Java Criteria.uniqueResult返回null，但未引发任何错误

我正在学习hibernate，我正在尝试查询一个表以查找具有特定id的人。当我调用criteria.unique result时，不会抛出错误，但它返回一个空值，有人能帮我找出其中的原因吗

@Entity
@Table(name="employees")
public class Employee {
private long emp_no; 
private Date birth_date; 
private String first_name;
private String last_name;
private char gender;
private Date hire_date; 

private Set<DepartmentEmp> department_emp = new HashSet<DepartmentEmp>(); 
private Set<DepartmentManager> department_manager = new HashSet<DepartmentManager>();
private Set<Projects> projects; 
private ContactInfo contactInfo; 


public Employee(){};
public Employee(long empt_no, Date birth_date, String first_name, String last_name, 
        char gender, Date hire_date){
        this.emp_no = emp_no;
        this.birth_date = birth_date; 
        this.first_name = first_name;
        this.last_name = last_name;
        this.gender = gender; 
        this.hire_date = hire_date; 
}

 @Id
 @Column(name="emp_no")
    public long getEmp_no() { return emp_no; }
    public void setEmp_no(long emp_no) {this.emp_no=emp_no;}


 @Column(name="birth_date")
    public Date getBirth_date() { return birth_date; }
    public void setBirth_date(Date birth_date) {this.birth_date = birth_date;}


 @Column(name="first_name")
    public String getFirst_name() { return first_name; } 
    public void setFirst_name(String name) { this.first_name = name; }



 @Column(name="last_name")
    public String getLast_name() { return last_name; } 
    public void setLast_name(String name) { this.last_name = name; }


 @Column(name="gender")
    public char getGender() { return gender; } 
    public void setGender(char gender) {this.gender = gender;}


  @Column(name="hire_date")
    public Date getHire_date() { return hire_date; } 
    public void setHire_date(Date hire_date) {this.hire_date = hire_date;}

    @OneToOne(cascade=CascadeType.ALL, fetch=FetchType.LAZY)
    @JoinColumn(name="emp_no")
    public ContactInfo getContactInfo() { return contactInfo; }
    public void setContactInfo(ContactInfo info) { this.contactInfo = info; }

  @OneToMany(mappedBy="employee", targetEntity=DepartmentEmp.class,
    cascade=CascadeType.ALL)
  @Fetch(value = FetchMode.SELECT) 
    public Set<DepartmentEmp> getDepartment_emp() { return department_emp; }
    public void setDepartment_emp(Set<DepartmentEmp> department_emp) { this.department_emp = department_emp; }



    @OneToMany(mappedBy="employee", targetEntity=DepartmentManager.class,
    cascade=CascadeType.ALL)
    @Fetch(value = FetchMode.SELECT) 

    public Set<DepartmentManager> getDepartment_manager() { return department_manager; }
    public void setDepartment_manager(Set<DepartmentManager> department_manager) { this.department_manager = department_manager; }

    @ManyToMany
      @JoinTable(name="employee_projects", 
            joinColumns={@JoinColumn(name="employee_emp_no")},
            inverseJoinColumns={@JoinColumn(name="projects_proj_num")})
    @Fetch(value = FetchMode.SELECT) 
        public Set<Projects> getProjects() { return projects; }
        public void setProjects(Set<Projects> projects) { this.projects = projects; }

    public void print() {
        System.out.printf("%d: %s %s %s %c %s\n", emp_no, birth_date.toString(), first_name, last_name, gender, hire_date.toString()); 
    } 

    public static void list()
    {
        Session session = HibernateContext.getSession();
        Criteria criteria = session.createCriteria(Employee.class);
        criteria.addOrder(Order.asc("emp_no"));

        List <Employee> employees = criteria.list();
        System.out.println("All Employees:");
        for (Employee employee : employees) {
            employee.print();
        }
        session.close();
    }

    //TODO finish 4 methods

    public static Employee find(long emp_no){
        Employee prototype = new Employee();
        prototype.setEmp_no(emp_no);
        Example ex = Example.create(prototype);

        Session sess = HibernateContext.getSession();
        Criteria criteria = sess.createCriteria(Employee.class);
        criteria.add(ex);

        Employee e = (Employee) criteria.uniqueResult();
        return e;
    }
    public static Employee find(String first, String last){
        Employee prototype = new Employee();
        prototype.setFirst_name(first);
        prototype.setLast_name(last);
        Example ex = Example.create(prototype);

        Session sess = HibernateContext.getSession();
        Criteria criteria = sess.createCriteria(Employee.class);
        criteria.add(ex);

        Employee e = (Employee) criteria.uniqueResult();

        sess.close();
        return e;
    }
    public static void main(String [] args){
           Class klasses [] = {Employee.class, DepartmentEmp.class, DepartmentManager.class, ContactInfo.class,
                                Salaries.class, Titles.class, Department.class, Projects.class};
            HibernateContext.addClasses(klasses);
            Employee.find((long)10001).print();
    }
  }

}

---

我已经编辑了其他类，因为它们不参与这个查询，但是当我在Employee中运行main函数时，堆栈跟踪只是在find.print（）处显示null指针异常；线路。我知道数据库已填充，正在运行Employee.list（）函数，所以我正在努力找出到底是什么地方出了问题。谢谢你的帮助

如果没有stacktrace，很难找到问题所在，以下是我的猜测：

如果我们看到你的打印方法

public void print() {
        System.out.printf("%d: %s %s %s %c %s\n", emp_no, birth_date.toString(),
 first_name, last_name, gender, 
hire_date.toString()); 
    } 

即使员工对象正确返回。。在

print

中，您将获得

hire\u date

如果未将

hire\u date

指定给

employee对象

并尝试调用hire\u date.toString（），则会出现

空指针异常

---

确保返回的Employee对象是否有

hire\u date

或没有

检查DB中id为10001的Employee对象。如果他已经有hire\u date，请忽略此答案。我会在你回家时输入确切的堆栈跟踪，但它所说的是在我调用print的行上出现空指针异常。在my employee.find方法中，criteria.unique result（）方法返回null，因此返回的employee对象中根本没有值。员工Id确实存在，因为我选择了DBOky中的一个，所以我意识到默认构造函数将char-gender初始化为“”，而不是null，这使得它也可以搜索查询的性别，我通过将其切换为字符串来修复它

public void print() {
        System.out.printf("%d: %s %s %s %c %s\n", emp_no, birth_date.toString(),
 first_name, last_name, gender, 
hire_date.toString()); 
    }