MySQL intersect on strings? - mysql

Question

I do have two tables:
Quest
- (int) id
- (text) characters
User
- (int) id
- (text) characters
Entries look like this:
Quest
id | characters
1 | abcdefgh
2 | mkorti
3 | afoxi
4 | bac
User
id | characters
1 | abcd
Now I want to select the easiest Quest for User. The easiest quest is the one if the most intersections of quest.characters and user.characters. So in this example the list would look like this (for user.id = 1):
questid | easiness
4 | 100
1 | 50
3 | 40
2 | 0
The easiness simply show how many percent was matched. Is it possible with MySQL to make intersections of columns like this? What's the performance like? In fact I do have relations as well (quest -> character and user -> characters), however I guess it's not very performant. As there are a few thousand quests and also a few thousand characters.
Update #1
Okay, relational still seems the way to go, okay. Now my tables look like this:
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.: Don't wonder why single_char has VARCHAR(10) as data type, but I will use multi-byte values and I don't know how MySQL handles them for char(1). So I just was generous there.
Update #2
My query for now is:
SELECT usercharacters.userid, questcharacters.questid
FROM `usercharacters`
LEFT OUTER JOIN questcharacters ON usercharacters.characterid = usercharacters.characterid
GROUP BY questcharacters.questid, usercharacters.userid;
But how to calculate the easiness/overlapping characters? On which field do I have to apply COUNT()?
Update #3
Okay, seems like I got it working with this query (uses a subselect):
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;
Update #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;
Only thing missing now: Selecting the easyiness. (As in the ORDER BY clause). Anyone knows how to do this?

Answer 1

So this is my final and working solution:
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;
Do I really need that many sub-selects?

Answer 2

If you actually have questcharacter and usercharacters tables, then that is the best way to go:
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
If you have them only as strings -- the SQL is not pretty. You have to do a cross join and lots of string manipulation. The best approach is to have things more normalized in the form of a relational database (one row per list element), rather than having lists embedded in strings.