Question

我刚刚在Ibm sprin mvc portlet上创建了一个小POC，它运行正常。但我有一个一直困扰着我的问题。在教程中它说我们可以拥有相同的DispatcherPortletor，我们可以使用自己的applicationContext和handler mapping来多个DispatcherPortlet。我的portlet.xml中有类似的东西：

<description>HelloWorld Portlet using Spring MVC portlet 1</description>
<portlet-name>HelloSpringPortletMVC1</portlet-name>
<display-name>Hello World Spring Portlet MVC Framework Portlet1</display-name>
<portlet-class>     org.springframework.web.portlet.DispatcherPortlet </portlet-class>
<supports>
    <mime-type>text/html</mime-type>
    <portlet-mode>VIEW</portlet-mode>
</supports>
<supported-locale>en</supported-locale>
<portlet-info>
<title> HelloSpringPortletMVC1 </title>
<short-title> HelloSpringPortletMVC1 </short-title>
<keywords> spring portlet </keywords>
</portlet-info>

<description>HelloWorld Portlet using Spring MVC portlet 2</description>
<portlet-name>HelloSpringPortletMVC2</portlet-name>
<display-name>Hello World Spring Portlet MVC Framework Portlet2</display-name>
<portlet-class>     org.springframework.web.portlet.DispatcherPortlet </portlet-class>
<supports>
    <mime-type>text/html</mime-type>
    <portlet-mode>VIEW</portlet-mode>
</supports>
<supported-locale>en</supported-locale>
<portlet-info>
<title> HelloSpringPortletMVC2 </title>
<short-title> HelloSpringPortletMVC2 </short-title>
<keywords> spring portlet </keywords>
</portlet-info>

我假设因为我们已经将portlet-class指定为org.springframework.web.portlet.DispatcherPortlet，每次呈现portlet时它都会创建一个DispatcherPortlet的新实例，那么我们怎么才能只有一个实例DispatcherPortlet将用于所有portlet。

Answer 1

我假设因为我们已经将portlet-class指定为org.springframework.web.portlet.DispatcherPortlet，每次呈现portlet时它都会创建一个DispatcherPortlet的新实例，那么我们怎么才能只有一个实例将用于所有portlet的DispatcherPortlet

不需要创建多个DispatcherPortlet实例，您已经定义了将所有portlet请求路由到控制器的DispatcherPortlet，例如Spring MVC应用程序DispatcherServlet将您的请求转发给控制器。您需要的只是创建Controller类（类）并描述spring配置

portlet.xml的一部分

    <portlet>
    <portlet-name>My Portlet</portlet-name>
    <display-name>My Portlet</display-name>
    <portlet-class>org.springframework.web.portlet.DispatcherPortlet</portlet-class>
    <init-param>
        <name>contextConfigLocation</name>
        <value>classpath:/my_portlet.xml</value>
    </init-param>
   ...
</portlet>

像这样的

my_portlet.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"
       xmlns:context="http://www.springframework.org/schema/context"
       xsi:schemaLocation="http://www.springframework.org/schema/beans
        http://www.springframework.org/schema/beans/spring-beans-3.1.xsd
        http://www.springframework.org/schema/context
        http://www.springframework.org/schema/context/spring-context-3.1.xsd>


    <context:annotation-config />
    <context:component-scan base-package="my.package.portlet.controllers" />

    <bean class="org.springframework.web.portlet.mvc.annotation.DefaultAnnotationHandlerMapping"/>

    <bean id="viewResolver" class="org.springframework.web.servlet.view.UrlBasedViewResolver">
        <property name="viewClass" value="org.springframework.web.servlet.view.JstlView"/>
        <property name="prefix" value="/WEB-INF/view/"/>
        <property name="suffix" value=".jsp"/>
    </bean>
    <bean name="myService" class="my.package.services.MyService"/> 
    </bean>
</beans>

控制器类是这样的（它可能包含错误，因为从内存中获取）

@Controller
@RequestMapping("view")
public class MyController {

    @Autowired
    private MyService myService;

    @RequestMapping
    public ModelAndView index(RenderRequest request, RenderResponse response) {
        ModelAndView modelAndView = new ModelAndView("index.jsp");
        modelAndView.addObject("attribute1", "Attribute 1 value");
        modelAndView.addObject("attribute1", new Object());
        return modelAndView;
    }

    @RenderMapping(params = "form=result")
    public ModelAndView doSomething(RenderRequest request, RenderResponse response) {
        ModelAndView modelAndView = new ModelAndView("result.jsp");
        ...
        return modelAndView;
    }

    @ActionMapping(params = "submitSomething=yes")
    public void submitSomething(@ModelAttribute("something") Something something, BindingResult bindingResult, ActionRequest request, ActionResponse response) {
        // do what you want with form data
        request.setAttribute("result", someResult);
        response.setRenderParameter("form", "result");
    }

}

您可以使用不同方法获得不同渲染/操作参数的多个Controller类的主要思想。也可以在控制器类中使用spring bean。

DispatcherPortlet的多个实例

1 个答案: