java - 如何在 spring mvc 应用程序中打开弹出窗口

Question

我有以下网络应用程序结构:

在addTerminal.jsp里面我写了以下内容:

....
<a href="#" class="open-map" onclick="showMap(55,22)">show map</a>
...
function showMap(lat,lng) {
        window.open('map.html?lat='+lat+'&lng='+lng, 'map', 'width=600,height=400');
    }
....

但是由于客户端无法访问 WEB-INF 文件夹，因此单击 href 时我看到 404

您能否建议如何解决我的问题？

我认为我可以将 map.html 放入 webapp 文件夹中，但我认为存在另一个修复。应用技术栈:

Tomcat + Hibernate + SpringMVC

web.xml:

<web-app id="WebApp_ID" version="2.4"
    xmlns="http://java.sun.com/xml/ns/j2ee" 
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee 
    http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">

    <display-name>Spring MVC Application</display-name>

     <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <listener>
        <listener-class>
            org.springframework.web.context.request.RequestContextListener
        </listener-class>
   </listener>

     <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/webContext.xml</param-value>
    </context-param> 
    <!-- Spring MVC -->
    <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/webContext.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>

    <!-- Spring Security -->
    <filter>
        <filter-name>springSecurityFilterChain</filter-name>
        <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
    </filter>

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

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

</web-app>

Answer 1

您只需创建一个 Controller 来处理对 map.html 的请求。

类似的东西:

@RequestMapping("/map.html")
public String popup(@RequestParam("lat") String lat, @RequestParam("lng") String lng){
    // your code here...
    return "map";
}