在 Spring MVC 中，我可以使用 JSTLView 的暴露上下文BeanNames(或暴露上下文BeansAsAttributes)在 JSP 中访问我的 bean。例如，然后，我可以在我的 JSP 中编写 (${properties.myProperty)。但是当同一个 JSP 是瓦片 View 的一部分时，这些属性就无法访问了。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:我已经实现了 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 Exception {
        AbstractUrlBasedView superView = super.buildView(viewName);
        if (superView instanceof TilesExposingBeansView) {
            TilesExposingBeansView view = (TilesExposingBeansView) superView;
            if (this.exposeContextBeansAsAttributes != null) view.setExposeContextBeansAsAttributes(this.exposeContextBeansAsAttributes);
            if (this.exposedContextBeanNames != null) view.setExposedContextBeanNames(this.exposedContextBeanNames);
        }
        return superView;
    }

}

TilesExposingBeansView.java:

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

import java.util.*;
import javax.servlet.http.*;
import org.springframework.web.context.support.ContextExposingHttpServletRequest;
import org.springframework.web.servlet.view.tiles2.TilesView;

public class TilesExposingBeansView extends TilesView {

    private boolean exposeContextBeansAsAttributes = false;
    private Set<String> exposedContextBeanNames;

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

    public void setExposedContextBeanNames(String[] exposedContextBeanNames) {
        this.exposedContextBeanNames = new HashSet<String>(Arrays.asList(exposedContextBeanNames));
    } 

    protected HttpServletRequest getRequestToExpose(HttpServletRequest originalRequest) {
        if (this.exposeContextBeansAsAttributes || this.exposedContextBeanNames != null) 
        return new ContextExposingHttpServletRequest(originalRequest, getWebApplicationContext(), this.exposedContextBeanNames);
        return originalRequest;
    }

    @Override
    protected void renderMergedOutputModel(Map<String, Object> model, HttpServletRequest request, HttpServletResponse response) throws Exception {
        HttpServletRequest requestToExpose = getRequestToExpose(request);
        exposeModelAsRequestAttributes(model, requestToExpose);
        super.renderMergedOutputModel(model, requestToExpose, response);
    }

}

Spring 配置:

    <bean id="tilesViewResolver" class="es.kcsolutions.util.spring.servlet.view.TilesExposingBeansViewResolver">
        <property name="order" value="1"/>
        <property name="viewClass" value="es.kcsolutions.util.spring.servlet.view.TilesExposingBeansView"/>
        <property name="exposedContextBeanNames">
            <list><value>properties</value></list>
        </property>
    </bean> 

如果您有问题，请查看 TilesExposingBeansView.renderMergedOutputModel。我做了一些测试，但可能需要做更多的测试。

最佳答案

你试过在xml配置文件中使用ServletContextAttributeExporter吗？

<bean
    class="org.springframework.web.context.support.ServletContextAttributeExporter">
    <property name="attributes">
        <map>
            <entry key="<bean key here>">
                <ref bean="<bean name here" />
            </entry>
        </map>
    </property>
</bean>

关于java - 从 Tiles View (JSP) 访问 Spring bean，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/2848415/