显示带有 url 模式“/*”的转发 JSP

发布于 2024-10-19 06:53:24 字数 1629 浏览 2 评论 0原文

为了提高我的 java 技能，我正在尝试构建一个简单的 j2ee 框架 (MVC)。

我构建它是为了处理 FrontServlet 中的每个请求。这是我使用的映射：

web.xml :
<servlet>
    <servlet-name>Front</servlet-name>
    <servlet-class>test.FrontServlet</servlet-class>
</servlet>
<servlet-mapping>
    <servlet-name>Front</servlet-name>
    <url-pattern>/*</url-pattern>
</servlet-mapping>

我的问题是，当我将请求从 FrontServlet 转发到 JSP 时，显然，JSP 请求由 FrontServlet 处理，并且视图未呈现。

< strong>如何通过保留 url-pattern“/*”来解决这个问题？
有没有一种方法可以在 Servlet 中呈现 JSP 而不会造成性能损失？

提前感谢您的帮助回复！

解决方案 1 (@Bryan Kyle)

我正在尝试遵循你的建议。我创建了这个过滤器：

public void doFilter(ServletRequest request,
   ServletResponse response, FilterChain chain) 
   throws IOException, ServletException 

   {
       HttpServletRequest req = (HttpServletRequest) request;

       if(!req.getRequestURL().toString().endsWith("jsp"))
       {
           // I changed the servlet url-pattern to "/front.controller"
           req.getRequestDispatcher("/front.controller").forward(req, response);
           /*chain.doFilter(req, resp);*/
       }
   }

<filter>
    <filter-name>Filter</filter-name>
    <filter-class>test.Filter</filter-class>
</filter>

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

是吗？

谢谢！

原文

To improve my java skills, I'm trying to build a simple j2ee framework (MVC).

I built it to handle every request in a FrontServlet. Here is the mapping that I used :

web.xml :
<servlet>
    <servlet-name>Front</servlet-name>
    <servlet-class>test.FrontServlet</servlet-class>
</servlet>
<servlet-mapping>
    <servlet-name>Front</servlet-name>
    <url-pattern>/*</url-pattern>
</servlet-mapping>

My problem is that when I forward the request from the FrontServlet to a JSP, obviously, the JSP request is handle by the FrontServlet and the view isn't rendered.

How can I resolve this problem by keeping the url-pattern "/*" ?
Is there a way to render a JSP in a Servlet without performance losses ?

Thanks in advance for your reply !

Solution 1 (@Bryan Kyle)

I'm trying to follow your advise. I created this filter :

public void doFilter(ServletRequest request,
   ServletResponse response, FilterChain chain) 
   throws IOException, ServletException 

   {
       HttpServletRequest req = (HttpServletRequest) request;

       if(!req.getRequestURL().toString().endsWith("jsp"))
       {
           // I changed the servlet url-pattern to "/front.controller"
           req.getRequestDispatcher("/front.controller").forward(req, response);
           /*chain.doFilter(req, resp);*/
       }
   }

<filter>
    <filter-name>Filter</filter-name>
    <filter-class>test.Filter</filter-class>
</filter>

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

Is it right?

Thanks !

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情话难免假 2024-10-26 06:53:24

对于前端控制器方法来说，Filter 是一个不合适的解决方案。

您想要优化 servlet 的 url-pattern，使其匹配 /pages/* 或 *.do。您不希望前端控制器介入不相关的请求，例如 CSS/JS/图像/等。以/pages/*为例，假设您在/WEB-INF/foo.jsp中有一个JSP，那么servlet中的以下内容

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    request.getRequestDispatcher("/WEB-INF" + request.getPathInfo() + ".jsp").forward(request, response);
}

应该显示： http://localhost:8080/contextname/pages/foo 上有问题的 JSP。

另请参阅：

A Filter is an inappropriate solution for a front controller approach.

You want to refine the url-pattern of your servlet so that it matches e.g. /pages/* or *.do. You don't want your front controller to kick in on irrelevant requests like CSS/JS/images/etc. To take /pages/* as an example, assuming that you've a JSP in /WEB-INF/foo.jsp, then the following in a servlet

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    request.getRequestDispatcher("/WEB-INF" + request.getPathInfo() + ".jsp").forward(request, response);
}

should display the JSP in question on http://localhost:8080/contextname/pages/foo.