Question

我有一个主要的spring hello webapp，它在web.xml中包含一个包含web片段的include。我似乎无法访问jar中的静态资源，即使用tomcat 7，servlet 3，Spring 3.2.3.RELEASE

我的jar结构是test.jar-＆gt; META-INF-＆gt; test.css

但是当我加载应用时，网址：

http://localhost:8686/spring-fragment-test/test.css

似乎不起作用，继续获得404

我配置如下：

主应用程序中的Spring.xml

<!-- 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.fragments" />

使用web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xmlns="http://java.sun.com/xml/ns/javaee" 
    xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" 
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" version="3.0">

    <!-- 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>

  <absolute-ordering>   
    <name>SampleFragment</name>
    </absolute-ordering>    

</web-app>

我的其他fragemt jar在meta-inf

中有以下xml

<?xml version="1.0" encoding="UTF-8"?>
<web-fragment id="WebFragment_ID" version="3.0" 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-fragment_3_0.xsd">

    <display-name>SampleFragment</display-name> 
    <name>SampleFragment</name> 

</web-fragment>

Answer 1

因为您将位置“/ resources /”放在xml中

/data/data

您必须将css文件放在

部分

网页内容/资源/（css文件）

您现在可以通过编写

来访问该文件

http://loclahost:8080/spring/resources/main.css

使用Web片段和弹簧时无法加载静态资源

1 个答案: