java - Spring Data Jpa不会自动获取下一个ID(PK)，它总是从1开始

Question

我正在开发 Spring Boot (2.1.7.RELEASE) +Data Jpa + Postgres 示例。在此示例中，我显式传递 EMP_ID value=100，接下来我允许 data-jpa 自动获取下一个 Id，即 101。我不确定为什么它不能以这种方式工作？

员工.java

@Data
@AllArgsConstructor
@NoArgsConstructor
@EqualsAndHashCode(callSuper = true)
@Entity
public class Employee extends BaseEntity{

    @Id
    @GeneratedValue(strategy = IDENTITY)
    @Column(name = "EMP_ID", unique = true, nullable = false)
    private Integer empId;

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

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


    @Builder(builderMethodName="eBuilder")
    public Employee(Integer empId, String empName, String empEmail,
            Instant createdDate, Instant lastUpdateDate,String createUser, String lastUpdateUser) {
        super(createdDate, lastUpdateDate, createUser, lastUpdateUser);
        this.empId = empId;
        this.empName = empName;
        this.empEmail = empEmail;
    }
}

BaseEntity.java

@Data
@MappedSuperclass
@NoArgsConstructor
@AllArgsConstructor
@EntityListeners(AuditingEntityListener.class)
public class BaseEntity {
    @CreatedDate
    @Column(name = "createdDate", nullable = false, updatable = false)
    private Instant createdDate;

    @Column(name = "lastUpdateDate", nullable = false)
    @LastModifiedDate
    private Instant lastUpdateDate;

    @Column(name = "createUser", nullable = false, length = 50)
    private String createUser;

    @Column(name = "lastUpdateUser", length = 50)
    private String lastUpdateUser;
}

MainApp.java

@SpringBootApplication
@EnableJpaAuditing
public class MyExampleApplication implements CommandLineRunner {

    public static void main(String[] args) {
        SpringApplication.run(MyExampleApplication.class, args);
    }

    @Autowired
    private EmployeeRepository employeeRepository;

    @Override
    public void run(String... args) throws Exception {
        Employee e = Employee.eBuilder().empId(100).empName("Shrutika")
                .empEmail("shrutika@hotmail.com")
                .createUser("Shrutika")
                .lastUpdateUser("Shrutika")
                .build();

        employeeRepository.save(e);

        Employee e1 = Employee.eBuilder().empName("Shantaram")
                .empEmail("shantaram@hotmail.com")
                .createUser("Shantaram")
                .lastUpdateUser("Shantaram")
                .build();
        employeeRepository.save(e1);
    }
}

即使我在下面使用，仍然效果不佳

@Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "emp_generator")
    @SequenceGenerator(name="emp_generator", sequenceName = "emp_seq", allocationSize=1)
    @Column(name = "EMP_ID", unique = true, nullable = false)
    private Integer empId;

Spring JIRA:https://jira.spring.io/browse/DATAJPA-1588

Answer 1

确保数据库的类型为EMP_ID:SERIAL 或Integer。要将 IDENTITY 与 postgres 一起使用，它必须是 SERIAL ( https://www.postgresql.org/docs/8.1/datatype.html#DATATYPE-SERIAL )。