-Re: Query a version of a column efficiently
Jerry Lam 2012-07-26, 23:34
Can you tell me which source code is responsible for the logic. The source code in the get and scan doesnt provide an indication of how the setTimeRange works.
Sent from my iPad (sorry for spelling mistakes)
On 2012-07-26, at 18:30, Stack <[EMAIL PROTECTED]> wrote:
> On Thu, Jul 26, 2012 at 11:40 PM, Jerry Lam <[EMAIL PROTECTED]> wrote:
>> Hi St.Ack:
>> Let say there are 5 versions for a column A with timestamp = [0, 1, 3, 6,
>> I want to execute an efficient query that returns one version of the column
>> that has a timestamp that is equal to 5 or less. So in this case, it should
>> return the value of the column A with timestamp = 3.
>> Using the setTimeRange(5, Long.MAX_VALUE) with setMaxVersion = 1, my guess
>> is that it will return the version 6 not version 3. Correct me if I'm
> What Tom says, try it. IIUC, it'll give you your 3. It won't give
> you 6 since that is outside of the timerange (try 0 instead of
> MAX_VALUE; I may have misled w/ MAX_VALUE... it might work but would
> have to check code).