在DispatcherServlet中找不到带有URI [/ myproject /]的HTTP请求的映射,名称为"appServlet"

Question

我绝对是Java和Spring的新手,我想从例子中学习.

我正在使用开箱即用的配置/安装

• Mac OSX
• Springsource工具套件作为IDE
• Spring 2.8.1.RELEASE
• 的vFabric-TC-服务器开发人员2.6.1.RELEASE

我试图基于"Spring Template Project"生成一个新项目.然后我选择了"Spring MVC Project".生成示例项目.之后,在没有修改任何内容的情况下,我尝试通过"Run As"执行de"home.jsp"页面.Web服务器启动,最后我在控制台选项卡中收到错误.

在DispatcherServlet中找不到带有URI [/ myproject /]的HTTP请求的映射,名称为"appServlet"

这些网页中的其他输出:

• http://localhost:8080/myproject/WEB-INF/views/home.jsp
• http://localhost:8080/myproject

在这里,您可以看到有关我的项目结构的图像(为STS自动生成):

怎么了？

在这里,您可以看到web.xml文件的内容:

<?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>

该根context.xml的文件中有这样的内容.

<?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-3.0.xsd">

    <!-- Root Context: defines shared resources visible to all other web components -->

</beans>

最后,servlet-context.xml具有此内容.

<?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="com.mycompany.myapp" />

</beans:beans>

有人有想法解决它吗？

Answer 1

WEB-INF 下的所有内容都无法从外部访问，MVC 框架的要点是在分派到视图之前调用控制器，因此无论如何都不应该直接调用 JSP。

而且您可能没有任何 Spring 控制器映射到根 URL，所以显然， URL 处没有任何内容http://localhost:8080/myproject/。