调用springmvc项目中创建的jsp文件时，在localhost上获取404错误消息

Question

在调用springmvc项目中创建的jsp文件时，我在localhost上收到一条404错误消息

我在eclipse中创建了动态web项目，并在其中添加了springjar。创建了控制器并通过它调用jsp文件，但得到404错误消息。服务器u

web.xml：

http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"id="WebApp_id"version="3.0">FirstMVPProject

<servlet>
<servlet-name>spring-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>spring-dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>

Answer 1

您没有在web.xml:中加载上下文

<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/applicationContext.xml</param-value>
</init-param>

正确的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>example</display-name>
<servlet>
<servlet-name>spring-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/applicationContext.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring-dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>