为什么“SpringSource Tool Suite”中的模板“Spring MVC Project”不能与Tomcat一起使用？

Question

当我在SpringSource Tool中创建Spring MVC模板项目并尝试在Tomcat服务器上运行时出现此错误：

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

这是默认值：/test/src/main/java/ru/test/test/HomeController.java

@Controller
public class HomeController {

    private static final Logger logger = LoggerFactory
            .getLogger(HomeController.class);

    /**
     * Simply selects the home view to render by returning its name.
     */
    @RequestMapping(value = "/", method = RequestMethod.GET)
    public String home(Locale locale, Model model) {
        logger.info("Welcome home! the client locale is " + locale.toString());

        Date date = new Date();
        DateFormat dateFormat = DateFormat.getDateTimeInstance(DateFormat.LONG,
                DateFormat.LONG, locale);

        String formattedDate = dateFormat.format(date);

        model.addAttribute("serverTime", formattedDate);

        return "home";
    }

}

这是默认值：/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">

    <!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/root-context.xml</param-value>
    </context-param>

    <!-- Creates the Spring Container shared by all Servlets and Filters -->
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <!-- Processes application requests -->
    <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>

</web-app>

这是默认的/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"
    xmlns:context="http://www.springframework.org/schema/context"
    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
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd">

    <!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->

    <!-- Enables the Spring MVC @Controller programming model -->
    <annotation-driven />

    <!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
    <resources mapping="/resources/**" location="/resources/" />

    <!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory -->
    <beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <beans:property name="prefix" value="/WEB-INF/views/" />
        <beans:property name="suffix" value=".jsp" />
    </beans:bean>

    <context:component-scan base-package="ru.test.test" />



</beans:beans>

默认/test/src/main/webapp/WEB-INF/spring/root-context.xml为空

Answer 1

我得到了完全相同的问题，我解决了它。基本上，当创建Spring MVC项目时，默认情况下Eclipse不会将src / main / webapp配置为源目录。它可能需要作为源目录，因为Eclipse会在构建文件时对文件进行不同的处理。

右键单击'webapp'文件夹并单击'Build Path - ＆gt;用作源文件夹'为我解决了这个问题。这里的其他注释是错误的：我没有必要更改我的RequestMapping或servlet url-pattern。两个'/'都适用于我的'localhost：8080 / test /'。

Answer 2

在我看来，你试图点击的网址是“/ test /”，而web.xml只是映射“/”。如果你想让spring处理所有网址，你可以将其更改为“/ *”，然后你还必须将你的家庭控制器更改为“/ test”。

或者您可以点击网址“http：// localhost：8080 /”，这是您已映射到家庭控制器的根网址。

Answer 3

在Eclipse中创建Spring模板MVC项目之后，必须手动构建它，然后再在服务器上运行它。

Answer 4

找到解决方案！

当您开始创建项目时，您必须定义它。 （注意第三级 superapp ）

为了访问应用程序，网址是 http://localhost:8080/的 superapp

这对我有用