相关疑难解决方法(0)

在Spring MVC中,我可以使用JstlView的exposedContextBeanNames(或exposeContextBeansAsAttributes)在JSP中访问我的bean.例如,然后,在我的JSP中我可以编写($ {properties.myProperty).但是,当相同的JSP是切片视图的一部分时,无法访问这些属性.是否可以正确配置Tiles或以其他方式访问这些属性？

我正在使用Spring MVC 3.0.2和Tiles 2.2.1.这是我的一些配置:

<bean id="tilesViewResolver" class="org.springframework.web.servlet.view.UrlBasedViewResolver">
    <property name="order" value="1"/>
    <property name="viewClass" value="org.springframework.web.servlet.view.tiles2.TilesView" />
</bean> 

<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
    <property name="viewClass" value="org.springframework.web.servlet.view.JstlView"/>
    <property name="order" value="2"/>
    <property name="prefix" value="/WEB-INF/views/"/>
    <property name="suffix" value=".jsp"/>
    <property name="exposedContextBeanNames">
        <list><value>properties</value></list>
    </property>
</bean>

编辑:我已经实施了Skaffman的解决方案.

TilesExposingBeansViewResolver.java:

package es.kcsolutions.util.spring.servlet.view;

import org.springframework.web.servlet.view.*;

public class TilesExposingBeansViewResolver extends UrlBasedViewResolver {

    private Boolean exposeContextBeansAsAttributes;
    private String[] exposedContextBeanNames;

    public void setExposeContextBeansAsAttributes(boolean exposeContextBeansAsAttributes) {
        this.exposeContextBeansAsAttributes = exposeContextBeansAsAttributes;
    }

    public void setExposedContextBeanNames(String[] exposedContextBeanNames) {
        this.exposedContextBeanNames = exposedContextBeanNames;
    }

    @Override
    protected AbstractUrlBasedView buildView(String viewName) throws …

在我的应用程序上下文中,我定义了属性文件

<context:property-placeholder  location="classpath:application.properties" />

我想获得JSP页面上该文件中定义的属性的值.有没有办法做到这一点

${something.myProperty}?