Hibernate Jpa多对多关系，生成错误的sql查询

我有一个房客对象，他可能有很多联系。同一联系人可与多个房客关联。这是一种多对多的关系

在我的房客中

@ManyToMany(mappedBy = "lodger")
private List<Contact> contactList;

编辑3

@Query("select c from Contact c where :lodgerId not member of c.lodger.lodgerId")
List<Contact> findByLodgerLodgerIdNot(@Param("lodgerId") Long lodger);

@Query（“从联系人c中选择c，其中：lodgerId不是c.lodger.lodgerId的成员”）
列出findByLodgerLodgerIdNot（@Param（“lodgerId”）长房客）；

org.hibernate.QueryException:非法尝试取消引用 具有元素属性的集合[contact0\u.contact\u id.lodger] 参考[寄宿者ID]

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从生成的JPQL可以看出，您的映射是错误的：Hibernate在

contact\u id=lodger\u id

上连接表（反之亦然），而不是在

contact\u id=contact\u id

上连接表（和

lodger\u id=lodger\u id

）

这是因为您使用inverseJoinColumns反转了joinColumns。映射应该是

@JoinTable(name = "lodger_contact", 
           joinColumns = @JoinColumn(name = "contact_id"), 
           inverseJoinColumns = @JoinColumn(name = "lodger_id"))

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在findByLodgerLodgerIdNot方法中有什么？Spring存储库中，使用name属性和操作通过其名称生成代码。如果您有一个具有firstname属性的类用户。您可以编写一个方法：findByFirstname（stringfirstname），spring将生成如下代码：select*from user where firstname=？更多信息：docs.spring.io/spring data/jpa/docs/current/reference/htmldon似乎不是解决方案。那么您在数据库中没有任何联系人，其中至少有一个房客未通过给定的房客ID识别。true。。。就像我说的，我有一个房客，一个联系人。我搜索与我的房客无关的联系人列表。但现在您的查询就是这样做的。它搜索至少有一个不是房客ID的房客的联系人。您需要以下查询：

从联系人c中选择c，其中：房客不是c的成员。房客

我更新、修改了您的查询，但希望知道是否有办法避免将对象发送到查询。

    select
    contact0_.contact_id as contact_1_9_,
    contact0_.address as address2_9_,
    contact0_.city_cityId as city_cit5_9_,
    contact0_.contact_sub_category_contactSubCategoryId as contact_6_9_,
    contact0_.first_name as first_na3_9_,
    contact0_.last_name as last_nam4_9_,
    contact0_.phone_phoneId as phone_ph7_9_ 
from
    contact contact0_ 
left outer join
    lodger_contact lodger1_ 
        on contact0_.contact_id=lodger1_.lodger_id 
left outer join
    lodger lodger2_ 
        on lodger1_.contact_id=lodger2_.lodger_id 
where
    lodger2_.lodger_id<>?

select
    contact0_.contact_id as contact_1_9_,
    contact0_.address as address2_9_,
    contact0_.city_cityId as city_cit5_9_,
    contact0_.contact_sub_category_contactSubCategoryId as contact_6_9_,
    contact0_.first_name as first_na3_9_,
    contact0_.last_name as last_nam4_9_,
    contact0_.phone_phoneId as phone_ph7_9_ 
from
    contact contact0_ 
left outer join
    lodger_contact lodger1_ 
        on contact0_.contact_id=lodger1_.contact_id 
left outer join
    lodger lodger2_ 
        on lodger1_.lodger_id=lodger2_.lodger_id 
where
    lodger2_.lodger_id<>?

@Query("select c from Contact c where :lodger not member of c.lodger")
List<Contact> findByLodgerLodgerIdNot(@Param("lodger") Lodger lodger);

select
    contact0_.contact_id as contact_1_9_,
    contact0_.address as address2_9_,
    contact0_.city_cityId as city_cit5_9_,
    contact0_.contact_sub_category_contactSubCategoryId as contact_6_9_,
    contact0_.first_name as first_na3_9_,
    contact0_.last_name as last_nam4_9_,
    contact0_.phone_phoneId as phone_ph7_9_ 
from
    contact contact0_ 
where
    ? not in  (
        select
            lodger1_.lodger_id 
        from
            lodger_contact lodger1_ 
        where
            contact0_.contact_id=lodger1_.contact_id

@Query("select c from Contact c where :lodgerId not member of c.lodger.lodgerId")
List<Contact> findByLodgerLodgerIdNot(@Param("lodgerId") Long lodger);

@JoinTable(name = "lodger_contact", 
           joinColumns = @JoinColumn(name = "contact_id"), 
           inverseJoinColumns = @JoinColumn(name = "lodger_id"))