Question

我的数据库中有3个表 - Booking，Restaurant和RestaurantTable。现在我正在尝试创建一个新的预订，其中一个步骤就是添加一个表格。但是当我尝试添加此表时，会出现以下错误：

org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role

这是我的餐厅课程：

@Entity
@Table(name="restaurant")
public class Restaurant {

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

    @Column(name="restaurant_name")
    private String restaurantName;

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

    @OneToMany(mappedBy = "restaurant")
    private Set<RestaurantTable> table;

    // Getters and setters

我可以改变＆＃34; table＆＃34;到FetchType.EAGER，但这会导致其他问题。我的餐厅课程：

@Entity
@Table(name="restaurant_table")
public class RestaurantTable {

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

    @Column(name="table_size")
    private Integer tableSize;

    @Column(name="table_number")
    private Integer tableNumber;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name="restaurant_id")
    private Restaurant restaurant;

    // Getters and setters.

我的BookingController.java：

@RequestMapping(value = "booking/create/{id}", method = RequestMethod.GET)
public String chooseTable(@PathVariable Long id, Model model) {
    Booking booking = bookingService.getBooking(id);
    Restaurant restaurant = booking.getRestaurant();
    Set<RestaurantTable> tableSet = restaurant.getTable();
    model.addAttribute("tables", tableSet);
    model.addAttribute("booking", booking);
    return "chooseTable";
}

.jsp文件发生错误：

<body>
<jsp:include page="../fragments/menu.jsp"/>
<div id="body">
    <h2>Create new booking</h2>

    <form:form method="POST" modelAttribute="booking" >
        <table>
            <tr>
                <td>Choose a table*:</td>
                <td><form:select path="tableNumber">
                        <form:option value="" label="--- Select ---" />
                        <form:options items="${tables}" itemValue="tableNumber" itemLabel="tableNumber"/>
                </form:select>
            </tr>
            <tr>
                <td colspan="3"><input type="submit" /></td>
            </tr>
        </table>
    </form:form>
    <div>
        <a href="/bookings">Back to List</a>
    </div>
</div>
<jsp:include page="../fragments/footer.jsp"/>

</body>

感谢任何帮助！

Answer 1

重构ResturantTable并删除此类中的提取类型

    @ManyToOne        
    @JoinColumn(name="restaurant_id")
    private Restaurant restaurant;

在Resturant类

中添加提取类型

@OneToMany(mappedBy = "restaurant",fetch = FetchType.LAZY)
private Set<RestaurantTable> table;

并添加此行bookingService类方法getBooking(id)以便将所有数据初始化

booking.getRestaurant().getTable().size();

booking您的服务方法getBooking(id)返回对象

Answer 2

以惰性模式加载相关实体意味着实体将在第一次访问时加载，前提是会话应该是开放且有效的。

会话结束时访问项目会引发LazyInitializationException。

确保在会话打开时访问项目或将获取类型模式从lazy更改为eager（默认）。

org.hibernate.LazyInitializationException（Spring / Hibernate）

2 个答案: