Oops!  I forgot that full and filtered cost maps return the same media-type.
Sorry about that.

Yes, media-type==alto-costmap+json && accepts == ""  =>  cost-type-names has
one entry.
media-type==alto-costmap+json && accepts == "alto-costmapfilter+json"  =>
cost-type-names has one or more entries.

Again, profuse apologies!

- Wendy

From:  "Y. Richard Yang" <[email protected]>
Date:  Wed, July 24, 2013 10:37
To:  Wendy Roome <[email protected]>
Cc:  "Y. Richard Yang" <[email protected]>, Richard Alimi
<[email protected]>, alto <[email protected]>
Subject:  Re: [alto] Cost-type names

Hi Wendy,

On Wednesday, July 24, 2013, Wendy Roome  wrote:
> Fine with me! 
> 
> Although I'd go one small step farther and say that if an entry has media-type
> alto-costmap+json, then it must not have an "accepts" entry.

Only specifying the predicate: media-type==alto-costmap+json && no accepts
=> #cost-type-names has one entry is safe.

Specifying media-type==alto-costmap+ json => no accepts will make it not
possible to list filtered cost maps. Right? Hence, we limit to the first
predicate. What do you think?
 
>  Similarly, a resource with media-type alto-networkmap+json cannot have
> "accepts".
> 
> Note that a server can provide full and filtered cost map services with the
> same uri, via GET and POST respectively. The IRD just needs two different
> entries, with the same uri but different media-types. Nothing says uris must
> be unique.

Yes. This is good comment and worthy of an explicit sentence to point it
out.

Thanks!

Richard

 
> 
> - Wendy Roome
> 
> From:  "Y. Richard Yang" <[email protected] <javascript:_e({}, 'cvml',
> '[email protected]');> >
> Date:  Wed, July 24, 2013 10:12
> Subject:  Re: [alto] Cost-type names
> 
> Hi Wendy,
> ŠŠ
> So, the rule is:
> 
> If media-type is alto-costmap+json, and no accepts (I.e., unfiltered map),
> then the cost-type-names in its capabilities can include only one entry. I
> believe that this is what you propose. If others do not see problems, we say
> that we reach consensus.
> 
> Richard 


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