web.xml中
<?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>springsecuritydemo</display-name>
<!-- <welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list> -->
<servlet>
<description></description>
<display-name>offers</display-name>
<servlet-name>offers</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>offers</servlet-name>
<url-pattern>/DispatcherServlet</url-pattern>
</servlet-mapping>
</web-app>
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报价-sevlet.xml
<?xml version="1.0" encoding="UTF-8"?>
<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"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation = "http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-4.3.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-4.3.xsd
http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc.xsd">
<context:component-scan base-package="com.spring.security.web"></context:component-scan>
<mvc:annotation-driven></mvc:annotation-driven>
<bean name="jspViewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/jsps/"></property>
<property name="suffix" value=".jsp"></property>
</bean>
</beans>
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这有什么不对?我无法访问home.jsp.我实际上是在春季3.0观看教程,我已经完成了视频中的显示.谁能在这里指出我的错误?
sag*_*mbu 10
问题出在servlet-mapping的url模式中.
<url-pattern>/DispatcherServlet</url-pattern>
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让我们说我们的控制器是
@Controller
public class HomeController {
@RequestMapping("/home")
public String home(){
return "home";
}
}
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当我们在浏览器上点击某个URL时.调度程序servlet将尝试映射此URL.
我们的serlvet目前的url模式/Dispatcher意味着资源的供应{contextpath}/Dispatcher
但是当我们要求时,我们http://localhost:8080/home实际上要求的资源/不可用.所以要么我们需要说调度员servlet /来做服务
<url-pattern>/</url-pattern>
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我们将它从/ Dispatcher服务 /Dispatcher/*
例如
<?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>springsecuritydemo</display-name>
<servlet>
<description></description>
<display-name>offers</display-name>
<servlet-name>offers</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>offers</servlet-name>
<url-pattern>/Dispatcher/*</url-pattern>
</servlet-mapping>
</web-app>
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并请求http://localhost:8080/Dispatcher/home
或只是/请求喜欢
http://localhost:8080/home
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