Spring mvc - 无法打开ServletContext资源[/WEB-INF/dispatcher-servlet.xml]

时间：2016-09-28 07:08:02

标签： java xml spring spring-mvc

enter image description here我的调度程序servlet xml被命名为cnc-dispatcher-servlet-xml。但为什么它指向WEB-INF / dispatcher-servlet.xml？我在WEB-INF下有config xml文件。由于这个原因，我无法呼叫控制器。它抛出了以下错误

cnccontroller 

package com.cnc.cnccontroller;

import org.springframework.web.bind.annotation.RequestMapping;   
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.servlet.ModelAndView;

public class CncController {
String message = "Welcome to Spring MVC!";

@RequestMapping("/hello")
public ModelAndView showMessage(
        @RequestParam(value = "name", required = false, defaultValue = "World") String name) {
    System.out.println("in controller");

    ModelAndView mv = new ModelAndView("helloworld");
    mv.addObject("welcomeMessage", "Welcome to first page");
    return mv;
}
}

CNC-调度-servlet.xml中

<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
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-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd">

<bean id="HandlerMapping" class="org.springframework.web.servlet.handler.BeanNameUrlHandlerMapping"></bean>
<context:component-scan base-package="com.cnc.cnccontroller.CncController" />

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


javax.servlet.ServletException: Servlet.init() for servlet dispatcher threw exception
org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:502)
org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:79)
org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:616)
org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:528)
org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1099)
org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:670)
org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.doRun(NioEndpoint.java:1520)
org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.run(NioEndpoint.java:1476)
java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142)
java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)
org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
java.lang.Thread.run(Thread.java:745)

root cause

org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from ServletContext resource [/WEB-INF/dispatcher-servlet.xml]; nested exception is java.io.FileNotFoundException: Could not open ServletContext resource [/WEB-INF/dispatcher-servlet.xml]
org.springframework.beans.factory.xml.XmlBeanDefinitionReader.loadBeanDefinitions(XmlBeanDefinitionReader.java:343)
org.springframework.beans.factory.xml.XmlBeanDefinitionReader.loadBeanDefinitions(XmlBeanDefinitionReader.java:303)
org.springframework.beans.factory.support.AbstractBeanDefinitionReader.loadBeanDefinitions(AbstractBeanDefinitionReader.java:180)
org.springframework.beans.factory.support.AbstractBeanDefinitionReader.loadBeanDefinitions(AbstractBeanDefinitionReader.java:216)
org.springframework.beans.factory.support.AbstractBeanDefinitionReader.loadBeanDefinitions(AbstractBeanDefinitionReader.java:187)
org.springframework.web.context.support.XmlWebApplicationContext.loadBeanDefinitions(XmlWebApplicationContext.java:125)
org.springframework.web.context.support.XmlWebApplicationContext.loadBeanDefinitions(XmlWebApplicationContext.java:94)
org.springframework.context.support.AbstractRefreshableApplicationContext.refreshBeanFactory(AbstractRefreshableApplicationContext.java:129)
org.springframework.context.support.AbstractApplicationContext.obtainFreshBeanFactory(AbstractApplicationContext.java:540)
org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:454)
org.springframework.web.servlet.FrameworkServlet.configureAndRefreshWebApplicationContext(FrameworkServlet.java:658)
org.springframework.web.servlet.FrameworkServlet.createWebApplicationContext(FrameworkServlet.java:624)
org.springframework.web.servlet.FrameworkServlet.createWebApplicationContext(FrameworkServlet.java:672)

我的web.xml是

 <?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"  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>CNC</display-name>
<servlet>
    <servlet-name>dispatcher</servlet-name>
    <servlet-class>
        org.springframework.web.servlet.DispatcherServlet
    </servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>

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

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/cnc-dispatcher-servlet.xml</param-value>
</context-param>

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


</web-app>

2 个答案:

答案 0 :(得分：0)

将您的servlet名称更改为cnc-dispatcher，如下所示。 您的调度程序servlet xml由<servlet-name>查看。

同样在你的contextConfigLocation中应该查找你的上下文xml或spring config xml，你可以在其中定义要加载和初始化的spring bean。 

<?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"  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>CNC</display-name>
<servlet>
    <servlet-name>cnc-dispatcher</servlet-name>
    <servlet-class>
        org.springframework.web.servlet.DispatcherServlet
    </servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>

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

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/<your application context/config xml></param-value>
</context-param>

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


</web-app>

答案 1 :(得分：0)

更改评论以回答..

  

1.您的servlet名称应该是cnc-dispatcher，因为您的文件名是cnc-dispathcer-servlet.xml.，无论您在-servlet之前提到什么，都应该提到servlet名称。

     

2.由于您提到new ModelAndView("helloworld");并且您的调度员 - servlet prefix和suffix只是/WEB-INF/和.jsp，请确保您拥有helloworld /WEB-INF下的.jsp，而不是任何其他子文件夹下（如果是，则更新前缀）

更新1：添加驱动到调度程序servlet的mvc注释。

<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:mvc="http://www.springframework.org/schema/mvc"
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/mvc/spring-mvc.xsd
http://www.springframework.org/schema/context/spring-context-3.0.xsd">

<bean id="HandlerMapping" class="org.springframework.web.servlet.handler.BeanNameUrlHandlerMapping"></bean>
<context:component-scan base-package="package to scan" /> 
 <mvc:annotation-driven/>

更新2 ：在jsp中将{welcomeMessage}更改为${welcomeMessage}。

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