一个女人/多个女人的无限循环?Spring/rest api

Question

我选择了从货架到仓库的多对一关系！我的问题是，当我将带有存储库的表的外键添加到带有存储架的表中时，我得到了一个带有数据的ifnite循环，我试图用jsonignore/jsonmanagereference/jsonbackreference来解决它，但后来我的服务器415出现了问题。你们中有人来帮我吗？我认为我的桌子之间的联系是罪魁祸首，但我不知道该怎么办。。

实体仓库：

@Entity
@Table(name="warehouse")
//@Access(AccessType.FIELD)
public class Warehouse {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name="id_warehouse")
private Long id_warehouse;

@Column(name="description_warehouse")
private String description_warehouse;

@OneToMany(mappedBy = "war_id")
private List<Shelve> shelves = new ArrayList<>();

@OneToMany(mappedBy = "war_id")
private List<Management> managements = new ArrayList<>();

public Warehouse() {
}
//getter setter

实体搁置：

@Entity
@Table(name = "shelve")
public class Shelve {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id_shelve")
private Long id_shelve;
@Column(name = "description_shelve")
private String description_shelve;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "war_id", referencedColumnName = "id_warehouse")
private Warehouse war_id;
public Shelve() {
}
//getter setter

错误：

Hibernate: select warehouse0_.id_warehouse as id_wareh1_2_0_, warehouse0_.description_warehouse as descript2_2_0_ from warehouse warehouse0_ where warehouse0_.id_warehouse=?
Hibernate: select shelves0_.war_id as war_id3_1_0_, shelves0_.id_shelve as id_shelv1_1_0_, shelves0_.id_shelve as id_shelv1_1_1_, shelves0_.description_shelve as descript2_1_1_, shelves0_.war_id as war_id3_1_1_ from shelve shelves0_ where shelves0_.war_id=?
2021-07-13 16:50:11.986  WARN 3968 --- [nio-8080-exec-7] .w.s.m.s.DefaultHandlerExceptionResolver : Failure while trying to resolve exception [org.springframework.http.converter.HttpMessageNotWritableException]
java.lang.IllegalStateException: Cannot call sendError() after the response has been committed
at org.apache.catalina.connector.ResponseFacade.sendError(ResponseFacade.java:472) ~[tomcat-embed-core-9.0.48.jar:9.0.48]
at org.springframework.web.servlet.mvc.support.DefaultHandlerExceptionResolver.sendServerError(DefaultHandlerExceptionResolver.java:553) ~[spring-webmvc-5.3.8.jar:5.3.8]
at org.springframework.web.servlet.mvc.support.DefaultHandlerExceptionResolver.handleHttpMessageNotWritable(DefaultHandlerExceptionResolver.java:443) ~[spring-webmvc-5.3.8.jar:5.3.8]
at org.springframework.web.servlet.mvc.support.DefaultHandlerExceptionResolver.doResolveException(DefaultHandlerExceptionResolver.java:210) ~[spring-webmvc-5.3.8.jar:5.3.8]
at org.springframework.web.servlet.handler.AbstractHandlerExceptionResolver.resolveException(AbstractHandlerExceptionResolver.java:141) ~[spring-webmvc-5.3.8.jar:5.3.8]
at org.springframework.web.servlet.handler.HandlerExceptionResolverComposite.resolveException(HandlerExceptionResolverComposite.java:80) ~[spring-webmvc-5.3.8.jar:5.3.8]
at org.springframework.web.servlet.DispatcherServlet.processHandlerException(DispatcherServlet.java:1323) ~[spring-webmvc-5.3.8.jar:5.3.8]
at org.springframework.web.servlet.DispatcherServlet.processDispatchResult(DispatcherServlet.java:1134) ~[spring-webmvc-5.3.8.jar:5.3.8]
at org.springframework.web.servlet.DispatcherServlet.doDispatch(DispatcherServlet.java:1080) ~[spring-webmvc-5.3.8.jar:5.3.8]
at org.springframework.web.servlet.DispatcherServlet.doService(DispatcherServlet.java:963) ~[spring-webmvc-5.3.8.jar:5.3.8]
at org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:1006) ~[spring-webmvc-5.3.8.jar:5.3.8]
at org.springframework.web.servlet.FrameworkServlet.doGet(FrameworkServlet.java:898) ~[spring-webmvc-5.3.8.jar:5.3.8]
at javax.servlet.http.HttpServlet.service(HttpServlet.java:655) ~[tomcat-embed-core-9.0.48.jar:4.0.FR]
at org.springframework.web.servlet.FrameworkServlet.service(FrameworkServlet.java:883) ~[spring-webmvc-5.3.8.jar:5.3.8]
at javax.servlet.http.HttpServlet.service(HttpServlet.java:764) ~[tomcat-embed-core-9.0.48.jar:4.0.FR]
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:228) ~[tomcat-embed-core-9.0.48.jar:9.0.48]
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:163) ~[tomcat-embed-core-9.0.48.jar:9.0.48]

Answer 1

您有错误的连接，您需要进行全面重构并了解关系的主要思想。请查看以下链接：

https://dev.to/jhonifaber/hibernate-onetoone-onetomany-manytoone-and-manytomany-8ba

Answer 2

如果有人有同样的问题，请尝试放入@JsonIgnore！并在应用程序属性中添加以下内容：spring.jackson.serialization.fail-on-emptybeans=false！

Answer 3

您可以将@JsonManagedReference放在实体仓库中，将@Json BackReference放在实体货架中。它可以很好地检索数据。

Answer 4

U可以使用三个@JsonManagedReference@JsonBackReference@JsonIgnore 之一

Answer 5

您可以在属性关系上使用@JsonIgnoreProperties，并将属性名称放入

@JsonIgnoreProperties("shelves")
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "war_id", referencedColumnName = "id_warehouse")
private Warehouse war_id;
@JsonIgnoreProperties("war_id")
@OneToMany(mappedBy = "war_id")
private List<Shelve> shelves = new ArrayList<>();

或者您可以在@OneToMany上使用@JsonManagedReference，在@ManyToOne上使用@JsonBackReference