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In applicationContext.xml or some other context file that's loaded by web.xml, add the following:

Code Block

	xml	xml
title	applicationContext.xml
	xml

<bean id="ajaxResponseController" class="org.jasig.web.servlet.mvc.AjaxResponseController"/>

    <bean class="org.springframework.web.servlet.view.XmlViewResolver"
    	p:order="0"/>

By default XmlViewResolver looks for WEB-INF/views.xml
jsonView automatically digests your model objects and represents them as JSON. You can probably tweak this by manipulating JsonConfig in the jsonView bean.

xml

Code Block

	xml
title	views.xml
	xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd">

    <bean id="jsonView" class="net.sf.json.spring.web.servlet.view.JsonView">
        <property name="contentType" value="application/json" />
    </bean>
</beans>

Set the order parameter for your existing view resolvers to something greater than 0.

xml

Code Block

	xml
title	Freemarker View Resolver example
	xml

<bean id="viewResolver" class="org.springframework.web.servlet.view.freemarker.FreeMarkerViewResolver"
		p:order="1"
		p:cache="true"
		p:prefix=""
		p:suffix=".ftl"/>

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The AJAX Servlet listens for AJAX requests. Add the following sections to your web.xml file in the appropriate locations.

xml

Code Block

	xml
title	web.xml excerpt
	xml

<servlet>
        <servlet-name>AjaxResponseServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/ajaxResponseServletContext.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

<servlet-mapping>
        <servlet-name>AjaxResponseServlet</servlet-name>
        <url-pattern>/ajaxResponse</url-pattern>
    </servlet-mapping>

Create this file in the location specified above.

xml

Code Block

	xml
title	ajaxResponseServletContext.xml
	xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd">

    <bean class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
        <property name="order" value="0"/>
        <property name="alwaysUseFullPath" value="true"/>
        <property name="defaultHandler" ref="ajaxResponseController" />
        <property name="urlMap">
            <map>
                <entry key="/ajaxResponse" value-ref="ajaxResponseController"/>
            </map>
        </property>
    </bean>

    <bean id="requiredAnnotationBeanPostProcessor" class="org.springframework.beans.factory.annotation.RequiredAnnotationBeanPostProcessor" />
</beans>

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With an annotated controller method, this looks something like this. (Note: This uses an alternate way of passing the model)

java

Code Block

	java
title	Annotated Controller Method Example
	java

@RequestMapping(params="action=ajaxTest")
public void ajaxTest(ActionRequest request,ActionResponse response) throws Exception {
   final Map<String, Object> model = new HashMap<String, Object>();
   model.put("testData", "foo");
   this.ajaxPortletSupportService.redirectAjaxResponse(this.ajaxServletName, model, request, response);
}

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