Tomcat、web.xml、排除路径

发布于 2024-12-05 13:33:09 字数 681 浏览 0 评论 0原文

在我的应用程序中，我需要在单个 servlet 中处理来自用户的所有请求，但是，我碰巧有一个包含静态内容的文件夹，我也希望静态提供该文件夹。

在我的 web.xml 文件中，我有以下内容：

    <servlet>
        <servlet-name>all</servlet-name>
        <servlet-class>a.b.c.WidgetlistXml</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>all</servlet-name>
        <url-pattern>/*</url-pattern>
    </servlet-mapping>

有没有办法可以排除所有 .zip 文件？

附言。我知道以前在 StackOverflow 上有人问过这个问题，但最新的帖子是 2006 年左右的，而且它们也涉及 Spring 或其他框架。我没有使用任何内容，而且自 2006 年以来，url 模式可能发生了一些变化。顺便说一句，在 web.xml 上查找文档也不是那么简单。

先感谢您。

原文

In my application, i need to process all the requests from users in a single servlet, however, i happen to have a folder with static content, which i would like to also serve statically.

In my web.xml file, i have the following:

    <servlet>
        <servlet-name>all</servlet-name>
        <servlet-class>a.b.c.WidgetlistXml</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>all</servlet-name>
        <url-pattern>/*</url-pattern>
    </servlet-mapping>

Is there a way i can exclude, say, all .zip files from this?

PS. I know this has been asked on StackOverflow before, but the latest posts were from about 2006, and they also concerned Spring or other frameworks. I use none, and since 2006 something could have changed in url-patterns. By the way, finding the docs on web.xml is not that simple either.

Thank you in advance.

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