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Spring Data JPA - 多对多查询

[英]Spring Data JPA - Many to many query

I have two entities Person and Movie.我有两个实体 Person 和 Movie。

@Entity
public class Person {
..some fields

@ManyToMany(fetch = FetchType.LAZY, mappedBy = "actors")
@OrderBy("id")
private Set<Movie> moviesActor = new TreeSet<>();

}

@Entity
public class Movie {
..fields
@JoinTable(name = "actor_movie",
            joinColumns = { @JoinColumn(name = "movie_id") },
            inverseJoinColumns = { @JoinColumn(name = "actor_id") })
    private Set<Person> actors = new TreeSet<>();
}

There is many to many relationship so there is new table actor_movie to keep it.存在多对多关系,因此有新表 actor_movie 来保留它。 And how can I get every person that has any movie in its set?我怎样才能让每个人都拥有任何电影? So what I want is to achieve is get every person that exists in actor_movie table.所以我想要实现的是获取 actor_movie 表中存在的每个人。 I tried used Spring data jpa but couldn't find right query.我尝试使用 Spring data jpa 但找不到正确的查询。

Best Practices in entity relations:实体关系的最佳实践:

  1. Always use fetch = FetchType.LAZY .始终使用fetch = FetchType.LAZY
  2. When you want to fetch another side of the relation too, use JOIN FETCH Query.当您也想获取关系的另一面时,请使用JOIN FETCH Query。
    This resolves LazyInitializationException of hibernate also.这也解决了休眠的LazyInitializationException
  3. Always use spring.jpa.open-in-view=false始终使用spring.jpa.open-in-view=false

Example:例子:
By Spring Data JPA with Hibernate as JPA Provider.由 Spring Data JPA 和 Hibernate 作为 JPA 提供者。

Entities:实体:

public class Blog{
   ...
   @ManyToMany(fetch = FetchType.LAZY) //default is LAZY in ManyToMany
    @JoinTable(name="blog_tag",
        joinColumns = @JoinColumn(name = "blog_id"),
        inverseJoinColumns = @JoinColumn(name = "tag_id"))
    @OrderBy //order by tag id
    private Set<Tag> tags = new HashSet<>();

    //2 utility methods in owner side
    public void addTag(Tag tag){
        tags.add(tag);
        tag.getBlogs().add(this);
    }
    public void removeTag(Tag tag){
        tags.remove(tag);
        tag.getBlogs().remove(this);
    }
    
    //override equals & hashcode 

}


public class Tag {
    ...
    @ManyToMany(mappedBy = "tags")
    private Set<Blog> blogs = new HashSet<>();

    //override equals & hashcode 
}

Now suppose, you want to fetch a Blog containing Tag items:现在假设,您想要获取一个包含标签项的博客:

Repository:存储库:

@Repository
public interface BlogRepository extends JpaRepository<Blog, Long> {
    @Query("select b from Blog b join fetch b.tags where b.name = :name")
    Blog getBlog(@Param("name") String blogName);
}

service:服务:

public interface BlogService {
    Blog getBlog(String blogName);
}

@Service
public class BlogServiceImpl implements BlogService{

    @Autowired
    private BlogRepository blogRepository;

    @Override
    public Blog getBlog(String blogName) {
        return blogRepository.getBlog(blogName);
    }

}

Since you are using Fetch type lazy, you need to use join fetch to get moviesActor.由于您使用的是 Fetch 类型lazy,因此您需要使用join fetch 来获取moviesActor。

You can use jpql with spring data.您可以将 jpql 与 spring 数据一起使用。 I have not tested the queries below , but should work.我没有测试下面的查询,但应该可以工作。

public interface PersonRepository extends JpaRepository<Person, Long> { //Long if Person.id is of type Long

  @Query("SELECT p FROM Person p LEFT JOIN FETCH p.moviesActor WHERE size(p.moviesActor) > 0");
  List<Person> findActors1();

  // Or

  @Query("SELECT p FROM Person p JOIN FETCH p.moviesActor");
  List<Person> findActors2();

}

More about jpql size() operator here: https://www.thoughts-on-java.org/jpql/有关 jpql size() 运算符的更多信息: https ://www.thoughts-on-java.org/jpql/

You can use join directely :您可以直接使用 join :

@Query("SELECT p FROM Person p  JOIN  p.moviesActor movie");

List findPersonHasMovie();列表 findPersonH​​asMovie();

You only need a single JOIN between Person and Movie.您只需要在 Person 和 Movie 之间进行一次JOIN As Hibernate abstracts the existence of the middle table, you don't need to worry about it.由于 Hibernate 抽象了中间表的存在,因此您无需担心它。

So, with Spring Data Repository:因此,使用 Spring Data Repository:

class PersonRepository extends CrudRepository<Person, Long> {

    List<Person> findByMoviesActor();
}

With Jpql:使用 Jpql:

SELECT person FROM Person person JOIN person.moviesActor movie

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