Spring JPA 生成无限查询

Question

我有以下实体：

用户：

@Data
@EqualsAndHashCode(callSuper = true)
@NoArgsConstructor
@AllArgsConstructor
@Builder
@Entity
@Audited
@Table(name="user_account")
public class User extends AbstractAuditableEntity implements UserDetails {
    private String email;
    private String password;
    private boolean enabled;
    private String name;

    @ManyToMany
    @Audited(targetAuditMode = NOT_AUDITED)
    private Set<Role> roles = new HashSet<>();

    @ManyToOne
    private Currency preferredCurrency;

货币：

@Data
@EqualsAndHashCode(callSuper = true)
@Builder
@AllArgsConstructor
@NoArgsConstructor
@Entity
@Audited
public class Currency extends AbstractAuditableEntity {
    private String isoCode;
    private String name;
    private BigDecimal euroConversion;
    private String symbol;
}

费用类型：

@Data
@EqualsAndHashCode(callSuper = true)
@Builder
@AllArgsConstructor
@NoArgsConstructor
@Entity
@Audited
public class ExpenseType extends AbstractAuditableEntity {
    private String name;
    private String description;

    @ManyToOne(fetch = FetchType.LAZY)
    private User user;

    @ManyToOne
    private ExpenseType parentExpenseType;

    @OneToMany(mappedBy = "parentExpenseType", fetch = FetchType.LAZY)
    private Set<ExpenseType> childExpenseTypes;
}

最后，费用：

@Data
@EqualsAndHashCode(callSuper = true)
@Builder
@AllArgsConstructor
@NoArgsConstructor
@Entity
public class Expense extends AbstractAuditableEntity {
    private String name;
    private String description;
    private LocalDateTime expenseDate;
    private BigDecimal amount;
    private boolean isIncome;

    @ManyToOne
    private Currency currency;

    @ManyToOne
    private ExpenseType expenseType;

    @ManyToOne
    private User user;
}

其父类AbstractAuditableEntity如下：

@MappedSuperclass
@EntityListeners(AuditingEntityListener.class)
@Getter
@Setter
@Audited
@JsonIgnoreProperties({"createdBy", "lastModifiedBy", "createdDate", "lastModifiedDate" })
public abstract class AbstractAuditableEntity extends AbstractPersistable<Long> implements Serializable {
    @JsonIgnore
    @Version
    protected Integer version;

    @CreatedDate
    @JoinColumn(name = "created_date", nullable = false, updatable = false)
    protected LocalDateTime createdDate;
    @LastModifiedDate
    @JoinColumn(name = "last_modified_date")
    protected LocalDateTime lastModifiedDate;

    @CreatedBy
    @ManyToOne
    @JoinColumn(name = "created_by_id", nullable = false, updatable = false)
    protected User createdBy;
    @LastModifiedBy
    @ManyToOne
    @JoinColumn(name = "last_modified_by_id")
    protected User lastModifiedBy;
}

当我尝试做的时候

repository.findById(id)

要获取“费用”实体，我收到以下错误： org.hibernate.exception.GenericJDBCException：执行 SQL 的 JDBC 异常

[选择 e1_0.id,e1_0.金额,c1_0.id,c2_0.id,c2_0.created_by_id,c2_0.created_date,c2_0.email,c2_0.enabled,l1_0.id,l1_0.created_by_id,l1_0.created_date,l1_0.email ,l1_0.enabled,l1_0.last_modified_by_id,l1_0.last_modified_date,l1_0.name,l1_0.password,p1_0.id,c3_0.id,c3_0.created_by_id,c3_0.created_date,c3_0.email,c3_0.enabled,c3_0.last_modified_by_id,c3_0 .last_modified_date,c3_0.name,c3_0.password,c3_0.preferred_currency_id,c3_0.version,p1_0.created_date,p1_0.euro_conversion,p1_0.iso_code,l2_0.id,l2_0.created_by_id,l2_0.created_date,l2_0.email,l2_0.enabled ,l2_0.last_modified_by_id... 等等

什么实体可能导致此问题？我找不到错误。

我尝试将一些实体放入延迟加载，但说实话我真的需要加载它们。

用户拥有货币，但货币没有“createdBy”和“lastModifiedBy”之外的用户。 Expense 具有以下各项之一：User、Currency 和 ExpenseType。 ExpenseType 可以有一个父级，我需要它来加载所有的父级。

Answer 1

它在测试环境之外执行时有效，因此它只是测试类的“@Transactional”注释。