如何在tomcat中发布java web项目

Question

要在tomcat中发布一个java-web应用程序，我将名为'demo-mvc‘的项目复制到tomcat的web应用程序中。启动tomcat后，我访问了chorme浏览器中的"http://localhost:8080/demo-mvc/xx.jsp“，但它提示”.I requested resource is not available“，并尝试按如下方式编辑server.xml

<Context docBase="D:\apache-tomcat-7.0.57\webapps\demo-mvc" path="/demo-mvc" reloadable="true" source="org.eclipse.jst.jee.server:website"/>

最后还是没有效果，我不知道问题出在哪里。

Answer 1

尝试从项目清理选项中清理项目，并确保映射所有资源

在你的web.xml中，代码应该是这样的：

    <?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>