Thanks that is really good to know and gives confidence to those of us that 
have hope for Cassandra. I still get comments from friends who wonder why I 
want to use it when Oracle is available. I tell them I am preparing for the 
future and willing to try and wait out for newer better releases.

Something like innovation, courage, determination, faith and wanting to be in 
the front and not behind drive me and good to see it here.

Best Regards,
Tony Anecito



On Wednesday, August 20, 2014 8:19 PM, Benedict Elliott Smith 
<belliottsm...@datastax.com> wrote:
 


The "run versions >= x.x.6" is IMO an out of date trope. Not only does 2.1 have 
more than twice as many engineers working full time on it, and five times the 
QA engineers (which is cumulative with prior QA), we are also seeing many more 
users in the wild running release candidates and providing valuable feedback 
before release. This is evidenced by the fact there have been six release 
candidates, instead of just two for 2.0.

As such, whilst no software is bug free, 2.1.0 is likely to be considerably 
more stable than 2.0.0.




On Thu, Aug 21, 2014 at 5:10 AM, Robert Coli <rc...@eventbrite.com> wrote:

On Wed, Aug 20, 2014 at 1:44 PM, Tony Anecito <adanec...@yahoo.com> wrote:
>
>When will Cassandra 2.1 be released as GA?
>>
>>
>>
>>Will there be a datastax CQL driver available then?
>>
>>
>>Will Cassandra 2.1 and Datastax CQL driver support nested selects? If not for 
>>a combination key table and detailed table (Star Schema) what is the fastest 
>>way to query without using heavy de-normalization?
>
>
>The form of your questions suggests you are planning to use a not-ready-for 
>production version of Cassandra inappropriately.  Probably don't do that.
>
>
>https://engineering.eventbrite.com/what-version-of-cassandra-should-i-run/
>
>
>
>In general the answer to complex query questions in Cassandra is "heavy 
>de-normalization", if you have a problem that you need to solve without "heavy 
>de-normalization" then perhaps Cassandra is not for you.
>
>
>=Rob
>
>

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