Java JSP / Servlet：控制器servlet抛出着名的堆栈溢出 [英] Java JSP/Servlet: controller servlet throwing the famous stack overflow

问题描述

我已经阅读了几篇文档但我没理解：我知道我做错了但我不明白。我有一个完全动态生成的网站：几乎没有任何静态内容。

I've read several docs and I don't get it: I know I'm doing something wrong but I don't understand what. I've got a website that is entirely dynamically generated: there's hardly any static content at all.

所以，试着理解JSP / Servlet，我写了自己的前端控制器拦截每一个查询，它看起来像这样：

So, trying to understand JSP/Servlet, I've written my own "front controller" intercepting every single query, it looks like this:

<servlet-mapping>
        <servlet-name>defaultservlet</servlet-name>
        <url-pattern>/*</url-pattern>
</servlet-mapping>

基本上我想要任何用户请求，例如：

Basically I want any user request, like:

• example.org


• example.org/bar


• example.org/foo.html

所有人都通过我编写的默认servlet。

to all go through a default servlet which I've written.

servlet然后检查URI并找到必须分派的 .jsp 请求，然后在正确设置所有属性之后，a：

The servlet then examines the URI and find to which .jsp the request must be dispatched, and then does, after having set all the attributes correctly, a:

RequestDispatcher dispatcher = getServletContext().getRequestDispatcher("/WEB-INF/jsp/index.jsp");
dispatcher.forward(req, resp);

当我使用url-pattern时（在 web.xml 中）比如说， * .html ，一切正常。但当我将其更改为 / * （为了真正拦截所有内容）时，我进入一个无限循环，最终得到一个... StackOverflow：）

When I'm using a url-pattern (in web.xml) like, say, *.html, everything works fine. But when I change it to /* (to really intercept everything), I enter an endless loop and it ends up with a... StackOverflow :)

当调度请求时，URI ... / WEB-INF / jsp / index.jsp本身是否与 web匹配。我设置的xml 过滤器/ *？

When the request is dispatched, is the URI ".../WEB-INF/jsp/index.jsp" itself matched by the web.xml filter /* that I set?

编辑显然，不，因为这是索引的精确映射.jsp 因此它绕过web.xml url-pattern。所以我仍然无法理解如何进入无限循环。

EDIT apparently, no, because this is an exact mapping to index.jsp and hence it bypasses the web.xml url-pattern. So I still don't get how I can enter that endless loop.

如果我想使用/ * url-pattern拦截所有内容，我该怎么办？能够派遣/转发/？

How should I do if I want to intercept everything using a /* url-pattern and yet be able to dispatch/forward/?

我不是在这里询问规格/ Javadocs：我对大局感到困惑，我需要解释一下可能发生的事情。

I'm not asking about specs/Javadocs here: I'm really confused about the bigger picture and I'd need some explanation as to what could be going on.

我不应该拦截真的一切吗？

Am I not supposed to intercept really everything?

如果我可以拦截所有内容，我应该注意哪些转发/调度？

If I can intercept everything, what should I be aware of regarding forwarding/dispatching?

推荐答案

不幸的是，Serlvet规范不允许创建servlet映射以仅匹配传入请求，而不是转发。但是，这可以用于过滤器映射（默认情况下，过滤器映射仅匹配传入请求）。

Unfortunately, Serlvet spec doesn't allow to create a servlet mapping to match only incoming request, not forwards. However, this can be done for filter mappings (and by default filter mappings match only incoming requests).

因此，使用单个servlet拦截所有内容的典型解决方案是使用 UrlRewriteFilter ：

So, the typical solution for intercepting everything with a single servlet is to use a UrlRewriteFilter:

<filter>
    <filter-name>urlRewrite</filter-name>
    <filter-class>org.tuckey.web.filters.urlrewrite.UrlRewriteFilter</filter-class>
</filter>

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

<servlet>
    <servlet-name>application</servlet-name>
    <servlet-class>...</servlet-class>
</servlet>

<servlet-mapping>
    <servlet-name>application</servlet-name>
    <url-pattern>/app/*</url-pattern>
</servlet-mapping>

/WEB-INF/urlrewrite.xml ：

<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE urlrewrite
    PUBLIC "-//tuckey.org//DTD UrlRewrite 3.0//EN"
    "http://tuckey.org/res/dtds/urlrewrite3.0.dtd">

<urlrewrite default-match-type="wildcard">
    <rule>
        <from>/**</from>
        <to>/app/$1</to>
    </rule>
    <outbound-rule>
        <from>/app/**</from>
        <to>/$1</to>
    </outbound-rule>    
</urlrewrite>

这种方式还允许您指定 / * 静态文件的映射。

This way also allows you to specify exceptions from /* mapping for static files.

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