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From Drew Kutcharian <d...@venarc.com>
Subject Re: How to reliably achieve unique constraints with Cassandra?
Date Sat, 07 Jan 2012 07:15:19 GMT
It makes great sense. You're a genius!!


On Jan 6, 2012, at 10:43 PM, Narendra Sharma wrote:

> Instead of trying to solve the generic problem of uniqueness, I would focus on the specific
problem. 
> 
> For eg lets consider your usecase of user registration with email address as key. You
can do following:
> 1. Create CF (Users) where row key is UUID and has user info specific columns.
> 2. Whenever user registers create a row in this CF with user status flag as waiting for
confirmation.
> 3. Send email to the user's email address with link that contains the UUID (or encrypted
UUID)
> 4. When user clicks on the link, use the UUID (or decrypted UUID) to lookup user
> 5. If the user exists with given UUID and status as waiting for confirmation then update
the status  and create a entry in another CF (EmailUUIDIndex) representing email address to
UUID mapping.
> 6. For authentication you can lookup in the index to get UUID and proceed.
> 7. If a malicious user registers with someone else's email id then he will never be able
to confirm and will never have an entry in EmailUUIDIndex. As a additional check if the entry
for email id exists in EmailUUIDIndex then the request for registration can be rejected right
away.
> 
> Make sense?
> 
> -Naren
> 
> On Fri, Jan 6, 2012 at 4:00 PM, Drew Kutcharian <drew@venarc.com> wrote:
> So what are the common RIGHT solutions/tools for this?
> 
> 
> On Jan 6, 2012, at 2:46 PM, Narendra Sharma wrote:
> 
>> >>>It's very surprising that no one seems to have solved such a common use
case.
>> I would say people have solved it using RIGHT tools for the task.
>> 
>> 
>> 
>> On Fri, Jan 6, 2012 at 2:35 PM, Drew Kutcharian <drew@venarc.com> wrote:
>> Thanks everyone for the replies. Seems like there is no easy way to handle this.
It's very surprising that no one seems to have solved such a common use case.
>> 
>> -- Drew
>> 
>> On Jan 6, 2012, at 2:11 PM, Bryce Allen wrote:
>> 
>> > That's a good question, and I'm not sure - I'm fairly new to both ZK
>> > and Cassandra. I found this wiki page:
>> > http://wiki.apache.org/hadoop/ZooKeeper/FailureScenarios
>> > and I think the lock recipe still works, even if a stale read happens.
>> > Assuming that wiki page is correct.
>> >
>> > There is still subtlety to locking with ZK though, see (Locks based
>> > on ephemeral nodes) from the zk mailing list in October:
>> > http://mail-archives.apache.org/mod_mbox/zookeeper-user/201110.mbox/thread?0
>> >
>> > -Bryce
>> >
>> > On Fri, 6 Jan 2012 13:36:52 -0800
>> > Drew Kutcharian <drew@venarc.com> wrote:
>> >> Bryce,
>> >>
>> >> I'm not sure about ZooKeeper, but I know if you have a partition
>> >> between HazelCast nodes, than the nodes can acquire the same lock
>> >> independently in each divided partition. How does ZooKeeper handle
>> >> this situation?
>> >>
>> >> -- Drew
>> >>
>> >>
>> >> On Jan 6, 2012, at 12:48 PM, Bryce Allen wrote:
>> >>
>> >>> On Fri, 6 Jan 2012 10:03:38 -0800
>> >>> Drew Kutcharian <drew@venarc.com> wrote:
>> >>>> I know that this can be done using a lock manager such as ZooKeeper
>> >>>> or HazelCast, but the issue with using either of them is that if
>> >>>> ZooKeeper or HazelCast is down, then you can't be sure about the
>> >>>> reliability of the lock. So this potentially, in the very rare
>> >>>> instance where the lock manager is down and two users are
>> >>>> registering with the same email, can cause major issues.
>> >>>
>> >>> For most applications, if the lock managers is down, you don't
>> >>> acquire the lock, so you don't enter the critical section. Rather
>> >>> than allowing inconsistency, you become unavailable (at least to
>> >>> writes that require a lock).
>> >>>
>> >>> -Bryce
>> >>
>> 
>> 
>> 
>> 
>> -- 
>> Narendra Sharma
>> Software Engineer
>> http://www.aeris.com
>> http://narendrasharma.blogspot.com/
>> 
>> 
> 
> 
> 
> 
> -- 
> Narendra Sharma
> Software Engineer
> http://www.aeris.com
> http://narendrasharma.blogspot.com/
> 
> 


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