在spring mvc中无法访问url

Question

我尝试使用spring mvc应用程序访问一个url，但是我收到错误404： 的 http://localhost:8181/201701241113-spring-pratique/welcome.htm

我有3个xml配置文件：

的applicationContext.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee 
    http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    id="WebApp_ID" version="2.5">

    </web-app>

法院service.xml中

<?xml version="1.0" encoding="UTF-8"?>

<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://www.springframework.org/schema/beans
    http://www.springframework.org/schema/beans/spring-beans-2.5.xsd">

    <bean id="reservationService" class="com.cozla.service.ReservationServiceImpl" />

法院servlet.xml中

<?xml version="1.0" encoding="UTF-8"?>

<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd">

    <bean name="/welcome.htm" class="com.cozla.web.WelcomeController" />

    <bean name="/reservationQuery.htm" class="com.cozla.web.ReservationQueryController">
        <property name="reservationService" ref="reservationService" />
    </bean>

    <bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix" value="/WEB-INF/jsp/" />
        <property name="suffix" value=".jsp" />
    </bean>

</beans>

的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee 
    http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    id="WebApp_ID" version="2.5">

    <display-name>Court Reservation System</display-name>
    <listener>
        <listener-class>
            org.springframework.web.context.ContextLoaderListener
        </listener-class>
    </listener>
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
            /WEB-INF/court-servlet.xml;/WEB-INF/court-service.xml
        </param-value>
    </context-param>

    <servlet>
        <servlet-name>201701241113-spring-pratique</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>201701241113-spring-pratique</servlet-name>
        <url-pattern>*.htm</url-pattern>
    </servlet-mapping>

</web-app>

我可以显示＆＃34;欢迎页面＆＃34;

控制器代码

package com.cozla.web;

import java.util.Date;

import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.AbstractController;
import org.springframework.web.servlet.mvc.Controller;

public class WelcomeController extends AbstractController {

    @Override
    protected ModelAndView handleRequestInternal(HttpServletRequest arg0, HttpServletResponse arg1) throws Exception {

        Date today = new Date();
        return new ModelAndView("welcome", "today", today);
    }
}

Answer 1

如果一切正常，那么在eclipse中有时需要在将tomcat服务器添加到新工作区时检查以下选项。确保在服务器位置下，您需要在下面的屏幕截图中选择使用Tomcat安装（第二个）：