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我在 Tomcat 中使用 Spring MVC，我认为这是标准配置。web.xml 如下所示：

<servlet>
<servlet-name>acme</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>
        /WEB-INF/spring/appServlet/acme-spring.xml
    </param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>acme</servlet-name>
    <url-pattern>/</url-pattern>
</servlet-mapping>

使用 Spring acme-spring.xml，我将 view-controller 标记设置为根路径：

<mvc:view-controller path="/" view-name="login.jsp"/

和站点解析器标签：

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

但是，当我点击 localhost:8080/acme 时，我会期待 login.jsp，但相反，我收到了 Invalid URL 错误。我查看了Spring 3.1 MVC 应用程序上的 HTTP 状态 404和Spring 3.1 MVC 应用程序上的 HTTP 状态 404，但没有任何乐趣。

关于我配置错误的想法？

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2 回答 2

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你的标签应该是这样的

<mvc:view-controller  path="/" view-name="login"/>

因为 veiwResolver 会处理前缀和后缀。

于 2012-05-06T21:00:35.480 回答
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if you are accessing it by using just context path , then you should either mention your welcome-file list in your web-xml or you should direcly access any welcome or jsp which you want to show on start-up. 1.First add below lines or hit localhost:8080/acme/Login.jsp instead of hitting localhost:8080/acme 

<welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
</welcome-file-list>

2. Add servlet name mapping like :

<servlet>
    <servlet-name>app_name</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>

  1. Add servlet URL mapping :

    app_name *.do //you can write url pattern as per your requir.

check and let me know if any other exception you are facing..!!

于 2012-05-07T12:01:12.237 回答