JPA OneToMany，如何在控制台或webapp中打印时忽略字段ManyToOne

问题描述 投票：0回答：1

我有2个表：流派（genre_id，genre_name）和电影（movie_id，movie_name，movie_score，genre_id）。 genre_id_fk来自类型中引用genre_id的电影。

@Entity
@Table(name = "genres", schema = "test")
public class Genre {

@Id
@Column(name = "genre_id")
private int id;
@Column(name = "genre_name")
private String name;
@OneToMany(mappedBy = "genre", fetch = FetchType.LAZY)
private List<Movie> movies = new ArrayList<>();
}

第二个电影实体

@Entity
@Table(name = "movies", schema = "test")
public class Movie {

@Id
@GeneratedValue
@Column(name = "movie_id")
private int id;
@Column(name = "movie_name")
private String name;
@Column(name = "movie_score")
private double score;

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "genre_id")
private Genre genre;
}

当我尝试使用此代码在控制台中打印它时：

public static void main(String[] args) {
    EntityManagerFactory emf = Persistence.createEntityManagerFactory("pu");
    EntityManager em = emf.createEntityManager();
    Genre genre = em.find(Genre.class, 1);
    System.out.println(genre);
}

接收Exception in thread "main" java.lang.StackOverflowError只删除电影类字段“genres”中的toString（）可以修复它。但是有可能避免它吗？和春季启动应用程序相同的问题

@RestController
public class GenreController {

@Autowired
private GenreService genreService;

@RequestMapping("/test/{id}")
public List<Genre> getGenreInfo(@PathVariable int id){
    return genreService.getGenreFilms(id);
}
}

在这里服务

@Service
public class GenreService {

public List<Genre> getGenreFilms(int id){
    EntityManagerFactory emf = Persistence.createEntityManagerFactory("pu");
    EntityManager em = emf.createEntityManager();
    List<Genre> genres = new ArrayList<>();
    Genre genre = em.find(Genre.class, id);
    genres.add(genre);
    return genres;
}
}

并收到这样的问题：[{“id”：1，“name”：“Thriller”，“movies”：[{“id”：1，“name”：“Any，你还好吗？”，“得分” ：5.45，“genre”：{“id”：1，“name”：“Thriller”，“movies”：[{“id”：1，“name”：“Any，你还好吗？”，“得分” ：5.45，“genre”：{“id”：1，“name”：“Thriller”，“movies”：[{“id”：1，“name”：“Any，你还好吗？”，“得分” ：5.45，“流派”：......和无限，有异常。控制台我可以通过忽略toString（）方法中的字段来修复。但是如何在webapplication中解决这个问题呢？

