Question

我有2个实体：用户和事件。每个映射到适当的表。我还有第三个表user_event，因为这两个实体具有多对多的关系。我需要从DB中选择用户参与的所有事件。

事件：

@Entity
@Table(name = "event")
public class Event extends AbstractPersistable<Long> {

@ManyToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)
@JoinTable(name = "user_event",
        joinColumns = @JoinColumn(name = "event_id", referencedColumnName = "id"),
        inverseJoinColumns = @JoinColumn(name = "user_id", referencedColumnName = "id"))
private Collection<User> participants;

用户：

@Entity
@Table(name = "user")
public class User extends AbstractPersistable<Long> {

    private String nickname;

user_event表在java代码中没有实体。我试过这个问题：

@Query("select e from Event e join user_event ue on ue.event_id = e.id where ue.user_id = :userId and e.startDate > CURRENT_TIMESTAMP")
Page<Event> findAllForUser(Pageable pageable, @Param("userId") Long userId);

但是这个查询会导致应用程序启动异常：

java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: Path expected for join! [select e from Event e join user_event ue on ue.event_id = e.id where ue.user_id = :userId and e.startDate > CURRENT_TIMESTAMP]

在MySQL Workbench中我尝试了这个：

select * from event e join user_event ue on e.id = ue.event_id where ue.user_id = 1 and e.startDate > now();

它有效。但是如何为spring数据创建良好的工作查询？

SQL转储：

select count(event0_.id) as col_0_0_ from event event0_ inner join   address address1_ on event0_.address_id=address1_.id 
cross join user_event participan2_, user user3_ where   event0_.id=participan2_.event_id and participan2_.user_id=user3_.id 
and (? in (.)) and event0_.startDate>CURRENT_TIMESTAMP

Answer 1

在@ManyToMany映射中，您有以下内容：

@JoinTable（name =“event_user_event”

但在您使用user_event的查询中。我想其中一个是拼写错误的？

在您的查询中

select e 
from Event e join user_event ue on ue.event_id = e.id 
where ue.user_id = :userId and e.startDate > CURRENT_TIMESTAMP"

您使用的user_event不是实体（因为它在异常消息中正确指出）。因此查询应如下所示：

select e 
from Event e join e.participants u 
where u.id = :userId and e.startDate > CURRENT_TIMESTAMP

假设您的User实体具有名为id的属性。此查询应返回与用户:userId关联的所有事件。

Spring Data ManyToMany选择查询

1 个答案: