LazyInitializationException-无法初始化代理-无会话

Question

我有一个LazyInitializationException问题，我不知道如何解决它。

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);
    }

之前的问题是我打电话的时候。getperson=null，但我修复了findProjectEmployeesWithinDates请求获取此人的调用。但当我调用“findProjectEmployeesWithinDates”时，我遇到了一个例外。查找项目员工的代码包括：

    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;
}

所以用debbug我看到：

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

它位于findProjectEmployeesWithinDates中for循环的第一行，但不在第二行，这就是异常发生的地方。

我得到的错误

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

org.hibernate.proxy.AbstractLazyInitializer.initialize（AbstractLazyInitializer.java:165）~[hibernate-core-4.3.6。Final.jar： 4.3.6。最终]在org.hibernate.Hibernate.initialize（Hibernate.java:75）~[hibernate-core-4.3.6。Final.jar： 4.3.6。最终]在com.xitee.ccpt.service.项目员工ervice.find项目员工有日期（项目员工ervice.java:105）~[类/： na]在com.xitee.ccpt.service.项目员工服务$$FastClassBySpringCGLIB$$63bfc6f9.invoke（）~[sping-core-4.1.1。RELEASE. jar： na]

项目员工类别：

@Entity

@table（name="雇员"， Schema="ccpt_data"）@NamedQuery（name="Employee.findAll"， query="SELECT e From雇员"）公共类雇员实现可序列化{私有静态最终长序列化VersionUID=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;
}

有人能帮我解决这个问题吗？

Answer 1

• 当您调用projectEmployeeRepository时。findProjectEmployeesWithinDates方法返回列表。此时，您的Hibernate会话已经关闭
• 因此，在读取ProjectEmployee对象时，只允许访问那些特定于对象的变量，而不允许访问特定于对象的子对象，因为您正在对子对象使用延迟初始化
• 因此，解决方法是保持hibernate会话处于打开状态，或者使用eager fetch或使用wrapper类对象将值从ProjectEmployee类传输到projectEmployeeRepository中的ProjectEmployeeWrapper。findProjectEmployeesWithinDates方法，然后返回ProjectEmployeeWrapper对象的列表

Answer 2

我建议采用以下方法之一：

1） 在您的存储库方法中，您可以

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将自动使用员工和人员对象填充执行结果