blocks|key|3222536|text|如果您在一个被映射到一个前缀模式(如/foo/*+)上的front+contoller+servlet中，那么您可以只使用HttpServletRequest#getPathInfo()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3222537|String+pathInfo+=+request.getPathInfo();
//+...|code-block|syntax|javascript|3222538|假设您的示例中的servlet被映射到/secure/*上，那么这将返回/users，这将是典型的前端控制器servlet中唯一感兴趣的信息。|3222539|但是，如果servlet被映射到后缀模式，比如*.foo+(然而，您的URL示例并没有表明是这种情况)，或者当您实际在一个过滤器中时(当需要调用的servlet还没有确定，所以getPathInfo()可以返回null)，那么您最好的选择是使用通常的String方法，根据上下文路径的长度自己子串请求URI：|3222540|HttpServletRequest+request+=+(HttpServletRequest)+req;
String+path+=+request.getRequestURI().substring(request.getContextPath().length());
//+...|3222541|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javaee/6/api/javax/servlet/http/HttpServletRequest.html#getPathInfo%2528%2529^0|I|6|1P|W|1P|W|0|0|0|J|9|10|6|0|N|5|2G|D|2X|4|3H|6|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1C|8|@$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|1L|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

If you're inside a front contoller servlet which is mapped on a prefix pattern such as <code>/foo/*</code>, then you can just use <a href="http://download.oracle.com/javaee/6/api/javax/servlet/http/HttpServletRequest.html#getPathInfo%28%29" rel="noreferrer"><code>HttpServletRequest#getPathInfo()</code></a>.
<pre><code>String pathInfo = request.getPathInfo();
// ...
</code></pre>
Assuming that the servlet in your example is mapped on <code>/secure/*</code>, then this will return <code>/users</code> which would be the information of sole interest inside a typical front controller servlet.
If the servlet is however mapped on a suffix pattern such as <code>*.foo</code> (your URL examples however does not indicate that this is the case), or when you're actually inside a filter (when the to-be-invoked servlet is not necessarily determined yet, so <code>getPathInfo()</code> could return <code>null</code>), then your best bet is to substring the request URI yourself based on the context path's length using the usual <code>String</code> method:
<pre><code>HttpServletRequest request = (HttpServletRequest) req;
String path = request.getRequestURI().substring(request.getContextPath().length());
// ...
</code></pre>

blocks|key|289841|text|request.getRequestURI().substring(request.getContextPath().length())|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|289842|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>request.getRequestURI().substring(request.getContextPath().length())
</code></pre>

blocks|key|288027|text|使用Spring，您可以做到：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|288028|String+path+=+new+UrlPathHelper().getPathWithinApplication(request);|code-block|syntax|javascript|288029|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

With Spring you can do:

<pre><code>String path = new UrlPathHelper().getPathWithinApplication(request);
</code></pre>

blocks|key|3008529|text|getPathInfo()有时返回null。在文档HttpServletRequest中|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3008530|3008531|如果没有额外的路径信息，则此方法返回null。|blockquote|3008532|3008533|3008534|我需要在过滤器中获取不带上下文路径的文件路径，getPathInfo()返回null。所以我使用了另一个方法:+httpRequest.getServletPath()|3008535|public+void+doFilter(ServletRequest+request,+ServletResponse+response,+FilterChain+chain)+throws+IOException,+ServletException
{
++++HttpServletRequest+httpRequest+=+(HttpServletRequest)+request;
++++HttpServletResponse+httpResponse+=+(HttpServletResponse)+response;

++++String+newPath+=+parsePathToFile(httpRequest.getServletPath());
++++...

}|code-block|syntax|javascript|3008536|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javaee/6/api/javax/servlet/http/HttpServletRequest.html#getPathInfo%2528%2529^0|P|I|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|-4|5|6|7|12|8|@]|9|@]|C|$]]|$1|E|3|F|5|G|7|13|8|@]|9|@]|C|$]]|$1|H|3|-4|5|6|7|14|8|@]|9|@]|C|$]]|$1|I|3|-4|5|6|7|15|8|@]|9|@]|C|$]]|$1|J|3|K|5|6|7|16|8|@]|9|@]|C|$]]|$1|L|3|M|5|N|7|17|8|@]|9|@]|C|$O|P]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

getPathInfo() sometimes return null. In documentation <a href="http://download.oracle.com/javaee/6/api/javax/servlet/http/HttpServletRequest.html#getPathInfo%28%29" rel="noreferrer">HttpServletRequest</a>

<blockquote>
 This method returns null if there was no extra path information. 
</blockquote>

I need get path to file without context path in Filter and getPathInfo() return me null. So I use another method: httpRequest.getServletPath()

<pre><code>public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException
{
 HttpServletRequest httpRequest = (HttpServletRequest) request;
 HttpServletResponse httpResponse = (HttpServletResponse) response;

 String newPath = parsePathToFile(httpRequest.getServletPath());
 ...

