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Posted to httpclient-users@hc.apache.org by MaGGE <ma...@magge.no> on 2008/01/06 13:50:15 UTC
Redirected - but where? HttpClient 4
Hello,
I've just started using HttpClient for a little util I'm making. I'm letting
it handle_redirects=true, but I need to process the URL it's been redirected
to. How can I find the "current URL" after I've executed a request that was
redirected?
I've tried a couple of things - which I feel is "wrong". First thing was to
disable handle_redirects, and look for the "Location" header in a loop of my
own. Challenge with this was that I could not see any way to replace the URI
for the request object, so I'd have to make a new one - which is a problem
because my method should be able to process any HttpUriRequest.
Second I tried to add a HttpRequestInterceptor to the client. But all I can
get from the HttpRequest object given to the process method is the
RequestLine. That one only has the info found behind GET/POST etc, that is
only the path and method. I'm still lacking the host, protocol and so on.
I'm sure there's an easy way to do this that I've just missed ... Please
advice. :)
Thanks for your time,
Magnus
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Re: Redirected - but where? HttpClient 4
Posted by MaGGE <ma...@magge.no>.
Aha! ...another neat way of doing it.
:)
Thanks,
Magnus
olegk wrote:
>
>
> On Sun, 2008-01-06 at 04:50 -0800, MaGGE wrote:
>> Hello,
>>
>> I've just started using HttpClient for a little util I'm making. I'm
>> letting
>> it handle_redirects=true, but I need to process the URL it's been
>> redirected
>> to. How can I find the "current URL" after I've executed a request that
>> was
>> redirected?
>>
>> I've tried a couple of things - which I feel is "wrong". First thing was
>> to
>> disable handle_redirects, and look for the "Location" header in a loop of
>> my
>> own. Challenge with this was that I could not see any way to replace the
>> URI
>> for the request object, so I'd have to make a new one - which is a
>> problem
>> because my method should be able to process any HttpUriRequest.
>>
>> Second I tried to add a HttpRequestInterceptor to the client. But all I
>> can
>> get from the HttpRequest object given to the process method is the
>> RequestLine. That one only has the info found behind GET/POST etc, that
>> is
>> only the path and method. I'm still lacking the host, protocol and so on.
>>
>> I'm sure there's an easy way to do this that I've just missed ... Please
>> advice. :)
>>
>> Thanks for your time,
>> Magnus
>
> Hi Magnus
>
> In addition to getting HttpHost out of the execution context, you may
> also
>
> (1) cast HttpRequest to HttpUriRequest in order to get the full request
> URI
>
> (2) extend or replace the default RedirectHandler with a custom one and
> implement whatever redirect handling logic you deem appropriate
>
> Hope this helps
>
> Oleg
>
>
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> To unsubscribe, e-mail: httpclient-users-unsubscribe@hc.apache.org
> For additional commands, e-mail: httpclient-users-help@hc.apache.org
>
>
>
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Re: Redirected - but where? HttpClient 4
Posted by Oleg Kalnichevski <ol...@apache.org>.
DrewC wrote:
> Ok, I guess I just needed to keep banging on this for another hour. I am now
> using a non-default context with the execute and then reconstructing the
> final require URL with:
>
> HttpUriRequest finalRequest =
> (HttpUriRequest) context.getAttribute(ExecutionContext.HTTP_REQUEST);
> HttpHost host = (HttpHost)
> context.getAttribute(ExecutionContext.HTTP_TARGET_HOST);
>
> StringBuilder uriString = new StringBuilder();
> uriString.append(host.getSchemeName() + "://");
> uriString.append(host.getHostName());
> if (host.getPort() != -1) {
> uriString.append(":" + host.getPort());
> }
> uriString.append(finalRequest.getURI().normalize().toString());
>
> If anyone could confirm this is the right method, that would be appreciated.
>
Yes, it is. HttpClient does not mutate the content of the original
request. Rather it creates a wrapper around it which does get changed in
the course of the request execution. The only way to get hold of that
object is by retrieving it from the execution context.
Another way of keeping track of all redirects is by providing a custom
redirect handler.
Hope this helps
Oleg
> Regards
> Drew
>
>
> DrewC wrote:
>> Hi, I am trying to do basically exactly what Magnus was attempting, to
>> access the final full URL to the resource after following
>> redirects. But I have had no success attempting the techniques seen here,
>> though I admit to being a bit confused by them. I have not had to deal
>> with the HttpContext previously and I am not exactly sure which
>> HttpGetMethod Olek suggests to cast. But with my simplistic approach,
>> after the request is executed (with httpClient.execute(httpGet) and I have
>> confirmed several redirects already occured and were followed, the
>>
>> HttpGet.getURI
>> (HttpUriRequest)httpGet).getURI()
>> (HttpUriRequest)httpGet).getRequestLine().getUri()
>>
>> all still retains the intial URL, not the final one. I am using a shared
>> HttpClient (created via ThreadSafeClientConnManager), not sure if that
>> makes any difference.
>>
>> Any help appreciated.
>>
>> Drew
>>
>>
>>
>> olegk wrote:
>>>
>>> On Sun, 2008-01-06 at 04:50 -0800, MaGGE wrote:
>>>> Hello,
>>>>
>>>> I've just started using HttpClient for a little util I'm making. I'm
>>>> letting
>>>> it handle_redirects=true, but I need to process the URL it's been
>>>> redirected
>>>> to. How can I find the "current URL" after I've executed a request that
>>>> was
>>>> redirected?
>>>>
>>>> I've tried a couple of things - which I feel is "wrong". First thing was
>>>> to
>>>> disable handle_redirects, and look for the "Location" header in a loop
>>>> of my
>>>> own. Challenge with this was that I could not see any way to replace the
>>>> URI
>>>> for the request object, so I'd have to make a new one - which is a
>>>> problem
>>>> because my method should be able to process any HttpUriRequest.
>>>>
>>>> Second I tried to add a HttpRequestInterceptor to the client. But all I
>>>> can
>>>> get from the HttpRequest object given to the process method is the
>>>> RequestLine. That one only has the info found behind GET/POST etc, that
>>>> is
>>>> only the path and method. I'm still lacking the host, protocol and so
>>>> on.
>>>>
>>>> I'm sure there's an easy way to do this that I've just missed ... Please
>>>> advice. :)
>>>>
>>>> Thanks for your time,
>>>> Magnus
>>> Hi Magnus
>>>
>>> In addition to getting HttpHost out of the execution context, you may
>>> also
>>>
>>> (1) cast HttpRequest to HttpUriRequest in order to get the full request
>>> URI
>>>
>>> (2) extend or replace the default RedirectHandler with a custom one and
>>> implement whatever redirect handling logic you deem appropriate
>>>
>>> Hope this helps
>>>
>>> Oleg
>>>
>>>
>>> ---------------------------------------------------------------------
>>> To unsubscribe, e-mail: httpclient-users-unsubscribe@hc.apache.org
>>> For additional commands, e-mail: httpclient-users-help@hc.apache.org
>>>
>>>
>>>
>>
>
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Re: Redirected - but where? HttpClient 4
Posted by DrewC <na...@drewcox.com>.
Ok, I guess I just needed to keep banging on this for another hour. I am now
using a non-default context with the execute and then reconstructing the
final require URL with:
HttpUriRequest finalRequest =
(HttpUriRequest) context.getAttribute(ExecutionContext.HTTP_REQUEST);
HttpHost host = (HttpHost)
context.getAttribute(ExecutionContext.HTTP_TARGET_HOST);
StringBuilder uriString = new StringBuilder();
uriString.append(host.getSchemeName() + "://");
uriString.append(host.getHostName());
if (host.getPort() != -1) {
uriString.append(":" + host.getPort());
}
uriString.append(finalRequest.getURI().normalize().toString());
If anyone could confirm this is the right method, that would be appreciated.
Regards
Drew
DrewC wrote:
>
> Hi, I am trying to do basically exactly what Magnus was attempting, to
> access the final full URL to the resource after following
> redirects. But I have had no success attempting the techniques seen here,
> though I admit to being a bit confused by them. I have not had to deal
> with the HttpContext previously and I am not exactly sure which
> HttpGetMethod Olek suggests to cast. But with my simplistic approach,
> after the request is executed (with httpClient.execute(httpGet) and I have
> confirmed several redirects already occured and were followed, the
>
> HttpGet.getURI
> (HttpUriRequest)httpGet).getURI()
> (HttpUriRequest)httpGet).getRequestLine().getUri()
>
> all still retains the intial URL, not the final one. I am using a shared
> HttpClient (created via ThreadSafeClientConnManager), not sure if that
> makes any difference.
>
> Any help appreciated.
>
> Drew
>
>
>
> olegk wrote:
>>
>>
>> On Sun, 2008-01-06 at 04:50 -0800, MaGGE wrote:
>>> Hello,
>>>
>>> I've just started using HttpClient for a little util I'm making. I'm
>>> letting
>>> it handle_redirects=true, but I need to process the URL it's been
>>> redirected
>>> to. How can I find the "current URL" after I've executed a request that
>>> was
>>> redirected?
>>>
>>> I've tried a couple of things - which I feel is "wrong". First thing was
>>> to
>>> disable handle_redirects, and look for the "Location" header in a loop
>>> of my
>>> own. Challenge with this was that I could not see any way to replace the
>>> URI
>>> for the request object, so I'd have to make a new one - which is a
>>> problem
>>> because my method should be able to process any HttpUriRequest.
>>>
>>> Second I tried to add a HttpRequestInterceptor to the client. But all I
>>> can
>>> get from the HttpRequest object given to the process method is the
>>> RequestLine. That one only has the info found behind GET/POST etc, that
>>> is
>>> only the path and method. I'm still lacking the host, protocol and so
>>> on.
>>>
>>> I'm sure there's an easy way to do this that I've just missed ... Please
>>> advice. :)
>>>
>>> Thanks for your time,
>>> Magnus
>>
>> Hi Magnus
>>
>> In addition to getting HttpHost out of the execution context, you may
>> also
>>
>> (1) cast HttpRequest to HttpUriRequest in order to get the full request
>> URI
>>
>> (2) extend or replace the default RedirectHandler with a custom one and
>> implement whatever redirect handling logic you deem appropriate
>>
>> Hope this helps
>>
>> Oleg
>>
>>
>> ---------------------------------------------------------------------
>> To unsubscribe, e-mail: httpclient-users-unsubscribe@hc.apache.org
>> For additional commands, e-mail: httpclient-users-help@hc.apache.org
>>
>>
>>
>
>
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Re: Redirected - but where? HttpClient 4
Posted by DrewC <na...@drewcox.com>.
Hi, I am trying to do basically exactly what Magnus was attempting, to access
the final full URL to the resource after following
redirects. But I have had no success attempting the techniques seen here,
though I admit to being a bit confused by them. I have not had to deal with
the HttpContext previously and I am not exactly sure which HttpGetMethod
Olek suggests to cast. But with my simplistic approach, after the request
is executed (with httpClient.execute(httpGet) and I have confirmed several
redirects already occured and were followed, the
HttpGet.getURI
(HttpUriRequest)httpGet).getURI()
(HttpUriRequest)httpGet).getRequestLine().getUri()
all still retains the intial URL, not the final one. I am using a shared
HttpClient (created via ThreadSafeClientConnManager), not sure if that makes
any difference.
Any help appreciated.
Drew
olegk wrote:
>
>
> On Sun, 2008-01-06 at 04:50 -0800, MaGGE wrote:
>> Hello,
>>
>> I've just started using HttpClient for a little util I'm making. I'm
>> letting
>> it handle_redirects=true, but I need to process the URL it's been
>> redirected
>> to. How can I find the "current URL" after I've executed a request that
>> was
>> redirected?
>>
>> I've tried a couple of things - which I feel is "wrong". First thing was
>> to
>> disable handle_redirects, and look for the "Location" header in a loop of
>> my
>> own. Challenge with this was that I could not see any way to replace the
>> URI
>> for the request object, so I'd have to make a new one - which is a
>> problem
>> because my method should be able to process any HttpUriRequest.
