我对Spring框架和Web应用程序还很陌生，尽管我对Java有经验。当我在本地tomcat服务器上运行网站时，URL为：http://localhost:8080/myApp/

现在，请求映射将我委托给我的主页，其中包括：

@RequestMapping(value = "/", method = RequestMethod.GET)
public String someMethod(Model model) { ... 
   return "index"; }

现在在文件中，index.xhtml我链接到另一个页面，<a href="apps/">link</a> 但是当我想链接回索引页面时，必须使用<a href="../index/">link</a>。我搜索了一个解决方案，发现：

<spring:url value='/apps' var="apps_url" />
<a href="${apps_url}">link</a>

但是spring：url总是解析为http://localhost:8080/myApp/-我当前所在的页面。另外，当我只使用像这样的链接时<a href="/otherSite">link</a>，它总是解析为，http://localhost:8080/otherSite而不是http://localhost:8080/myApp/otherSite我期望的那样。如何获得我的工作链接？是否http://localhost:8080/myApp隐式定义为我的上下文，或者可以/应该将其更改为http://localhost:8080/？

另外，本地tomcat服务器上的URL和Web应用程序发布时将具有的URL之间是否存在任何连接？

这是我的一些应用程序文件：

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"
    xmlns:tx="http://www.springframework.org/schema/tx"
    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/tx http://www.springframework.org/schema/tx/spring-tx-3.1.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 />

    <context:component-scan base-package="myApp" />

    <tx:annotation-driven transaction-manager="transactionManager"/>

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

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

</beans:beans>

来自web.xml的摘录：

<!-- 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>