}
</code></pre>

blocks|key|288020|text|如果在过滤器中使用request.getPathInfo()，似乎总是得到null+(至少对于jetty是这样)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|288021|这个简洁的无效bug+%2B响应暗示了我认为的问题：|288022|https://issues.apache.org/bugzilla/show_bug.cgi?id=28323|offset|length|288023|我怀疑这与过滤器在servlet获得请求之前运行的事实有关。这可能是一个容器错误，或者是我无法识别的预期行为。|288024|但是contextPath是可用的，所以fforws解决方案即使在过滤器中也可以使用。我不喜欢手工完成，但是实现被破坏了或者|288025|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|1K|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@$F|V|G|W|1|X]]|A|$]]|$1|H|3|I|5|6|7|Y|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|Z|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|E]]]]

If you use request.getPathInfo() inside a Filter, you always seem to get null (at least with jetty). 

This terse invalid bug + response alludes to the issue I think:

<a href="https://issues.apache.org/bugzilla/show_bug.cgi?id=28323" rel="noreferrer">https://issues.apache.org/bugzilla/show_bug.cgi?id=28323</a>

I suspect it is related to the fact that filters run before the servlet gets the request. It may be a container bug, or expected behaviour that I haven't been able to identify. 

The contextPath is available though, so fforws solution works even in filters. I don't like having to do it by hand, but the implementation is broken or

blocks|key|3008444|text|要做到这一点，一种方法是从请求URI中获取servelet上下文路径。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3008445|String+p+=+request.getRequestURI();
String+cp+=+getServletContext().getContextPath();

if+(p.startsWith(cp))+{
++String.err.println(p.substring(cp.length());
}|code-block|syntax|javascript|3008446|阅读here。|offset|length|3008447|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javaee/5/api/javax/servlet/ServletContext.html#getContextPath%2528%2529^0|0|0|2|4|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

A way to do this is to rest the servelet context path from request URI.

<pre><code>String p = request.getRequestURI();
String cp = getServletContext().getContextPath();

if (p.startsWith(cp)) {
 String.err.println(p.substring(cp.length());
}
</code></pre>

Read <a href="http://download.oracle.com/javaee/5/api/javax/servlet/ServletContext.html#getContextPath%28%29" rel="noreferrer">here</a> .

blocks|key|293067|text|也许您可以只使用split方法来消除'/myapp‘，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|293068|string[]+uris=request.getRequestURI().split("/");
string+uri="/"%2Buri[1]%2B"/"%2Buris[2];|code-block|syntax|javascript|293069|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

May be you can just use the split method to eliminate the '/myapp'
for example:

<pre><code>string[] uris=request.getRequestURI().split("/");
string uri="/"+uri[1]+"/"+uris[2];
</code></pre>

The Method <a href="http://docs.oracle.com/javaee/6/api/javax/servlet/http/HttpServletRequest.html#getRequestURI%28%29" rel="noreferrer">request.getRequestURI()</a> returns URI with context path. 

For example, if the base URL of an application is <code>http://localhost:8080/myapp/</code> (i.e. the context path is myapp), and I call <code>request.getRequestURI()</code> for <code>http://localhost:8080/myapp/secure/users</code>, it will return <code>/myapp/secure/users</code>. 

Is there any way we can get only this part <code>/secure/users</code>, i.e. the URI without context path?

How to get request URI without context path?

容器

方法返回带有上下文路径的URI。例如，如果应用程序的基URL是http://localhost:8080/myapp/ (即上下文路径是myapp)，并且我为http://localhost:8080/myapp/secure/users调用request.getRequestURI()，它将返回/myapp/secure/users。有没有办法只得到这个部分的/secure/users，也就是没

问如何获取没有上下文路径的请求URI？
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问如何获取没有上下文路径的请求URI？
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