如何在Spring MVC控制器中映射动态URL / prj / noticeOpen / 2

Question

我在以下网址上遇到了麻烦：

<a href="/noticeOpen/2">dynamicLink</a>

使用以下控制器方法进行映射：

@RequestMapping(value="/noticeOpen/{quesId}") 
public ModelAndView noticeOpen(@ModelAttribute("NoticeForm") NoticeForm noticeForm,
                               ModelMap model,
                               @PathVariable("quesId") Integer quesId){
    System.out.println(quesId);
    return new ModelAndView("/noticeOpen","noticeForm",noticeForm);

}

当我单击锚定dynamicLink时，问题开始了，它没有将控制权转移到我的控制器，而是在浏览器的地址栏中显示了以下内容：

http://127.0.0.1:8080/prj/noticeOpen/2/WEB-INF/pages/noticeOpen.jsp

而且我在applicationContext.xml中有以下映射

 <bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix">
            <value>WEB-INF/pages/</value>
        </property>
        <property name="suffix">
            <value>.jsp</value>
        </property>
</bean>

如果我从控制器的@RequestMapping中删除{quesId}和从方法签名中删除@PathParam（也从锚中删除问题ID），这一切都很好

http://127.0.0.1:8080/prj/noticeOpen 

但这听起来并不动态，无法满足我的要求。

web.xml

    <?xml version="1.0" encoding="UTF-8"?>
    <web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"     
    xmlns="http://java.sun.com/xml/ns/javaee" xmlns:jsp="http://java.sun.com/xml/ns/javaee/jsp" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" 
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" version="2.5">

   <display-name>Spring Web MVC Application</display-name>

  <servlet>
    <servlet-name>mvc-dispatcher</servlet-name>
        <servlet-class>
               org.springframework.web.servlet.DispatcherServlet
        </servlet-class>
        <load-on-startup>1</load-on-startup>
  </servlet>

  <servlet-mapping>
    <servlet-name>mvc-dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
  </servlet-mapping>

  <context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/applicationContext.xml</param-value>
  </context-param>
  <listener>
        <listener-class>
        org.springframework.web.context.ContextLoaderListener
        </listener-class>
  </listener>   
</web-app>

更新资料

我/noticeOpen/{quesId}为其创建了新的Controller，并且现在获得了该控件，但是我无法理解以下方法的行为。请在NoticeController上查看以下内容，然后得出以下结果：

  @Controller
public class NoticeController {


    @RequestMapping(value="/noticeOpen/{quesId}") 
    public ModelAndView noticeOpen(@ModelAttribute("NoticeForm") NoticeForm noticeForm,ModelMap model,@PathVariable("quesId") Integer quesId){

        return new ModelAndView("noticeOpen","noticeForm",noticeForm);

    }

    @RequestMapping(value="/noticeOpen") 
    public ModelAndView noticeOpen(@ModelAttribute("NoticeForm") NoticeForm noticeForm,ModelMap model){

        return new ModelAndView("noticeOpen","noticeForm",noticeForm);

    }

@RequestMapping(value="/noticeOpen")它将我重定向到正确的noticeOpen.jsp @RequestMapping(value="/noticeOpen/{quesId}")，将我重定向到以下错误页面

HTTP Status 404 - /prj/noticeOpen/WEB-INF/pages/noticeOpen.jsp
type Status report
message /prj/noticeOpen/WEB-INF/pages/noticeOpen.jsp
description The requested resource is not available.
Apache Tomcat/6.0.36

Answer 1

如下更改你的applicationContext.xml中的前缀值

<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix">
            <value>/WEB-INF/pages/</value>
        </property>
        <property name="suffix">
            <value>.jsp</value>
        </property>
</bean>

WEB-INF之前的斜线将起作用。