如何让OpenSessionInViewInterceptor在Spring mvc中工作

Question

我正在尝试在Springmvc中设置OpenSessionInViewInterceptor来修复：org.hibernate.LazyInitializationException：无法初始化代理 - 没有会话。

下面是我已经拥有的代码以及错误的来源。

AppConfig.java

@Configuration
@PropertySource("classpath:db.properties")
@EnableTransactionManagement
@ComponentScans(value = { @ComponentScan("com.debugger.spring.web.tests"),  @ComponentScan("com.debugger.spring.web.service"), @ComponentScan("com.debugger.spring.web.dao"),
@ComponentScan("com.debugger.spring.web.controllers") })
public class AppConfig implements WebMvcConfigurer {

@Autowired
private Environment env;

@Bean
public LocalSessionFactoryBean getSessionFactory() {
    LocalSessionFactoryBean factoryBean = new LocalSessionFactoryBean();

    Properties props = new Properties();

    // Setting JDBC properties
    ...

    // Setting Hibernate properties
    ...

    // Setting C3P0 properties
        ...

    return factoryBean;
}

@Bean
public OpenSessionInViewInterceptor openSessionInViewInterceptor() {
    OpenSessionInViewInterceptor openSessionInViewInterceptor = new OpenSessionInViewInterceptor();
    openSessionInViewInterceptor.setSessionFactory(getSessionFactory().getObject());
    return openSessionInViewInterceptor;
}
}

featured.jsp

<c:choose>
                            <c:when
                                test='${article.user.isSubscribed() and article.user.subscription.type eq "silver" }'>
                                <a class="bold"
                                    href='${pageContext.request.contextPath}/u/${article.user.username}'><span
                                    class="silvername"> <c:out value="${article.user.name}"></c:out></span></a>
                            </c:when>
                            <c:when
                                test='${article.user.isSubscribed() and article.user.subscription.type eq "gold" }'>
                                <a class="bold"
                                    href='${pageContext.request.contextPath}/u/${article.user.username}'><span
                                    class="goldname"> <c:out value="${article.user.name}"></c:out></span></a>
                            </c:when>
                            <c:when
                                test='${article.user.isSubscribed() and article.user.subscription.type eq "premium" }'>
                                <a class="bold"
                                    href='${pageContext.request.contextPath}/u/${article.user.username}'><span
                                    class="premiumname"> <c:out
                                            value="${article.user.name}"></c:out></span></a>
                            </c:when>
                            <c:otherwise>
                                <a class="bold"
                                    href='${pageContext.request.contextPath}/u/${article.user.username}'><span>
                                        <c:out value="${article.user.name}"></c:out>
                                </span></a>
                            </c:otherwise>
                        </c:choose>

${article.user.isSubscribed()}抛出该错误很可能是因为无法提取用户。我希望它运行时不使用渴望获取，我想我可以通过正确设置OpenSessionInViewInterceptor来实现。

Answer 1

在配置类中重写WebMvcConfigurer # add interceptors(InterceptorRegistry ):

@Override
public void addInterceptors(InterceptorRegistry registry) {
    OpenSessionInViewInterceptor openSessionInViewInterceptor = new OpenSessionInViewInterceptor();
    openSessionInViewInterceptor.setSessionFactory(getSessionFactory().getObject());

    registry.addWebRequestInterceptor(openSessionInViewInterceptor).addPathPatterns("/**");
}

另外，在配置类上添加＜code＞@EnableWebMvc＜/code＞。

针对OP的评论：

我不确定为什么它不起作用。对我来说，一切都很好。还有另一种方法可以实现此目的：

将hibernate.enable_lazy_load_no_trans属性设置为 true。

见23.9.1。获取Hibernate用户指南中的属性。

但正如《指南》所述，这不是一个很好的选择：

虽然启用此配置可以使< code > LazyInitializationException 消失，但最好使用一个提取计划来保证在会话关闭之前所有属性都已正确初始化。

In reality, you shouldn’t probably enable this setting anyway.