java - Spring + hibernate : No property findAddresses found for type UserProfile while fetching a list

Question

我在使用 Hibernate 访问数据库上的实体的数据列表时遇到问题。

我有一个实体UserProfile，其中包含一些字段，但最重要的是:

@OneToMany(cascade = CascadeType.ALL)
private Set<Address> addresses;

该类有lombok的@Getter，以及添加地址的方法:

public void addProfileAddress(Address address) {
    addresses.add(address);
}

现在，因为我们只需要单向关系，所以 Address 类如下所示:

@Getter
@Entity
@Table(name = "addresses")
@Builder
@NoArgsConstructor
@AllArgsConstructor
@SQLDelete(sql = "UPDATE addresses SET deleted = true, updated_at = current_date WHERE id = ?")
public class Address extends Auditable implements Serializable {

    private static final long serialVersionUID = -5034153190245766924L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Setter
    @Column(name = "country", nullable = false)
    private String country;

    @Setter
    @Column(name = "state", nullable = false)
    private String state;

    @Setter
    @Column(name = "city", nullable = false)
    private String city;

    @SuppressWarnings("PMD.AvoidFieldNameMatchingTypeName")
    @Setter
    @Column(name = "address", nullable = false)
    private String address;


    public Address(Address address) {}
}

我有一个 UserProfileRepository:

@Repository
public interface UserProfileRepository extends AuditableRepository<UserProfile> {

    List<Address> findAddresses(Long id);

}

但是当我尝试运行代码时(使用 GraphQL 请求:

extend type Query {
    userProfiles: [UserProfile]
    addresses(id: ID): [Address]
    userProfile(id: ID): UserProfile
}

使用 GraphQL 查询解析器方法:

@PreAuthorize("hasAuthority('ACCOUNT_OWNER', 'USER')")
public Set<Address> getAddresses(Long id) {
    return userProfileRepository.findAddresses(id).stream().map(Address::new).collect(Collectors.toSet());
}

我得到:

Caused by: org.springframework.beans.factory.BeanCreationException: Error creating bean with 
name 'userProfileRepository': Invocation of init method failed; nested exception is 
java.lang.IllegalArgumentException: Failed to create query for method public abstract 
java.util.List 
com.l4a.app.infrastructure.database.repository.UserProfileRepository.findAddresses(java.
lang.Long)! No property findAddresses found for type UserProfile!

有人可以帮我解决这个问题吗？我被困住了...

重要编辑:

当我将 Repository 方法重命名为 findByAddresses 时，我得到:

Caused by: org.hibernate.LazyInitializationException: failed to lazily 
initialize a collection of role: 
com.l4a.app.domain.userprofile.UserProfile.addresses, could 
not initialize proxy - no Session

Answer 1

默认情况下，所有集合都是延迟加载的，这意味着 spring/hibernate 在查询期间不会加载模型。您必须将 JOIN FETCH 添加到您的查询中:

@Query("SELECT addr FROM UserProfile user JOIN FETCH user.addresses addr WHERE user.id = ?1")
Set<Address> findAddresses(Long id);

如果您希望建立单向关系，其中 UserProfile 可以访问 Address，但反之则不然，您可以执行以下操作:

将userId添加到Address表和相应的实体:

@Getter
@Entity
@Table(name = "addresses")
@Builder
@NoArgsConstructor
@AllArgsConstructor
@SQLDelete(sql = "UPDATE addresses SET deleted = true, updated_at = current_date WHERE id = ?")
public class Address extends Auditable implements Serializable {

    private static final long serialVersionUID = -5034153190245766924L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Setter
    @Column(name ="userId", nullable = true)
    private Long userId;

    @Setter
    @Column(name = "country", nullable = false)
    private String country;

    @Setter
    @Column(name = "state", nullable = false)
    private String state;

    @Setter
    @Column(name = "city", nullable = false)
    private String city;

    @SuppressWarnings("PMD.AvoidFieldNameMatchingTypeName")
    @Setter
    @Column(name = "address", nullable = false)
    private String address;


    public Address(Address address) {}
}

创建一个AddressRepository:

@Repository
public interface AddressRepository extends AuditableRepository<Address> {

    Set<Address> findByUserId(Long userId);
}

然后您可以从您的方法中获取地址集，如下所示:

@PreAuthorize("hasAuthority('ACCOUNT_OWNER', 'USER')")
public Set<Address> getAddresses(Long id) {
  return addressRepository.findByUserId(id);
}