>>
>> Second I tried to add a HttpRequestInterceptor to the client. But all I
>> can
>> get from the HttpRequest object given to the process method is the
>> RequestLine. That one only has the info found behind GET/POST etc, that
>> is
>> only the path and method. I'm still lacking the host, protocol and so on.
>>
>> I'm sure there's an easy way to do this that I've just missed ... Please
>> advice. :)
>>
>> Thanks for your time,
>> Magnus
>
> Hi Magnus
>
> In addition to getting HttpHost out of the execution context, you may
> also
>
> (1) cast HttpRequest to HttpUriRequest in order to get the full request
> URI
>
> (2) extend or replace the default RedirectHandler with a custom one and
> implement whatever redirect handling logic you deem appropriate
>
> Hope this helps
>
> Oleg
>
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: httpclient-users-unsubscribe@hc.apache.org
> For additional commands, e-mail: httpclient-users-help@hc.apache.org
>
>
>
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Re: Redirected - but where? HttpClient 4
Posted by Oleg Kalnichevski <ol...@apache.org>.
On Sun, 2008-01-06 at 04:50 -0800, MaGGE wrote:
> Hello,
>
> I've just started using HttpClient for a little util I'm making. I'm letting
> it handle_redirects=true, but I need to process the URL it's been redirected
> to. How can I find the "current URL" after I've executed a request that was
> redirected?
>
> I've tried a couple of things - which I feel is "wrong". First thing was to
> disable handle_redirects, and look for the "Location" header in a loop of my
> own. Challenge with this was that I could not see any way to replace the URI
> for the request object, so I'd have to make a new one - which is a problem
> because my method should be able to process any HttpUriRequest.
>
> Second I tried to add a HttpRequestInterceptor to the client. But all I can
> get from the HttpRequest object given to the process method is the
> RequestLine. That one only has the info found behind GET/POST etc, that is
> only the path and method. I'm still lacking the host, protocol and so on.
>
> I'm sure there's an easy way to do this that I've just missed ... Please
> advice. :)
>
> Thanks for your time,
> Magnus
Hi Magnus
In addition to getting HttpHost out of the execution context, you may
also
(1) cast HttpRequest to HttpUriRequest in order to get the full request
URI
(2) extend or replace the default RedirectHandler with a custom one and
implement whatever redirect handling logic you deem appropriate
Hope this helps
Oleg
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Re: Redirected - but where? HttpClient 4
Posted by MaGGE <ma...@magge.no>.
Aha - I knew there was a way... :)
So the correct way of doing what I described is to use the
HttpRequestInterceptor and in it I'd use the HttpContext objects
getAttribute(ExecutionContext.HTTP_TARGET_HOST) followed by the HttpRequest
object getRequestLine().getUri(), I guess.
My code now look like this (if anyone wants to do the same):
client.addRequestInterceptor(new HttpRequestInterceptor() {
public void process(HttpRequest request, HttpContext context)
throws HttpException, IOException {
newuri =
((HttpHost)context.getAttribute(ExecutionContext.HTTP_TARGET_HOST)).toURI()
+ request.getRequestLine().getUri();
}
});
Thanks a bunch!
Magnus
Roland Weber wrote:
>
> Hello Magnus,
>
> the target host of the request is available after execution from
> the HttpContext. The attribute is ExecutionContext.HTTP_TARGET_HOST:
>
> http://hc.apache.org/httpcomponents-core/httpcore/apidocs/org/apache/http/protocol/ExecutionContext.html#HTTP_TARGET_HOST
>
> The value is an instance of HttpHost.
>
> cheers,
> Roland
>
>
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>
>
>
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Re: Redirected - but where? HttpClient 4
Posted by Roland Weber <os...@dubioso.net>.
Hello Magnus,
the target host of the request is available after execution from
the HttpContext. The attribute is ExecutionContext.HTTP_TARGET_HOST:
http://hc.apache.org/httpcomponents-core/httpcore/apidocs/org/apache/http/protocol/ExecutionContext.html#HTTP_TARGET_HOST
The value is an instance of HttpHost.
cheers,
Roland
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