Re: Do partition keys create skinny or wide rows?

[Vladimir Yudovin]

querying them would be inefficient (impossible?
Impossible. In the case of multi-column partition key all of them must be 
restricted in WHERE clause:

CREATE TABLE data.table (id1 int, id2 int, primary KEY ((id1,id2)));
SELECT * FROM data.table WHERE id1 = 0;
InvalidRequest: Error from server: code=2200 [Invalid query] message="Partition 
key parts: id2 must be restricted as other parts are"



Best regards, Vladimir Yudovin, 
Winguzone - Hosted Cloud Cassandra on Azure and SoftLayer.
Launch your cluster in minutes.




 On Sun, 09 Oct 2016 00:27:12 -0400 Graham Sandersongra...@vast.com 
wrote  

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.rac...@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 gra...@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.rac...@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.rac...@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.

 


 










 

Re: Do partition keys create skinny or wide rows?

[Graham Sanderson]

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  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  > 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 > > wrote:
>> 
>> the last '4 billion rows' should say '4 billion columns / cells'
>> 
>> On Sun, Oct 9, 2016 at 6:34 AM, Ali Akhtar > > 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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Description: S/MIME cryptographic signature

Re: Do partition keys create skinny or wide rows?

[Ali Akhtar]

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  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  wrote:
>
> the last '4 billion rows' should say '4 billion columns / cells'
>
> On Sun, Oct 9, 2016 at 6:34 AM, Ali Akhtar  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.
>>
>
>
>

Re: Do partition keys create skinny or wide rows?

[Graham Sanderson]

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  wrote:
> 
> the last '4 billion rows' should say '4 billion columns / cells'
> 
> On Sun, Oct 9, 2016 at 6:34 AM, Ali Akhtar  > 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.
> 



smime.p7s
Description: S/MIME cryptographic signature

Re: Do partition keys create skinny or wide rows?

[Ali Akhtar]

the last '4 billion rows' should say '4 billion columns / cells'

On Sun, Oct 9, 2016 at 6:34 AM, Ali Akhtar  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.
>

Do partition keys create skinny or wide rows?

[Ali Akhtar]

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.