Question

我遇到了LazyInitializationException的问题，我不知道如何解决。

for (Long id : employeeIds)
    {
        List<ProjectEmployee> projectEmployeeList = projectEmployeeService.findProjectEmployeesWithinDates(id,
                startDate, endDate);

        // if no data, then continue with next employee
        if (projectEmployeeList.isEmpty())
        {
            continue;
        }

        gridCreated = true;

        Employee employee = projectEmployeeList.get(0).getEmployee();
        Label titleLabel = new Label(employee.getPerson().getSurname() + " " + employee.getPerson().getName() + " ["
                                     + employee.getRole().getHumanizedRole() + "]");
        titleLabel.setStyleName("header-bold");

        ProjectEmployeePanel projectEmployeePanel = new ProjectEmployeePanel(id, startDate, endDate);
        gridPanelsLayout.addComponents(titleLabel, projectEmployeePanel);
    }

问题出在我打电话给.getperson = null之前，但我修复了findProjectEmployeesWithinDates呼叫，要求在那里找人。但是当我调用'findProjectEmployeesWithinDates'时，出现了异常。代码findProjectEmployeesWithinDates：

    public List<ProjectEmployee> findProjectEmployeesWithinDates(Long employeeId, LocalDate startDate, LocalDate endDate) {
    List<Long> list = new ArrayList<>();
    list.add(employeeId);
    List<ProjectEmployee> listProjectEmployees = projectEmployeeRepository.findProjectEmployeesWithinDates(list,
            LocaleUtils.getDateFromLocalDate(startDate, LocaleUtils.APPLICATION_DEFAULT_ZONE_ID),
            LocaleUtils.getDateFromLocalDate(endDate, LocaleUtils.APPLICATION_DEFAULT_ZONE_ID));
    for (ProjectEmployee pe : listProjectEmployees)
    {
        Hibernate.initialize(pe.getEmployee());
        Hibernate.initialize(pe.getEmployee().getPerson());

    }
    return listProjectEmployees;
}

因此，使用debbug我看到了：

 Hibernate.initialize(pe.getEmployee()); ----line 105
Hibernate.initialize(pe.getEmployee().getPerson()); ---line 106

它在findProjectEmployeesWithinDates中的for循环的第一行，但不在第二行，这是发生异常的地方。

我得到的错误

Caused by: org.hibernate.LazyInitializationException: could not initialize proxy - no Session

at org.hibernate.proxy.AbstractLazyInitializer.initialize（AbstractLazyInitializer.java:165）〜[hibernate-core-4.3.6.Final.jar：4.3.6.Final] 在org.hibernate.Hibernate.initialize（Hibernate.java:75）〜[hibernate-core-4.3.6.Final.jar：4.3.6.Final] 在com.xitee.ccpt.service.ProjectEmployeeService.findProjectEmployeesWithinDates（ProjectEmployeeService.java:105）〜[classes /：na] 在com.xitee.ccpt.service.ProjectEmployeeService $$ FastClassBySpringCGLIB $$ 63bfc6f9.invoke（）〜[spring-core-4.1.1.RELEASE.jar：na]

Project Employee类：

@Entity

@Table（name =“ employee”，schema =“ ccpt_data”） @NamedQuery（name =“ Employee.findAll”，query =“ SELECT e FROM Employee e”）公共类Employee实现Serializable { 私有静态最终长serialVersionUID = 1L;

@Id
@Column(name = "employee_id")
@SequenceGenerator(name = "EMPLOYEE_ID_GENERATOR", sequenceName = "employee_id_seq", allocationSize = 1)
@GeneratedValue(strategy = GenerationType.AUTO, generator = "EMPLOYEE_ID_GENERATOR")
private Long employeeId;

@OneToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "person_id")
private Person person;

@Column(name = "monthly_cost")
private String monthlyCost;

@Enumerated(EnumType.STRING)
@Column(name = "role")
private EmployeeRole role;

@Column(name = "employee_manager")
private String employeeManager;

@Column(name = "obsolete")
private Boolean obsolete;

@Column(name = "bank_account_number")
private String bankAccountNumber;

@Column(name = "last_employer")
private String lastEmployer;

@Column(name = "starting_day")
private String startingDay;

@Column(name = "hours")
private Short hours;

@OneToMany(mappedBy = "employee", cascade =
    {
     CascadeType.ALL
    }, orphanRemoval = true)
private Set<EmployeeWorkload> employeeWorkloads;

@OneToMany(mappedBy = "employee", fetch = FetchType.LAZY, orphanRemoval = true)
private Set<ProjectEmployee> projectEmployee;

@OneToMany(mappedBy = "employee", cascade =
    {
     CascadeType.PERSIST
    }, orphanRemoval = true)
private Set<Qualification> qualifications;

@OneToMany(mappedBy = "employee", cascade =
    {
     CascadeType.PERSIST
    }, orphanRemoval = true)
private List<CareerExperience> careerExperiences;

@Transient
private Map<Integer, String> exportOptions;

@OneToMany(mappedBy = "employee", fetch = FetchType.LAZY, orphanRemoval = true)
private Set<ProjectEmployeeRejection> projectEmployeeRejections;

@Transient
private boolean decrypted = true; // allows editing and viewing for users with no encryption rights

/**
 * Initialization vector used for encryption of this employee or NO_KEY if no encryption was used
 *
 * @since 0.4.0
 */
@Column(name = "iv")
private String iv;

@Column(name = "cis_employee_id")
private Long cISEmployeeId;

@Column(name = "experience_summary")
private String experienceSummary;

@Enumerated(EnumType.STRING)
@Column(name = "employee_job_type")
private EmployeeJobType employeeJobType;

@Column(name = "ending_day")
@Type(type = "date")
private Date endingDay;

@Column(name = "main_skill")
private String mainSkill;

public Employee()
{
}

public Long getEmployeeId()
{
    return employeeId;
}

public void setEmployeeId(Long employeeId)
{
    this.employeeId = employeeId;
}

public Person getPerson()
{
    return person;
}

public void setPerson(Person person)
{
    this.person = person;
}

public String getExperienceSummary()
{
    return experienceSummary;
}

public void setExperienceSummary(String experienceSummary)
{
    this.experienceSummary = experienceSummary;
}

有人可以帮助我解决这个问题吗？

Answer 1

我建议采用以下方法之一：

1）您可以在存储库方法findProjectEmployeesWithinDates中

for (ProjectEmployee pe : listProjectEmployees)
{
    pe.getEmployee().getPerson();

}

因此它将在会话打开时初始化对象

2）您可以使用查询获取数据

  SELECT * FROM ProjectEmployee pe JOIN FETCH pe.employee e JOIN FETCH e.person

通过这种方式，Hibernates将自动用员工和人员对象填充执行结果

Answer 2

致电时 projectEmployeeRepository.findProjectEmployeesWithinDates方法返回列表。此时，您的休眠会话已关闭。
因此，当您读取ProjectEmployee对象时，由于您正在对子对象使用惰性初始化，因此您只能访问那些特定于对象的变量，而不能访问特定于对象的子对象。
因此，解决方法是保持休眠会话打开或使用急切的获取或使用包装器类对象将值从ProjectEmployee类中的值传输到projectEmployeeRepository.findProjectEmployeesWithinDates方法中的ProjectEmployee类，然后返回ProjectEmployeeWrapper对象的列表。

LazyInitializationException-无法初始化代理-没有会话

2 个答案: