无法访问spring mvc中的url

Question

我尝试使用 spring mvc 应用程序访问 url，但出现错误 404： http://localhost:8181/201701241113-spring-pratique/welcome.htm

我有 3 个 xml 配置文件：

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>

court-service.xml

<?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" />

court-servlet.xml

<?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>

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“欢迎页面”

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，有时您需要在将 tomcat 服务器添加到新工作区时检查以下选项。确保在 Server Locations 下，您需要在下面的屏幕截图中选择 Use Tomcat installation（第二个）：