Question

我正在java中开发一个小的Spring和hibernate基础应用程序，我的应用程序与Employee和Unit 有一个很强的关系，Employee有一个单元，Unit有很多Employee。在这个小应用程序genarate错误像这样`

com.mysql.jdbc.exceptions.jdbc4.MySQLIntegrityConstraintViolationException：列'unit_id'不能为空

如果允许null为unit_id的其他数据插入插入unit_id

这是我的实体类

@Entity
@Table(name = "employee")
public class Employee implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)

    @Column(name = "id")
    private Integer id;

    @Column(name = "epf")
    private int epf;

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

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

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

    @JoinColumn(name = "unit_id", referencedColumnName = "id")
    @ManyToOne//(optional = false)
    private Unit unit;


@Entity
@Table(name = "unit")
public class Unit implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private Integer id;

    @Column(name = "name")

我的控制器类

@RequestMapping(value="/employee.do", method=RequestMethod.POST)
    public ModelAndView doActions(@ModelAttribute Employee emp, BindingResult result, @RequestParam String action, Map<String, Object> map){
        ModelAndView modelAndView = new ModelAndView("employee");

        Employee employeetResult = new Employee();
        switch(action.toLowerCase()){   //only in Java7 can put String in switch
        case "add":
            employeeService.addEmployee(emp);
            employeetResult = emp;
            break;
        case "edit":
            employeeService.updateEmployee(emp);
            employeetResult = emp;
            break;
        case "delete":
            employeeService.deleteEmployee(emp.getId());
            employeetResult = new Employee();
            break;
        case "search":
            Employee searchedStudent = employeeService.getEmployee(emp.getId());
            employeetResult = searchedStudent!=null ? searchedStudent : new Employee();
            break;
        }
        map.put("employee", employeetResult);
        map.put("employeeList", employeeService.getAllEmployee());
        return modelAndView;
    }

我的DAO课程

> @Repository public class EmployeeDaoImp implements EmployeeDao {
> 
>   @Autowired  private SessionFactory sessionfactory;
>           public void addEmployee(Employee emp) {         sessionfactory.getCurrentSession().save(emp);
> 
>   }
> 
>       public void updateEmployee(Employee emp) {
>       sessionfactory.getCurrentSession().update(emp);
> 
>   }
> 
>       public void deleteEmployee(int id) {
>               sessionfactory.getCurrentSession().delete(getEmployee(id));     }
> 
>    Employee   public Employee getEmployee(int empId) {
>               return (Employee) sessionfactory.getCurrentSession().get(Employee.class,empId);     }
> 
>       public List getAllEmployee() {
>               return sessionfactory.getCurrentSession().createQuery("from Employee").list();  }

Answer 1

使用@JoinColumn(name = "unit_id", referencedColumnName = "id")，您在Unit表上定义了一个不能为空的FK。在将代码保存到数据库之前，您需要使用代码中的employee.setUnit(someUnit)将您的员工链接到某个单元。

Answer 2

列ID不应为null。您将从guruzon.com获得更好的答案。他们已经包含了示例代码和程序的详细说明。除了hibernate他们已经用示例程序，Sping，面试问题解释了Hadoop技术

Spring-Hibernate插入错误与异常

2 个答案: