How to get request URI without context path?

问题

The Method request.getRequestURI() returns URI with context path.

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

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

回答1:

If you're inside a front contoller servlet which is mapped on a prefix pattern, then you can just use HttpServletRequest#getPathInfo().

String pathInfo = request.getPathInfo();
// ...

Assuming that the servlet in your example is mapped on /secure, then this will return /users which would be the information of sole interest inside a typical front controller servlet.

If the servlet is however mapped on a suffix pattern (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 getPathInfo() could return null), then your best bet is to substring the request URI yourself based on the context path's length using the usual String method:

HttpServletRequest request = (HttpServletRequest) req;
String path = request.getRequestURI().substring(request.getContextPath().length());
// ...

回答2:

request.getRequestURI().substring(request.getContextPath().length())

回答3:

With Spring you can do:

String path = new UrlPathHelper().getPathWithinApplication(request);

回答4:

getPathInfo() sometimes return null. In documentation HttpServletRequest

This method returns null if there was no extra path information.

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

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());
    ...

}

回答5:

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:

https://issues.apache.org/bugzilla/show_bug.cgi?id=28323

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

回答6:

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

String p = request.getRequestURI();
String cp = getServletContext().getContextPath();

if (p.startsWith(cp)) {
  String.err.println(p.substring(cp.length());
}

Read here .

回答7:

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

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

来源：https://stackoverflow.com/questions/4278083/how-to-get-request-uri-without-context-path

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java

servlets