Question

我是Spring-MVC的新手并且正在学习它。我正在使用Spring-MVC尝试一个简单的hello-world程序，我的目标（任务）是将servlet-name-servlet.xml文件名更改为spring.xml（用户定义的名称）我已经尝试使用init-param标记，通过在param-value标记中指定路径，它的工作正常！ 但是，我关心的是使用context-param和listener标签将文件名从servlet-name-servlet.xml文件名更改为spring.xml。

我正在接受R.E - IOException从ServletContext资源[/WEB-INF/hello-servlet.xml]解析XML文档;嵌套异常是java.io.FileNotFoundException：无法打开ServletContext资源[/WEB-INF/hello-servlet.xml]

的index.jsp

<html>
<body>
<h2>Hello World!</h2>
<form action="./hello.htm">
<pre>
Enter Name: <input type="text" name="name">
<input type="submit" name="submit">
</pre>
</form>
</body>
</html>

的success.jsp

<%@page isELIgnored="false" %>
${msg}

HelloController.java

public class HelloController  extends AbstractController{

    @Override
    protected ModelAndView handleRequestInternal(HttpServletRequest req, HttpServletResponse resp) throws Exception {
        String name = req.getParameter("name");
        Map m = new HashMap();
        m.put("msg","Hello..."+ name);
        ModelAndView mav = new ModelAndView("success",m);
        return mav;
    }

}

spring.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-3.0.xsd
   http://www.springframework.org/schema/context
   http://www.springframework.org/schema/context/spring-context-3.0.xsd">

<!-- Handler -->
<bean class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<props>
<prop key="/hello.htm">h</prop>
</props>
</property>
</bean>

<!-- Controller -->
<bean id="h" class="com.controller.HelloController">
</bean>

<!-- ViewResolver -->
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/output/"></property>
<property name="suffix" value=".jsp"></property>
</bean>
</beans>

的web.xml

<!DOCTYPE web-app PUBLIC
 "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
 "http://java.sun.com/dtd/web-app_2_3.dtd" >

<web-app>
    <display-name>Archetype Created Web Application</display-name>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring.xml</param-value>
    </context-param>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <servlet>
        <servlet-name>hello</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
         <!-- using init-param the code is working fine -->
          <!--<init-param> 
              <param-name>contextConfigLocation</param-name> 
              <param-value>/WEB-INF/spring.xml</param-value> 
            </init-param> -->
    </servlet>

    <servlet-mapping>
        <servlet-name>hello</servlet-name>
        <url-pattern>*.htm</url-pattern>
    </servlet-mapping>

</web-app>

我已经使用maven依赖项添加了所有必需的jar。我可以使用context-param和listener标签将预定义的文件名（servlet-name-servlet.xml）更改为用户定义的文件名（spring.xml）吗？我的代码出了什么问题？

使用context-param和listener标记将文件名从servlet-name-servlet.xml文件名更改为spring.xml。

0 个答案: