我见过很多类似的线程,但没有解决我的'问题?'。请告诉我我错过了什么?当我尝试转到http://localhost:8080/helloworld.html时,我会收到“HTTP状态404”。
HelloWorldController.java
@Controller
public class HelloWorldController {
@RequestMapping(value="/helloworld.html")
public ModelAndView helloWorld() {
return new ModelAndView("HelloWorldPage"); }
}
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">
<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>*.html</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/dispatcher-servlet.xml</param-value>
</context-param>
</web-app>
调度-servlet.xml中
<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"
xsi:schemaLocation="
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context.xsd">
<context:component-scan base-package="com.site.controller"/>
<bean
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/pages/" />
<property name="suffix" value=".jsp"/>
</bean>
</beans>
文件结构: https://s18.postimg.org/4j699vn6x/stack.png
的pom.xml
...
<dependencies>
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-context</artifactId>
<version>3.1.0.RELEASE</version>
</dependency>
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-webmvc</artifactId>
<version>3.1.0.RELEASE</version>
</dependency>
</dependencies>
...
答案 0 :(得分:0)
这是您应该如何配置应用程序:
在您的web.xml上,您应该将调度程序servlet简化为:
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/dispatcher-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
控制器应如下所示:
@Controller
public class HelloWorldController {
@RequestMapping(value="/")
public ModelAndView helloWorld() {
return new ModelAndView("HelloWorldPage"); }
}
现在,根据你的形象我猜测你的tomcat上下文是untitled7
所以在tomcat启动之后&amp;正在尝试点击下一个网址: