Question

我有一张桌子＆＃34;人物＆＃34;。

另一个＆＃34; hobbiesConnection＆＃34;，它们作为OneToMany连接。

我想建立一个搜索，在那里可以搜索有业余爱好并且没有兴趣的人。

类似的东西：

SELECT * FROM people
LEFT JOIN hobbiesConnection ON hobbiesConnection.people_id = 
hobbiesConnection.hobby_id
WHERE hobby_id = 4 AND hobby_id != 5

现在我得到了爱好者4，但我希望得到一个空洞的结果，因为这两个条件都必须适合这个人。

有人知道如何做到这一点吗？

Answer 1

首先，您的 JOIN 不正确。这两个表应该与外键/公共列链接。请尝试以下查询;

SELECT * FROM people
LEFT JOIN hobbiesConnection 
ON hobbiesConnection.people_id = people.people_id
And (hobbiesConnection.hobby_id = 4 AND hobbiesConnection.hobby_id != 5)

Answer 2

使用（NOT）EXISTS条款：

SELECT * FROM people p
WHERE TRUE
AND EXISTS (
  SELECT TRUE FROM hobbiesConnection hc 
  WHERE hc.people_id = p.people_id -- or p.id? you didnt mention
  AND hobby_id = 4
)
AND NOT EXISTS (
  SELECT TRUE FROM hobbiesConnection hc 
  WHERE hc.people_id = p.people_id -- or p.id? you didnt mention
  AND hobby_id = 5
)

使用连接

可以实现相同的效果

SELECT p.* 
FROM people p
JOIN hobbiesConnection h4 ON TRUE
    AND h4.people_id = p.people_id -- or p.id? you didnt mention
    AND h4.hobby_id = 4
LEFT JOIN hobbiesConnection h5 ON TRUE
    AND h5.people_id = p.people_id -- or p.id? you didnt mention
    AND h5.hobby_id = 5
WHERE h5.id IS NULL -- or whatever column from that table

看看小提琴：http://sqlfiddle.com/#!9/d9926/4

SELECT如果左连接IS AND IS NOT（MYSQL）？

2 个答案: