多对多/连接表-Spring JPA Hibernate-无法通过反射设置值

Question

因此，我试图通过Spring JPA实现多对多的应用。

{
    "message": "Internal server error",
    "details": "Could not set field value [26] value by reflection : [class com.domain.configuredview.model.RelationshipViewPersonPK.personId] setter of com.domain.configuredview.model.RelationshipViewPersonPK.personId; nested exception is org.hibernate.PropertyAccessException: Could not set field value [26] value by reflection : [class com.domain.configuredview.model.RelationshipViewPersonPK.personId] setter of com.domain.configuredview.model.RelationshipViewPersonPK.personId"
}

import com.domain.person.model.Person;
import java.io.Serializable;
import javax.persistence.*;
import lombok.Data;
import lombok.NoArgsConstructor;

@Data
@NoArgsConstructor
@Entity
@Table(name = "relationship_view_person")
public class RelationshipViewPerson implements Serializable {

  @EmbeddedId
  private RelationshipViewPersonPK entryId;

  @ManyToOne(fetch = FetchType.LAZY)
  @MapsId("viewId")
  @JoinColumn(name = "view_id", nullable = false)
  private ConfiguredView view;

  @ManyToOne(fetch = FetchType.LAZY)
  @MapsId("personId")
  @JoinColumn(name = "person_id", nullable = false)
  private Person person;

  public RelationshipViewPerson(ConfiguredView view, Person person) {
    this.view = view;
    this.person = person;
    this.entryId = new RelationshipViewPersonPK(view.getId(), person.getWsGlobalId());
  }

  public void setEntryId(Long viewId, String personId) {
    this.entryId.setViewId(viewId);
    this.entryId.setPersonId(personId);
  }

  public void setView(ConfiguredView view) {
    this.view = view;
  }

  public void setPerson(Person person) {
    this.person = person;
  }

  public Person getPerson() {
    return this.person;
  }

  public ConfiguredView getView() {
    return this.view;
  }
}

import java.io.Serializable;
import javax.persistence.*;
import lombok.NoArgsConstructor;

@Embeddable
@NoArgsConstructor
public class RelationshipViewPersonPK implements Serializable {

  @Column(name = "view_id")
  private Long viewId;

  @Column(name = "person_id")
  private String personId;

  public RelationshipViewPersonPK(Long viewId, String personId) {
    this.viewId = viewId;
    this.personId = personId;
  }

  // setters & getters
  public void setViewId(Long viewId) {
    this.viewId = viewId;
  }

  public void setPersonId(String personId) {
    this.personId = personId;
  }

  public String getPersonId() {
    return this.personId;
  }

  public Long getViewId() {
    return this.viewId;
  }

  // Override equals and hashCode

  @Override
  public boolean equals(Object o) {
    if (this == o) {
      return true;
    }

    if (o == null || getClass() != o.getClass()) {
      return false;
    }

    RelationshipViewPersonPK that = (RelationshipViewPersonPK) o;

    return this.viewId.equals(that.viewId) && this.personId.equals(that.personId);
  }

  @Override
  public int hashCode() {
    return this.viewId.hashCode() + this.personId.hashCode();
  }
}

ConfiguredView模型的PK为ID，Person模型的字段为WSGlobalID。Person表上的键不是PK。

我有设置器、获取器、构造器和几乎所有的东西。不确定Hibernate在这里发生了什么，它找不到getter或其他什么？不知道这是怎么回事。

Answer 1

您应该有一个如下所示的setter：

public void setEntryId(RelationshipViewPersonPK entryId) {
    this.entryId.setViewId(entryId.getViewId());
    this.entryId.setPersonId(entryId.getPersonId());
}

代替（或除此之外）：

public void setEntryId(Long viewId, String personId) {
    this.entryId.setViewId(viewId);
    this.entryId.setPersonId(personId);
}