可选地连接多对多表

Question

我有以下表格

用户

user_id  name
101      Tony
102      Skyle
103      Kenne

兴趣

intrest_id intrest_name
201          Eating
202          Sleeping
203          Drinking

爱好

hobby_id  hobby_name
301         Smoking
302         Hiking
303         Browsing

User_Intrest

user_id  intrest_id
101      201
102      201
102      202
103      201
103      202
103      203

User_Hobby

user_id  hobby_id
101      301
102      301
102      302
103      301
103      302
103      303

现在要查找既有兴趣Eating又Sleeping的用户id我已经写了

select u.user_id
from user u, intrest i, user_intrest ui
where u.user_id = ui.user_id 
and i.intrest_id = ui.intrest_id and i.intrest_name in ('Eating', 'Sleeping')
group by u.user_id
having count(i.intrest_name) = 2

输出

user_id
102
103

同上我也可以找到爱好Smoking,Hiking,Browsing的用户id如下

select u.user_id
from user u, hobby h, user_hobby uh
where u.user_id = uh.user_id 
and h.hobby_id = uh.hobby_id and h.hobby_name in ('Smoking', 'Hiking', 'Browsing')
group by u.user_id
having count(i.intrest_name) = 3

输出

user_id
    103

现在我想以一种可选的方式混合这两者，如果只传递兴趣，那么就会找到具有这些兴趣的用户，或者如果只传递兴趣爱好，那么就会找到具有这些爱好的用户，或者如果兴趣和爱好都是passed 然后找到具有这些兴趣爱好的用户

更新：这是 spring 数据剩余 api 的一部分，带有本机查询，其中查询参数是可选的。下面是一个寻找兴趣用户的工作示例。现在我想扩展它以包括爱好，所以如果两者都在查找用户的请求中传递，那么两者都将被使用

@RestResource(path = "getUserByIntrestsAndHobbies", rel = "getUserByIntrestsAndHobbies")
    @Query(value = "SELECT u FROM User u WHERE (COALESCE(:intrests, NULL) IS NOT NULL AND u.userId IN (SELECT u.userId " +
            "FROM User u, Intrest i, UserIntrest ui " +
            "WHERE u.userId = ui.userId AND i.intrestId = ui.intrestId " +
            "AND i.intrestName IN (:intrests) " +
            "GROUP BY u.userId " +
            "HAVING (:intrestsSize IS NULL OR :intrestsSize = count(i.intrestName))))"
    )
    Page<User> getUsersByIntrestsAndHobbies(@Param("intrests") List<String> intrests,
                                                                           @Param("intrestsSize") Long intrestsSize,
                                                                           @Param("hobbies") List<String> hobbies,
                                                                           @Param("hobbiesSize") Long hobbiesSize,
                                                                           Pageable pageable);

Answer 1

好的，当您使用显式连接（而不是隐式、折旧的 comma-separated 连接）时，问题会变得更加清晰。然后您会看到您需要使用 left 联接以允许在匹配时返回记录。

select u.[user_id]
from [user] u
left join user_intrest ui on ui.[user_id] = u.[user_id]
left join intrest i on i.intrest_id = ui.intrest_id and i.intrest_name in ('Eating', 'Sleeping')
left join user_hobby uh on uh.[user_id] = u.[user_id]
left join hobby h on h.hobby_id = uh.hobby_id and h.hobby_name in ('Smoking', 'Hiking', 'Browsing')
group by u.[user_id]
having count(distinct i.intrest_name) = 2 and count(distinct h.hobby_name) = 3;

最好不要使用 user 或 user_id 等保留字，因为您总是需要对它们进行转义。
除了 id 列之外，不需要将 table 名称添加到列名称前面，例如hobby_name 应该只是 name - 你只是给自己更多的输入。
Interest 不拼 Intrest :)
如果您提供DDL/DML作为示例数据，那么我们测试起来会容易得多。

Answer 2

我建议：

select i.userid
from (select ui.userid
      from user_interest ui join
           interest i
           on i.interest_id = ui.interest_id 
      where i.interest_name in ('Eating', 'Sleeping')
      group by ui.userid
      having count(*) = 2
     ) i join
     (select uh.userid
      from user_hobby uh join
           hobby h
           on uh.hobby_id = i.hobby_id 
      where h.hobby_name in ('Smoking', 'Hiking', 'Browsing')
      group by uh.userid
      having count(*) = 3
     ) h
     on i.userid = h.userid;

这会分别沿每个维度进行计数，因此计数是准确的。

可选地连接多对多表

Join many to many tables optionally

sql

sql-server

join

spring-data

spring-data-rest