在调用在spring mvc项目中创建的jsp文件时,我在localhost上收到404错误消息。
我已经在eclipse中创建了动态web项目,并在其中添加了spring jar。创建了控制器并通过它调用jsp文件,但得到404错误消息。服务器使用
web.xml:
http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd“id="WebApp_ID”version="3.0"> FirstMVCProject
<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>
发布于 2019-07-16 02:41:02
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>
https://stackoverflow.com/questions/57044902
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