1

在 STS 中运行我的项目会在 Tomcat 上给出错误：

Mai 19, 2016 4:44:52 PM org.springframework.web.servlet.PageNotFound noHandlerFound
WARNUNG: No mapping found for HTTP request with URI [/CloudService/] in DispatcherServlet with name 'HelloWeb'
Mai 19, 2016 4:45:35 PM org.springframework.web.servlet.PageNotFound noHandlerFound
WARNUNG: No mapping found for HTTP request with URI [/CloudService/HelloWeb/hello] in DispatcherServlet with name 'HelloWeb'
Mai 19, 2016 4:48:13 PM org.springframework.web.servlet.PageNotFound noHandlerFound
WARNUNG: No mapping found for HTTP request with URI [/CloudService/] in DispatcherServlet with name 'HelloWeb'

我已经检查了几乎所有关于 stackoverflow 的相关文章，但我无法解决我的问题。

src/main/java/HelloController.java

package com.stackoverflow;

import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.ui.ModelMap;

@Controller
@RequestMapping("/hello")
public class HelloController{

   @RequestMapping(method = RequestMethod.GET)
   public String printHello(ModelMap model) {
      model.addAttribute("message", "Hello Spring MVC Framework!");

      return "hello";
   }

}

WebContent/WEB-INF/HelloWeb-servlet.xml

<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">

   <context:component-scan base-package="com.stackoverflow" />

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

</beans>

网页内容/WEB-INF/web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" id="WebApp_ID" version="3.1">
  <display-name>CloudService</display-name>

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

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


</web-app>

网页内容/WEB-INF/jsp/hello.jsp

<%@ page contentType="text/html; charset=UTF-8" %>
<html>
<head>
<title>Hello World</title>
</head>
<body>
   <h2>${message}</h2>
</body>
</html>
4

2 回答 2

1

您的配置看起来不错。但是，您正在请求未映射到任何控制器处理程序方法的 /CloudService/ url。你的控制器有一个 /CloudService/hello 的处理程序。您可以为 / 包含一个处理程序，或者如果 / 和 /hello 都应该由相同的方法处理，您可以这样做

@Controller
@RequestMapping({"/hello" , "/"})
public class HelloController{

   @RequestMapping(method = RequestMethod.GET)
   public String printHello(ModelMap model) {
      model.addAttribute("message", "Hello Spring MVC Framework!");

      return "hello";
   }

}

注意您的应用程序有一个应用程序上下文 HelloWeb-servlet.xml，因此您不需要 ContextLoaderListener 配置来创建父上下文。但是，一旦您有必须将服务 bean 分离到不同的上下文中，您就需要对其进行配置

于 2016-05-19T16:21:27.250 回答
0

这是一个示例 web.xml

    <?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee
  http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
         version="3.1">

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

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

    <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>

    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>

</web-app>

这是 dispatcher-servlet.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"
       xmlns:mvc="http://www.springframework.org/schema/mvc"
       xmlns:util="http://www.springframework.org/schema/util"
       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
   http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc-3.1.xsd
http://www.springframework.org/schema/util
http://www.springframework.org/schema/util/spring-util-3.0.xsd
   ">

    <context:component-scan base-package="controllers"/>
    <mvc:annotation-driven/>

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

    <bean name="shape" class="shapes.Shape">
        <constructor-arg index="0" value ="12" />
        <constructor-arg index="1" value ="2" />
    </bean>
    <import resource="spring-dao.xml"/>
 </beans>

我相信这也应该对你有用。

于 2016-05-19T16:05:27.860 回答