SQL從表1中選擇一行加入表2中的多行

Question

所以我有兩個表，第一個是 users_

Name
------
Carol
Sue

第二個是興趣_

Name        Interest
----------------------
Carol       Books
Carol       Dancing
Carol       Sports    
Sue         Books
Sue         Dancing

用戶將看到復選框以根據類似興趣的標准選擇匹配項

因此，如果用戶選擇書籍和舞蹈作為他們的比賽興趣，我將構建什么類型的 sql 查詢來組合多行興趣並確保結果是 Sue，因為她的興趣是書籍和舞蹈而不是體育？

任何幫助都會大有幫助，謝謝！

Answer 1

所以基本上要做到這一點，請建立一個負面列表，列出擁有超過這兩個的用戶，然后選擇其他用戶

SELECT u.name 
FROM users_ u     
JOIN interests_ i ON i.name = u.name
JOIN
(   SELECT u.name 
    FROM users_ u
    JOIN interests_ i ON i.name = u.name
    WHERE i.interest NOT IN('Books', 'Dancing')
) t
WHERE u.name <> t.name 
  AND i.interest IN('Books', 'Dancing')
GROUP BY u.name
HAVING COUNT(u.name) = 2;

演示

Answer 2

這是執行此操作的簡單方法

select
i.name 
from interests i
where i.interest in ('Books','Dancing')
and not exists
(
  select 1 from interests i1
  where interest not in ('Books','Dancing')
  AND i.name = i1.name
)
group by i.name
having count(*) = 2

演示

Answer 3

實現此結果的一種方法：

SELECT u.name
  FROM users_ u
  JOIN interests_ n01 ON n01.name = u.name AND n01.interest = 'Books'
  JOIN interests_ n02 ON n02.name = u.name AND n02.interest = 'Dancing'
  LEFT JOIN interests_ x01 ON x01.name = u.name AND x01.interest = 'Sports'
  LEFT JOIN interests_ x02 ON x02.name = u.name AND x01.interest = 'Wine'
 WHERE x01.name IS NULL
   AND x02.name IS NULL

這種方法要求“包含”每個興趣的 JOIN，以及“排除”每個興趣的反聯接。 這種方法是靈活的，但對於大量利益來說可能變得笨拙。

要僅查找指定興趣的匹配項，而不必列出所有不匹配的興趣，我們可以使用反聯接查找不匹配的興趣行：

SELECT u.name
  FROM users_ u
  JOIN interests_ n01 ON n01.name = u.name AND n01.interest = 'Books'
  JOIN interests_ n02 ON n02.name = u.name AND n02.interest = 'Dancing'
  LEFT
  JOIN interests_ o 
    ON o.name = u.name
   AND o.interest NOT IN ('Books','Dancing')
 WHERE o.name IS NULL

另一種方法是利用 JOIN 操作和COUNT()聚合，例如

SELECT u.name
  FROM users_ u
  JOIN interests_ n 
    ON n.name = u.name 
   AND n.interest IN ('Books','Dancing')
  LEFT
  JOIN interests_ o
    ON o.name = u.name
   AND o.interest NOT IN ('Books','Dancing')
 WHERE o.name IS NULL
 GROUP BY u.name
HAVING COUNT(DISTINCT n.interest) = 2

還有一些其他方法（這些只是一些示例）。

Answer 4

in操作符中使用 SQL

select * from  interests_ where Interest in ('Books', 'Dancing') and Interest not in ('Sports');

編輯 1

這對我有用

SELECT * FROM user_ WHERE EXISTS ( 
    SELECT i.Name, count(distinct i.Interest) FROM interests_ as i
    WHERE i.Name=user_.Name
    AND i.Interest IN ('Books','Dancing')
    GROUP BY i.Name
    HAVING count(distinct i.Interest) = 2
);

REF : SQL 相當於 IN 運算符，它充當 AND 而不是 OR？

Answer 5

左連接非常有用。 你可以試試這個。

SELECT users_.*
FROM users_
    LEFT JOIN interests_
    ON users_.Name = Interests_.Name
            WHERE interests_.Interests = 'Books'
            AND interests_.Interests = 'Dancing'