这里是控制台打印时的hibernate调试日志

22:15:37.154 [main] DEBUG org.hibernate.engine.internal.TwoPhaseLoad - 
Resolving associations for [com.company.Genre#1]
22:15:37.164 [main] DEBUG org.hibernate.engine.internal.TwoPhaseLoad - Done 
materializing entity [com.company.Genre#1]
22:15:37.164 [main] DEBUG 
org.hibernate.resource.jdbc.internal.ResourceRegistryStandardImpl - 
HHH000387: ResultSet's statement was not registered
22:15:37.165 [main] DEBUG 
org.hibernate.loader.entity.plan.AbstractLoadPlanBasedEntityLoader - Done 
entity load : com.company.Genre#1
22:15:37.165 [main] DEBUG 
org.hibernate.resource.jdbc.internal.LogicalConnectionManagedImpl - 
Initiating JDBC connection release from afterTransaction
22:15:37.167 [main] DEBUG org.hibernate.loader.collection.plan.AbstractLoadPlanBasedCollectionInitializer 
- Loading collection: [com.company.Genre.movies#1]
22:15:37.167 [main] DEBUG org.hibernate.SQL - select movies0_.genre_id as 
genre_id4_1_0_, movies0_.movie_id as movie_id1_1_0_, movies0_.movie_id as 
movie_id1_1_1_, movies0_.genre_id as genre_id4_1_1_, movies0_.movie_name as 
movie_na2_1_1_, movies0_.movie_score as movie_sc3_1_1_ from test.movies 
movies0_ where movies0_.genre_id=?
Hibernate: select movies0_.genre_id as genre_id4_1_0_, movies0_.movie_id as 
movie_id1_1_0_, movies0_.movie_id as movie_id1_1_1_, movies0_.genre_id as 
genre_id4_1_1_, movies0_.movie_name as movie_na2_1_1_, movies0_.movie_score 
as movie_sc3_1_1_ from test.movies movies0_ where movies0_.genre_id=?
22:15:37.168 [main] DEBUG 
org.hibernate.loader.plan.exec.process.internal.ResultSetProcessorImpl - 
Preparing collection intializer : [com.company.Genre.movies#1]
22:15:37.170 [main] DEBUG 
org.hibernate.loader.plan.exec.process.internal.ResultSetProcessorImpl - 
Starting ResultSet row #0
22:15:37.171 [main] DEBUG org.hibernate.loader.plan.exec.process.internal.CollectionReferenceInitializerIm 
pl - Found row of collection: [com.company.Genre.movies#1]
22:15:37.171 [main] DEBUG 
org.hibernate.loader.plan.exec.process.internal.ResultSetProcessorImpl - 
Starting ResultSet row #1
22:15:37.172 [main] DEBUG org.hibernate.loader.plan.exec.process.internal.CollectionReferenceInitializerIm 
pl - Found row of collection: [com.company.Genre.movies#1]
22:15:37.172 [main] DEBUG 
org.hibernate.loader.plan.exec.process.internal.ResultSetProcessorImpl - 
Starting ResultSet row #2
22:15:37.172 [main] DEBUG org.hibernate.loader.plan.exec.process.internal.CollectionReferenceInitializerIm 
pl - Found row of collection: [com.company.Genre.movies#1]
22:15:37.172 [main] DEBUG org.hibernate.engine.internal.TwoPhaseLoad - 
Resolving associations for [com.company.Movie#1]
22:15:37.172 [main] DEBUG org.hibernate.engine.internal.TwoPhaseLoad - Done 
materializing entity [com.company.Movie#1]
22:15:37.173 [main] DEBUG org.hibernate.engine.internal.TwoPhaseLoad - 
Resolving associations for [com.company.Movie#2]
22:15:37.173 [main] DEBUG org.hibernate.engine.internal.TwoPhaseLoad - Done 
materializing entity [com.company.Movie#2]
22:15:37.173 [main] DEBUG org.hibernate.engine.internal.TwoPhaseLoad - 
Resolving associations for [com.company.Movie#3]
22:15:37.173 [main] DEBUG org.hibernate.engine.internal.TwoPhaseLoad - Done 
materializing entity [com.company.Movie#3]
22:15:37.173 [main] DEBUG 
org.hibernate.engine.loading.internal.CollectionLoadContext - 1 collections 
were found in result set for role: com.company.Genre.movies
22:15:37.173 [main] DEBUG 
org.hibernate.engine.loading.internal.CollectionLoadContext - Collection 
fully initialized: [com.company.Genre.movies#1]
22:15:37.173 [main] DEBUG 
org.hibernate.engine.loading.internal.CollectionLoadContext - 1 collections 
initialized for role: com.company.Genre.movies
22:15:37.173 [main] DEBUG 
org.hibernate.resource.jdbc.internal.ResourceRegistryStandardImpl - 
HHH000387: ResultSet's statement was not registered
22:15:37.173 [main] DEBUG org.hibernate.loader.collection.plan.AbstractLoadPlanBasedCollectionInitializer 
- Done loading collection
Exception in thread "main" java.lang.StackOverflowError

告诉我我在哪里做错了以及如何解决或谷歌这个问题？只是不能通过localhost：8080 / genre / id打印这个对象？制作特定的印刷品或什么？

java spring-boot jpa
1个回答
2
投票

当你尝试toString你的Genre实体时，你似乎有无限的递归。您的代码首先按id加载您的Genre实体，然后调用Genre.toString()。因为你与@OneToManys有Movie关系，所以lazy loads列表然后为每个与该类型相关的电影调用Movie.toString()。然后，对于每部电影，你都会与流派有@ManyToOne关系。这就是问题所在。它将再次为列表中的每部电影调用Genre.toString()。

可能的解决方案

  1. 如果您只想在控制台中打印它，请不要在Genre.toString()中包含电影列表
  2. 如果您正在使用杰克逊，请将@JsonBackReference添加到您在@ManyToOne的Movie关系中，这样杰克逊在映射到Json Annotation documentation here时会忽略它
  3. 如果使用DTO，请不要在DTO中包含Movie属性。

希望这可以帮助。

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