java.lang.StackOverflowError 异常怎么解决

发布网友 发布时间：2022-04-30 16:02

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懂视网 时间：2022-04-10 03:45

@Entity
@Table(name = "app",schema = "", catalog = "game")
public class AppEntity {
 @Id
 @Column(name = "appId")
 private String appId;

 @Basic
 @Column(name = "appName")
 private String appName;
 @Basic
 @Column(name = "appdesc")
 private String appdesc;
 @Basic
 @Column(name = "provider")
 private String provider;
 @Basic
 @Column(name = "version")
 private String version;
 @Basic
 @Column(name="postTime")
 private Long postTime;
 @Basic
 @Column(name = "updateTime")
 private Long updateTime;

  @OneToMany(mappedBy = "appEntity",fetch = FetchType.EAGER)
 private Set<ProductEntity> productEntities;
 @OneToMany(mappedBy = "appEntity",fetch = FetchType.LAZY)
 private Set<OrdersEntity> ordersEntities;

多的一方
@Entity
@Table(name = "product", schema = "" , catalog = "game")
@IdClass(ProductEntityPK.class)
public class ProductEntity {
 @Id
 @Column(name = "appId")
 private String appId;
 @Id
 @GeneratedValue
 @Column(name = "productId")
 private long productId;
 @Basic
 @Column(name = "productName")
 private String productName;
 @Basic
 @Column(name = "productDesc")
 private String productDesc;
 @Basic
 @Column(name = "price")
 private long price;
 
 @ManyToOne(fetch = FetchType.EAGER)
  @JoinColumn(name="appId",insertable = false, updatable = false)
 private AppEntity appEntity;

@Override

public String toString() {
 return "ProductEntity{" +
  "appId=‘" + appId + ‘‘‘ +
  ", productId=" + productId +
  ", productName=‘" + productName + ‘‘‘ +
  ", productDesc=‘" + productDesc + ‘‘‘ +
  ", price=" + price 
  ", appEntity=" + appEntity +
  ‘}‘;
}

解决办法：将productEntity的toString方法中去掉 ", appEntity=" + appEntity +

hibernate+mysql关联映射师遇到java.lang.StackOverflowError异常

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