gpt4 book ai didi

java - 在名称为 'appServlet' 的 DispatcherServlet 中找不到具有 URI [/myproject/] 的 HTTP 请求的映射

转载 作者:搜寻专家 更新时间:2023-11-01 00:52:22 24 4
gpt4 key购买 nike

<分区>

我是 Java 和 Spring 的新手,我想从示例中学习。

我正在使用开箱即用的配置/安装

  • Mac 操作系统
  • 作为 IDE 的 Springsource 工具套件
  • Spring 2.8.1.RELEASE
  • vfabric-tc-server-developer-2.6.1.RELEASE

我尝试基于“Spring Template Project”生成一个新项目。然后我选择了“Spring MVC 项目”。示例项目已生成。之后,在不修改任何内容的情况下,我尝试通过“运行方式”执行“home.jsp”页面。 Web 服务器启动,最后我在控制台选项卡中收到错误。

No mapping found for HTTP request with URI [/myproject/] in DispatcherServlet with name 'appServlet'

以及这些网页中的其他输出:

  • http://localhost:8080/myproject/WEB-INF/views/home.jsp
  • http://localhost:8080/myproject

enter image description here

在这里您可以看到关于我的项目结构的图像(为 STS 自动生成):

enter image description here

怎么了?

这里可以看到web.xml文件的内容:

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

<!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>

<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<!-- Processes application requests -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>

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

</web-app>

root-context.xml 文件包含此内容。

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
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">

<!-- Root Context: defines shared resources visible to all other web components -->

</beans>

最后 servlet-context.xml 有这个内容。

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
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">

<!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->

<!-- Enables the Spring MVC @Controller programming model -->
<annotation-driven />

<!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />

<!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory -->
<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".jsp" />
</beans:bean>

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

</beans:beans>

有人有解决办法吗?

24 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com