> Forgive me if I am a little insistent, but it's important for us and I'm sure 
> we are not the only ones interested in this feature...

Not an issue, it's how things get done on :)

Create a jira ticket https://issues.apache.org/jira/browse/CASSANDRA with your 
ideas to start the process and ask others to vote if they would also like to 
see it. If you have time to donate for the feature include that on the ticket. 

Thanks 
 
-----------------
Aaron Morton
Freelance Cassandra Developer
@aaronmorton
http://www.thelastpickle.com

On 9 Jun 2011, at 06:55, Jeremiah Jordan wrote:

> I am pretty sure this would cut down on network traffic, but not on Disk IO 
> or CPU use.  I think Cassandra would still have to deserialize the whole 
> column to get to the name.  So if you really have a use case where you just 
> want the name, it would be better to store a separate "name with no data" 
> column.
> 
> From: Patrick de Torcy [mailto:pdeto...@gmail.com] 
> Sent: Wednesday, June 08, 2011 4:00 AM
> To: user@cassandra.apache.org
> Subject: Re: how to know there are some columns in a row
> 
> There is no reason for ambiguities...
> We could add in the api another method call (similar to get_count) :
> 
> get_columnNames
> 
> list<string> get_columnNames(key, column_parent, predicate, consistency_level)
> 
> Get the columns names present in column_parent within the predicate.
> 
> The method is not O(1). It takes all the columns from disk to calculate the 
> answer. The only benefit of the method is that you do not need to pull all 
> their values over Thrift interface to get their names
> 
> 
> (just to get the idea...)
> 
> In fact column names can really be data in themselves, so there should be a 
> way to retrieve them (without their values). When you have big values, it's a 
> real show stopper to use get_slice, since a lot of unnecessary traffic would 
> be generated...
> 
> Forgive me if I am a little insistent, but it's important for us and I'm sure 
> we are not the only ones interested in this feature...
> 
> cheers

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