小编典典

如何使OpenSessionInViewInterceptor在Spring MVC中工作

spring-mvc

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

以下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()}最有可能引发错误,因为无法提取用户。我希望它能在不使用急切获取的情况下运行,并且我认为可以通过正确设置OpenSessionInViewInterceptor来实现它。


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2020-06-01

共1个答案

小编典典

在配置类中重写WebMvcConfigurer#addInterceptors(InterceptorRegistry)

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

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

还要添加@EnableWebMvcconfig类。

针对OP的评论:

我不确定为什么它不起作用。在我看来一切都很好。还有另一种方法可以实现此目的:

设置hibernate.enable_lazy_load_no_trans属性true


23.9.1。
有关更多信息,请参见《 Hibernate用户指南》中的“
获取属性

但是,如指南中所述,这 不是 一个很好的选择:

尽管启用此配置可​​以
LazyInitializationException取消,但最好使用获取计划,该计划可确保在关闭会话之前正确初始化所有属性。

In reality, you shouldn’t probably enable this setting anyway.
2020-06-01