使用Hibernate保存对象时收到以下错误

object references an unsaved transient instance - save the transient instance before flushing

当前回答

此错误的可能性非常多,其他一些可能性也出现在添加页或编辑页上。在我的案例中,我试图保存一个对象AdvanceSalary。问题是,在编辑AdvanceSalary employee.employee_id时为空,因为在编辑时我没有设置employee。

    @Entity(name = "ic_advance_salary")
    @Table(name = "ic_advance_salary")
    public class AdvanceSalary extends BaseDO{

        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
        @Column(name = "id")
        private Integer id;

        @ManyToOne(fetch = FetchType.EAGER)
        @JoinColumn(name = "employee_id", nullable = false)
        private Employee employee;

        @Column(name = "employee_id", insertable=false, updatable=false)
        @NotNull(message="Please enter employee Id")
        private Long employee_id;

        @Column(name = "advance_date")
        @DateTimeFormat(pattern = "dd-MMM-yyyy")
        @NotNull(message="Please enter advance date")
        private Date advance_date;

        @Column(name = "amount")
        @NotNull(message="Please enter Paid Amount")
        private Double amount;

        @Column(name = "cheque_date")
        @DateTimeFormat(pattern = "dd-MMM-yyyy")
        private Date cheque_date;

        @Column(name = "cheque_no")
        private String cheque_no;

        @Column(name = "remarks")
        private String remarks;

        public AdvanceSalary() {
        }

        public AdvanceSalary(Integer advance_salary_id) {
            this.id = advance_salary_id;
        }

        public Integer getId() {
            return id;
        }

        public void setId(Integer id) {
            this.id = id;
        }

        public Employee getEmployee() {
            return employee;
        }

        public void setEmployee(Employee employee) {
            this.employee = employee;
        }


        public Long getEmployee_id() {
            return employee_id;
        }

        public void setEmployee_id(Long employee_id) {
            this.employee_id = employee_id;
        }

    }

其他回答

如果您使用的是SpringDataJPA,那么在服务实现中添加@Transactional注释可以解决这个问题。

在引入乐观锁定(@Version)之后,我对所有PUT HTTP事务都面临同样的错误

在更新实体时,必须发送该实体的id和版本。如果任何实体字段与其他实体相关,那么对于该字段,我们也应该提供id和版本值,而不是JPA首先将相关实体作为新实体持久化

示例:我们有两个实体-->Vehicle(id、Car、version);汽车(id、版本、品牌);要更新/保存车辆实体,请确保车辆实体中的“车辆”字段已提供id和版本字段

在我的例子中,当我试图使用对具有空id的实体的引用来检索相关实体时,发生了这种情况。

@Entity
public class User {
@Id
private Long id;
}

@Entity
public class Address {
@Id
private Long id;
@JoinColumn(name="user_id")
@OneToOne
private User user;
}

interface AddressRepository extends JpaRepository<Address, Long> {
Address findByUser(User user);
}

User user = new User(); // this is transient, does not have id populated
// user.setId(1L) // works fine if this is uncommented

addressRepository.findByUser(user); // throws exception

只需在基类中创建映射的构造函数。就像你想要实体A、实体B中的一对一关系一样。如果你将A作为基类,那么A必须有一个构造函数,B作为参数。

解决这个问题的简单方法是保存这两个实体。首先保存子实体,然后保存父实体。因为父实体依赖于外键值的子实体。

下面是一对一关系的简单检查

insert into Department (name, numOfemp, Depno) values (?, ?, ?)
Hibernate: insert into Employee (SSN, dep_Depno, firstName, lastName, middleName, empno) values (?, ?, ?, ?, ?, ?)

Session session=sf.openSession();
        session.beginTransaction();
        session.save(dep);
        session.save(emp);