDER BY子句中）。谁知道怎么做？

Answer 1

如果您实际拥有questcharacter和usercharacters表，那么这是最好的方法：

SELECT uc.id AS userid, 
       qc.id AS qcid, 
       COUNT(*) AS NumCharacters,
       COUNT(qc.char) AS Nummatches,
       COUNT(qc.char) / count(*) AS Easiness
FROM UserCharacters uc 
   LEFT OUTER JOIN QuestCharacters qc ON uc.char = qc.char
WHERE uc.id = 1
   GROUP BY uc.id, qc.id
   ORDER BY easiness DESC
LIMIT 1

如果你只将它们作为字符串 - 那么SQL并不漂亮。你必须进行交叉连接和大量的字符串操作。最好的方法是以关系数据库的形式（每个列表元素一行）更加规范化，而不是将列表嵌入字符串中。

Answer 2

所以这是我最后的工作解决方案：

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;

我真的需要那么多次选择吗？

MySQL在字符串上相交？

更新＃1

更新＃2

更新＃3

更新＃4

2 个答案: