Question

以下情况：我正在使用Spring Roo 1.1（Apache Tiles＆amp; Spring MVC）开发Web应用程序。我想要一个管理部分，通过路径可以访问每个实体（/ admin / users，/ admin / roles，...）

到目前为止一切正常。唯一的问题是，我想在/ admin上有一个静态页面。因为我不想创建一个自己的控制器，我在webmvc-config.xml中添加了：

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

同样在目录WEB-INF / views / admin / views.xml中：

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<!DOCTYPE tiles-definitions PUBLIC "-//Apache Software Foundation//DTD Tiles Configuration 2.1//EN" "http://tiles.apache.org/dtds/tiles-config_2_1.dtd">
<tiles-definitions>
    <definition extends="default" name="admin/index">
        <put-attribute name="body" value="/WEB-INF/views/admin/index.jspx"/>
    </definition>
</tiles-definitions>

我也没有忘记创建index.jspx。

web.xml的内容：

<display-name>reservation</display-name>

<description>Roo generated reservation application</description>


<!-- Enable escaping of form submission contents -->
<context-param>
    <param-name>defaultHtmlEscape</param-name>
    <param-value>true</param-value>
</context-param>

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>classpath*:META-INF/spring/applicationContext*.xml</param-value>
</context-param>

<filter>
    <filter-name>CharacterEncodingFilter</filter-name>
    <filter-class>org.springframework.web.filter.CharacterEncodingFilter</filter-class>
    <init-param>
        <param-name>encoding</param-name>
        <param-value>UTF-8</param-value>
    </init-param>
    <init-param>
        <param-name>forceEncoding</param-name>
        <param-value>true</param-value>
    </init-param>
</filter>



<filter>
    <filter-name>HttpMethodFilter</filter-name>
    <filter-class>org.springframework.web.filter.HiddenHttpMethodFilter</filter-class>
</filter>



<filter>
    <filter-name>springSecurityFilterChain</filter-name>
    <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
</filter>
<filter>
    <filter-name>Spring OpenEntityManagerInViewFilter</filter-name>
    <filter-class>org.springframework.orm.jpa.support.OpenEntityManagerInViewFilter</filter-class>
</filter>
<filter-mapping>
    <filter-name>CharacterEncodingFilter</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>

<filter-mapping>
    <filter-name>HttpMethodFilter</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>



<filter-mapping>
    <filter-name>springSecurityFilterChain</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>
<filter-mapping>
    <filter-name>Spring OpenEntityManagerInViewFilter</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>

<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<!-- Handles Spring requests -->
<servlet>
    <servlet-name>reservation</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/webmvc-config.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

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

<session-config>
    <session-timeout>10</session-timeout>
</session-config>

<error-page>
    <exception-type>java.lang.Exception</exception-type>
    <location>/uncaughtException</location>
</error-page>

<error-page>
    <error-code>404</error-code>
    <location>/resourceNotFound</location>
</error-page>

不幸的是，在请求/ admin时，我收到了资源未找到错误。

有人可以给我一个提示吗？

Answer 1

请确保您的XML配置中有<mvc:annotation-driven />个定义。

Spring MVC静态页面

1 个答案: