在tomcat上更改Spring MVC应用程序的应用程序根 [英] changing app root for spring mvc app on tomcat

问题描述

我正在Spring MVC 3.0上使用示例RESTEasy 2.0资源并依赖于Tomcat6.我可以通过http://localhost:8080/examples-resteasy-2.1-SNAPSHOT/contacts来获取资源，但是我想通过http://localhost:8080/contacts甚至http://localhost:8080/myservice/contacts来访问

I am working with a sample RESTEasy 2.0 resource on Spring MVC 3.0 and deplying to Tomcat 6. I can get to my resource through http: //localhost:8080/examples-resteasy-2.1-SNAPSHOT/contacts but I would like to access through http: //localhost:8080/contacts or even http: //localhost:8080/myservice/contacts

我的应用程序映射到路径的方式是否需要更改?

Is there something I need to change in the way my application is mapped to the path?

Web.xml

<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>

springmvc-servlet.xml

<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资源类别

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

推荐答案

您可以在Tomcat server.xml中进行设置.

You can set these in your Tomcat server.xml.

在<Host>内添加一个<Context>元素，如下所示，该元素将您的examples-resteasy-2.1-SNAPSHOT设置为默认Web应用.

Add a <Context> element within the <Host> like below which sets your examples-resteasy-2.1-SNAPSHOT as the default web app.

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

这应该允许您以http://localhost:8080/contacts

This should allow you to access it as http: //localhost:8080/contacts

将路径设置为"myservice"，如下所示

Set the path to "myservice" like below

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

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

should allow you to access it as http: //localhost:8080/myservice/contacts

这篇关于在tomcat上更改Spring MVC应用程序的应用程序根的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！