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Re: HashInURI

From: Julian Reschke <julian.reschke@gmx.de>
Date: Fri, 11 Feb 2011 09:36:57 +0100
Message-ID: <4D54F529.6070102@gmx.de>
To: "Eric J. Bowman" <eric@bisonsystems.net>
CC: nathan@webr3.org, Yves Lafon <ylafon@w3.org>, Karl Dubost <karld@opera.com>, Ashok Malhotra <ashok.malhotra@oracle.com>, "www-tag@w3.org List" <www-tag@w3.org>
On 11.02.2011 06:12, Eric J. Bowman wrote:
> Nathan wrote:
>>
>> In reality everything is a different resource? well without getting
>> in to the semantics of this, because that statement is provably
>> false, I'll take it at face value to mean that everything which the
>> application provides a view of (or things which comprise a composite
>> view), in which case I think you'll find every "resource" is defined
>> by it's own unique URI, unless of course it's only available as part
>> of a collection.
>>
>
> Here are two different resources:
>
> http://twitter.com/#!/webr3
> http://twitter.com/#!/ericbow

"3.5. Fragment

The fragment identifier component of a URI allows indirect 
identification of a secondary resource by reference to a primary 
resource and additional identifying information. The identified 
secondary resource may be some portion or subset of the primary 
resource, some view on representations of the primary resource, or some 
other resource defined or described by those representations. A fragment 
identifier component is indicated by the presence of a number sign ("#") 
character and terminated by the end of the URI." -- 
<http://greenbytes.de/tech/webdav/rfc3986.html#rfc.section.3.5>

So only one *primary* resource.

> We know they're different resources because URIs are opaque, and these
> strings aren't character-for-character identical.  This is reality --
> each user's feed is a separate resource, is it not?
>
> OTOH, since # has special meaning, any HTTP client will treat those as
> being the same resource, i.e. dereferencing either one will dereference
> http://twitter.com/ .  This makes http://twitter.com/ the *one* resource
> being dereferenced for *everything*.

Yes.

> ...

+1

Best regards, Julian
Received on Friday, 11 February 2011 08:37:50 GMT

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