Question

我正在使用Spring MVC 3.0上的示例RESTEasy 2.0资源并使用Tomcat 6.我可以通过http：//localhost:8080/examples-resteasy-2.1-SNAPSHOT/contacts访问我的资源但我想通过http：// localhost：8080 / contacts甚至http：// localhost：8080 / myservice / contacts

访问

我的应用程序映射到路径的方式是否需要更改？

Web.xml中

<web-app>

    <servlet>
        <servlet-name>springmvc</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>classpath:springmvc-servlet.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>springmvc</servlet-name>
        <url-pattern>/contacts/*</url-pattern>
    </servlet-mapping>

</web-app>

用SpringMVC-servlet.xml中

<beans xmlns="http: //www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="
        http: //www.springframework.org/schema/context http: //www.springframework.org/schema/context/spring-context-3.0.xsd
        http: //www.springframework.org/schema/beans http: //www.springframework.org/schema/beans/spring-beans.xsd
    ">
    <context:component-scan base-package="org.jboss.resteasy.examples.springmvc" />
    <context:annotation-config />
    <import resource="classpath:springmvc-resteasy.xml" />  <!-- this is included in the resteasy-spring library-->

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

</beans>

我的RESTEasy资源类

@Controller
@Path("/contacts")
public class ContactsResource {
...

Answer 1

您可以在Tomcat server.xml中设置它们。

在<Context>中添加<Host>元素，如下所示，将examples-resteasy-2.1-SNAPSHOT设置为默认网络应用。

<Context docBase="examples-resteasy-2.1-SNAPSHOT" path="" reloadable="true" />

这应该允许您以http：// localhost：8080 / contacts

的形式访问它

将路径设置为“myservice”，如下所示

<Context docBase="examples-resteasy-2.1-SNAPSHOT" path="/myservice" reloadable="true" />

应允许您以http：// localhost：8080 / myservice / contacts

的形式访问它

在tomcat上更改spring mvc app的app root

1 个答案: