查询SELECT选择说所有指定语言的所有用户

Question

我有一个表，我们有人的用户名和他们可以说的语言。只有两列，用户标识和语言。

我想找到所有可以说印地语和英语，德语和法语的用户。 我应该如何编写这个查询？我无法使用Inner Join 4次。问题是检查语言的数量可能会增加，我可能想要检查更多的语言。

userid |语言
1 |英语
4 |英文
1 |法语
1 |德国
.................

Answer 1

如果使用MySQL，你可以做这样的事情（指被调试，未测试）：

SELECT userid FROM (
    SELECT userid, GROUP_CONCAT(language SEPARATOR ',') AS languages 
    FROM UserLanguage 
    ORDER BY userid ASC, language ASC 
    GROUP BY userid) 
WHERE languages LIKE '%english%french%german%hindi%'; 

（LIKE子句中的语言都进行排序）

How to use GROUP BY to concatenate strings in MySQL?

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或者，也许更快：

SELECT userid 
FROM UserLanguage 
WHERE language IN ('fr', 'en, 'de', 'hi') 
GROUP BY userid 
HAVING COUNT(DISTINCT(language)) >= 4 

Answer 2

你可以使用一个correlated subquery，像这样：

SELECT UserId 
FROM UserLanguage UL1 
WHERE 
    EXISTS (SELECT * FROM UserLanguage UL2 WHERE UL2.UserId = UL1.UserId 
     AND UL2.Language = 'english') AND 
    EXISTS (SELECT * FROM UserLanguage UL2 WHERE UL2.UserId = UL1.UserId 
     AND UL2.Language = 'hindi') AND 
    EXISTS (SELECT * FROM UserLanguage UL2 WHERE UL2.UserId = UL1.UserId 
     AND UL2.Language = 'german') AND 
    EXISTS (SELECT * FROM UserLanguage UL2 WHERE UL2.UserId = UL1.UserId 
     AND UL2.Language = 'french') 

Answer 3

您可以使用子查询：

SELECT userid 
FROM (
    SELECT userid, COUNT(*) AS cnt 
    FROM mytable 
    WHERE language IN ('hindi','german','french','english') 
    GROUP BY userid 
) AS t 
WHERE cnt = 4; 

Answer 4

我想，你没有语言的数量和名称，然后您应该尝试以下查询...

select userid 
from UserLanguage 
group by userid 
having count(language)= (select count(distinct language) from userid)