在springboot中找不到类型错误的属性

Question

在尝试创建一对多（以及多对一）关系时，我遇到以下错误：

Caused by: org.springframework.data.mapping.PropertyReferenceException: No property 'id' found for type 'UserRole'!

以下是完整的故事。目前在这些课程中，我有以下几点：

LoginUser.java

@Entity
@Table(name = "table_users")
public class LoginUser {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "user_id", nullable = false, unique = true, updatable = false)
    private int userId;

    @Column(name = "first_name", nullable = false)
    private String firstName;

    @Column(name = "last_name", nullable = false)
    private String lastName;

    @Column(name = "username", nullable = false, unique = true)
    private String username;

    @JsonIgnore
    @Column(name = "password", nullable = false)
    private String password;

    //@Column(name = "user_role", nullable = false)
    @ManyToOne
    @JoinColumn(name = "role_id", referencedColumnName = "role_id", nullable = false)
    private UserRole userRole;

    @Column(name = "is_active", nullable = false)
    private boolean isActive;

    public LoginUser() {}

    public LoginUser(
            String firstName,
            String lastName,
            String username,
            String password,
            UserRole userRole,
            boolean isActive
    ) {
        this.firstName = firstName;
        this.lastName = lastName;
        this.username = username;
        this.password = password;
        this.userRole = userRole;
        this.isActive = isActive;
    }

... Getters and Setters ...

用户角色。JAVA

@Entity
@Table(name = "lu_user_roles")
public class UserRole {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @PrimaryKeyJoinColumn
    @Column(name = "role_id")
    private int roleId;

    @Column(name = "role_title", nullable = false, unique = true, updatable = false)
    private String roleTitle;

    @OneToMany(mappedBy = "userRole")
    private List<LoginUser> loginUsers;

    public UserRole() {}

    public UserRole(String roleTitle) {
        this.roleTitle = roleTitle;
    }

... Getters and Setters ...

用户角色存储。JAVA

public interface UserRolesRepository extends CrudRepository<UserRole, Integer> {
    UserRole findUserRoleById(@Param("role_id") Integer roleId);
}

开始atabase.java

@Configuration
public class StartDatabase {
    private static final Logger LOG = LoggerFactory.getLogger(StartDatabase.class);

    @Autowired UserRolesRepository userRolesRepository;

    @Bean
    CommandLineRunner initUserRolesTable(UserRolesRepository repository) {
        return args -> {
            if(((List<UserRole>) repository.findAll()).size() == 0) {
                LOG.info("Adding " + repository.save(new UserRole("Developer")));
                LOG.info("Adding " + repository.save(new UserRole("Admin")));
                LOG.info("Adding " + repository.save(new UserRole("User")));
            }
        };
    }

    @Bean
    CommandLineRunner initUserTable(LoginUserRepository repository) {
        return args -> {
            if(((List<LoginUser>) repository.findAll()).size() == 0) {
                LOG.info("Preloading " + repository.save(
                        new LoginUser(
                                "Jamie",
                                "Edwards",
                                "jamie.edwards",
                                "password1",
                                userRolesRepository.findUserRoleById(1),
                                true
                        )
                ));
            }
        };
    }
}

我得到的全部错误是：

Caused by: org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'userRolesRepository' defined in com.jre.assetregister.database.repositories.loginuser.UserRolesRepository defined in @EnableJpaRepositories declared on JpaRepositoriesRegistrar.EnableJpaRepositoriesConfiguration: Invocation of init method failed; nested exception is org.springframework.data.repository.query.QueryCreationException: Could not create query for public abstract com.jre.assetregister.database.entities.loginuser.UserRole com.jre.assetregister.database.repositories.loginuser.UserRolesRepository.findUserRoleById(java.lang.Integer)! Reason: Failed to create query for method public abstract com.jre.assetregister.database.entities.loginuser.UserRole com.jre.assetregister.database.repositories.loginuser.UserRolesRepository.findUserRoleById(java.lang.Integer)! No property 'id' found for type 'UserRole'!; nested exception is java.lang.IllegalArgumentException: Failed to create query for method public abstract com.jre.assetregister.database.entities.loginuser.UserRole com.jre.assetregister.database.repositories.loginuser.UserRolesRepository.findUserRoleById(java.lang.Integer)! No property 'id' found for type 'UserRole'!


Caused by: org.springframework.data.repository.query.QueryCreationException: Could not create query for public abstract com.jre.assetregister.database.entities.loginuser.UserRole com.jre.assetregister.database.repositories.loginuser.UserRolesRepository.findUserRoleById(java.lang.Integer)! Reason: Failed to create query for method public abstract com.jre.assetregister.database.entities.loginuser.UserRole com.jre.assetregister.database.repositories.loginuser.UserRolesRepository.findUserRoleById(java.lang.Integer)! No property 'id' found for type 'UserRole'!; nested exception is java.lang.IllegalArgumentException: Failed to create query for method public abstract com.jre.assetregister.database.entities.loginuser.UserRole com.jre.assetregister.database.repositories.loginuser.UserRolesRepository.findUserRoleById(java.lang.Integer)! No property 'id' found for type 'UserRole'!


Caused by: java.lang.IllegalArgumentException: Failed to create query for method public abstract com.jre.assetregister.database.entities.loginuser.UserRole com.jre.assetregister.database.repositories.loginuser.UserRolesRepository.findUserRoleById(java.lang.Integer)! No property 'id' found for type 'UserRole'!


Caused by: org.springframework.data.mapping.PropertyReferenceException: No property 'id' found for type 'UserRole'!

我不完全确定在哪里或为什么它试图找到一个属性的id时，我显然已经设置了所有类型的UserRole。

Answer 1

在UserRoleRepository接口中，正如您所提到的，方法名为findUserRoleById，根据JPA惯例，此方法将在UserRole中查找id列，并尝试将其与作为@Param传递的roleId匹配。但是，UserRole类没有id属性，因此您将面临错误。

你有两个选择

>

可以将方法名更改为

findUserRolebyRoleId(@Param("role_id") Integer roleId)

因为您在UserRole@Entity中定义的属性名是roleId，而不是id。

如果您不想更改方法名称，您可以使用@Query注释像这样注释方法

@Query("FROM UserRole ur WHERE ur.roleId = :role_id")
public interface UserRolesRepository extends CrudRepository<UserRole, Integer> {
 UserRole findUserRoleById(@Param("role_id") Integer roleId);
}

Answer 2

问题在于findUserRoleById，因为spring与确切的字段名匹配，所以必须输入roleId-