gpt4 book ai didi

spring - 如何在 Spring MVC 和 Tomcat 中配置映射

转载 作者:行者123 更新时间:2023-11-28 22:22:52 25 4
gpt4 key购买 nike

我正在尝试配置 SpringMVC 项目,但在编译时出现错误

WARN : org.springframework.web.servlet.PageNotFound - No mapping found for HTTP request with URI [/test/home.jsp] in DispatcherServlet with name 'appServlet'

我这里有一个jsp页面:/test/src/main/webapp/WEB-INF/views/home.jsp

这是我的代码:/test/src/main/webapp/WEB-INF/web.xml

<?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">
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/spring/root-context.xml
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<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>
<filter>
<filter-name>charsetFilter</filter-name>
<filter-class>org.springframework.web.filter.CharacterEncodingFilter</filter-class>
<init-param>
<param-name>encoding</param-name>
<param-value>UTF-8</param-value>
</init-param>
<init-param>
<param-name>forceEncoding</param-name>
<param-value>true</param-value>
</init-param>
</filter>
<filter-mapping>
<filter-name>charsetFilter</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
</web-app>

/test/src/main/webapp/WEB-INF/spring/appServlet/servlet-context.xml

<?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"
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">
<annotation-driven />
<resources mapping="/resources/**" location="/resources/" />
<beans:bean
name="test"
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".jsp" />
</beans:bean>
<beans:import resource="controllers.xml" />
</beans:beans>

最佳答案

不要使用 jsp 扩展直接将其映射为 html 例如

<display-name>appServlet</display-name>
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>*.html</url-pattern>
</servlet-mapping>

然后在没有扩展名的 Controller 映射文件中:

@RequestMapping(value="/home.html", method=RequestMethod.GET)
public ModelAndView indexView(){
ModelAndView mv = new ModelAndView("test/home");
return mv;
}

在此示例中,只要用户点击/home.html,Dispatcher servlet 就会解析它并检索/test/home.jsp

希望对您有所帮助。

关于spring - 如何在 Spring MVC 和 Tomcat 中配置映射,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7206504/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com