Hi Anshum,

What if you have more nodes than shards*replicationFactor.
In the example below, originally I created the collection to use 6 shards* 2 replicationFactor = 12 nodes total. Now I added 6 more nodes, 18 nodes total. I just want to add 1 extra replica per shard.
How will it get evenly distributed, what is the determining criteria.

Thanks,

Rishi.



-----Original Message-----
From: Anshum Gupta <ans...@anshumgupta.net>
To: solr-user <solr-user@lucene.apache.org>
Sent: Tue, May 14, 2013 9:42 pm
Subject: Re: SOLR Cloud Collection Management quesiotn.


Hi Rishi,

If you have your cluster up and running, just add the nodes and they will get evenly assigned to the shards. As of now, the replication factor is not
persisted.


On Wed, May 15, 2013 at 1:07 AM, Rishi Easwaran <rishi.easwa...@aol.com>wrote:

Ok looks like...I have to go to every node, add a replica
individually,
create the cores and add them to the collection.

ex:
http://newNode1:port/solr/**admin/cores?action=CREATE&**

name=testCloud1_shard1_**replica3&collection=**testCloud1&shard=shard1&**
collection.configName=myconf

http://newNode2:port/solr/**admin/cores?action=CREATE&**

name=testCloud1_shard2_**replica3&collection=**testCloud1&shard=shard2&**
collection.configName=myconf


Is there an easier way to do this.
Any ideas.

Thanks,

Rishi.


-----Original Message-----
From: Rishi Easwaran <rishi.easwa...@aol.com>
To: solr-user <solr-user@lucene.apache.org>
Sent: Tue, May 14, 2013 2:58 pm
Subject: SOLR Cloud Collection Management quesiotn.


Hi,

I am beginning to work on SOLR cloud implementation.
I created a collection using the collections API

http://myhost:port/solr/admin/**collections?action=CREATE&**
name=testCloud1&numShards=6&**replicationFactor=2&**
collection.configName=myconf&**maxShardsPerNode=1

My cluster now has 6 shards and 2 replicas  (1 leader & 1 replica) for
each shard.
Now I want to add extra replicas to each shard in my cluster without
out changing the replicationFactor used to create the collection.
Any ideas on how to go about doing that.

Thanks,

Rishi.








--

Anshum Gupta
http://www.anshumgupta.net

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