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From Graham Sanderson <gra...@vast.com>
Subject Re: Do partition keys create skinny or wide rows?
Date Sun, 09 Oct 2016 04:27:12 GMT
No the employees would end up in arbitrary partitions, and querying them would be inefficient
(impossible? - I am levels back on C* so don’t know if ALLOW FILTERING even works for this).

I would be tempted to use organization_id only or organization_Id and maybe a few shard bits
(if you are worried about huge orgs) from the employee_Id to make the partition key, but it
really depends what other queries you will be making
> On Oct 8, 2016, at 11:19 PM, Ali Akhtar <ali.rac200@gmail.com> wrote:
> 
> In the case of PRIMARY KEY((organization_id, employee_id)), could I still do a query
like Select ... where organization_id = x, to get all employees in a particular organization?
> 
> And, this will put all those employees in the same node, right?
> 
> On Sun, Oct 9, 2016 at 9:17 AM, Graham Sanderson <graham@vast.com <mailto:graham@vast.com>>
wrote:
> Nomenclature is tricky, but PRIMARY KEY((organization_id, employee_id)) will make organization_id,
employee_id the partition key which equates roughly to your latter sentence (I’m not sure
about the 4 billion limit - that may be the new actual limit, but probably not a good idea).
> 
>> On Oct 8, 2016, at 8:35 PM, Ali Akhtar <ali.rac200@gmail.com <mailto:ali.rac200@gmail.com>>
wrote:
>> 
>> the last '4 billion rows' should say '4 billion columns / cells'
>> 
>> On Sun, Oct 9, 2016 at 6:34 AM, Ali Akhtar <ali.rac200@gmail.com <mailto:ali.rac200@gmail.com>>
wrote:
>> Say I have the following primary key:
>> PRIMARY KEY((organization_id, employee_id))
>> 
>> Will this create 1 row whose primary key is the organization id, but it has a 4 billion
column / cell limit?
>> 
>> Or will this create 1 row for each employee in the same organization, so if i have
5 employees, they will each have their own 5 rows, and each of those 5 rows will have their
own 4 billion rows?
>> 
>> Thank you.
>> 
> 
> 


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