How to get the full url from HttpServletRequest?

Question

Ok, i have these codes:

HttpServletRequest httpRequest = (HttpServletRequest) request;
String requestQueryString = httpRequest.getQueryString();
System.out.println(requestQueryString);

When open the url "http://127.0.0.1:8888/MyProject.html?gwt.codesvr=127.0.0.1:9997?_escaped_fragment_=home",

it printed out:

 gwt.codesvr=127.0.0.1:9997?_escaped_fragment_=home

That url is missing the http://127.0.0.1:8888/MyProject.html? part.

How to fix it?

it doesn't have getFullURL, & we have to construct separate part, but I looking for 1 line of code that can do the job — Tum, May 16 '14 at 05:01
Check this post this question apppears to be a duplicate ,.. http://stackoverflow.com/questions/2222238/httpservletrequest-to-complete-url — Kenneth Clark, May 16 '14 at 05:01

score 5 · Accepted Answer · answered May 16 '14 at 05:01

5

It would be

httpRequest.getRequestURL() + "?" + httpRequest.getQueryString()

prints

http://127.0.0.1:8888/MyProject.html?gwt.codesvr=127.0.0.1:9997?_escaped_fragment_=home

answered May 16 '14 at 05:01

Alex Pakka

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if the url is "xcd.com#!br" then will your code work ok? – Tum May 16 '14 at 05:02
@tum The `#...` is not something that should be sent in a request. It's for use in the client only. – Sotirios Delimanolis May 16 '14 at 05:04
Fragments are not sent in HTTP headers: http://blog.httpwatch.com/2011/03/01/6-things-you-should-know-about-fragment-urls/ – Alex Pakka May 16 '14 at 05:04

How to get the full url from HttpServletRequest?

1 Answers1