如何使 OpenSessionInViewInterceptor 在 Spring mvc 中工作

Question

我正在尝试在 spring mvc 中设置 OpenSessionInViewInterceptor 来修复：org.hibernate.LazyInitializationException: could not initialize proxy - no Session。

以下是我已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()} 最有可能因为无法获取用户而引发错误。我希望它在不使用 Eager fetch 的情况下运行，我认为我可以通过正确设置 OpenSessionInViewInterceptor 来实现它。

Answer 1

在您的配置类中覆盖 WebMvcConfigurer#addInterceptors(InterceptorRegistry)：

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

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

还在配置类中添加@EnableWebMvc。

回应 OP 的评论：

我不确定它为什么不起作用。在我看来一切都很好。还有另一种方法可以实现：

设置hibernate.enable_lazy_load_no_trans属性true。

请参阅23.9.1. Fetching properties in Hibernate User Guide 了解更多信息。

但这不是指南中所述的一个非常好的选择：

虽然启用此配置可​​以使 LazyInitializationException走开，最好用一个fetch计划 保证所有属性在之前正确初始化 会话已关闭。

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