On 2018/07/17 8:17, Alvaro Herrera wrote:
> On 2018-Jul-16, Ashutosh Bapat wrote:
>
>>> Hmm, let me reword this comment completely. How about the attached?
>
>> That looks much better. However it took me a small while to understand
>> that (1), (2) and (3) correspond to strategies.
>
> You're r
On 2018-Jul-16, Ashutosh Bapat wrote:
> > Hmm, let me reword this comment completely. How about the attached?
> That looks much better. However it took me a small while to understand
> that (1), (2) and (3) correspond to strategies.
You're right. Amended again and pushed. I also marked the op
On Sat, Jul 14, 2018 at 2:41 AM, Alvaro Herrera
wrote:
> On 2018-Jul-13, Ashutosh Bapat wrote:
>
>> On Fri, Jul 13, 2018 at 1:15 PM, Amit Langote
>> wrote:
>> >>
>> >> I don't think this is true. When equality conditions and IS NULL clauses
>> >> cover
>> >> all partition keys of a hash partitio
On 2018-Jul-13, Ashutosh Bapat wrote:
> On Fri, Jul 13, 2018 at 1:15 PM, Amit Langote
> wrote:
> >>
> >> I don't think this is true. When equality conditions and IS NULL clauses
> >> cover
> >> all partition keys of a hash partitioned table and do not have
> >> contradictory
> >> clauses, we sh
On Fri, Jul 13, 2018 at 1:15 PM, Amit Langote
wrote:
>>
>> I don't think this is true. When equality conditions and IS NULL clauses
>> cover
>> all partition keys of a hash partitioned table and do not have contradictory
>> clauses, we should be able to find the partition which will remain unprun
Thanks for the review.
On 2018/07/12 22:01, Ashutosh Bapat wrote:
> On Thu, Jul 12, 2018 at 11:10 AM, Amit Langote
> wrote:
>>>
>>> I think your fix is correct. I slightly modified it along with updating
>>> nearby comments and added regression tests.
>>
>> I updated regression tests to reduce l
I think we should add this to open items list so that it gets tracked.
On Thu, Jul 12, 2018 at 6:31 PM, Ashutosh Bapat
wrote:
> On Thu, Jul 12, 2018 at 11:10 AM, Amit Langote
> wrote:
>>>
>>> I think your fix is correct. I slightly modified it along with updating
>>> nearby comments and added r
On Thu, Jul 12, 2018 at 11:10 AM, Amit Langote
wrote:
>>
>> I think your fix is correct. I slightly modified it along with updating
>> nearby comments and added regression tests.
>
> I updated regression tests to reduce lines. There is no point in
> repeating tests like v2 patch did.
+ *
+
On Thu, Jul 12, 2018 at 11:10 AM, Amit Langote
wrote:
> On 2018/07/12 14:32, Amit Langote wrote:
>> Thanks Ashutosh for reporting and Dilip for the analysis and the patch.
>>
>> I think your fix is correct. I slightly modified it along with updating
>> nearby comments and added regression tests.
On 2018/07/12 14:32, Amit Langote wrote:
> Thanks Ashutosh for reporting and Dilip for the analysis and the patch.
>
> On 2018/07/11 21:39, Dilip Kumar wrote:
>> On Wed, Jul 11, 2018 at 5:36 PM, amul sul wrote:
>>> On Wed, Jul 11, 2018 at 5:10 PM Dilip Kumar wrote:
>>
>>> I am not sure that
Thanks Ashutosh for reporting and Dilip for the analysis and the patch.
On 2018/07/11 21:39, Dilip Kumar wrote:
> On Wed, Jul 11, 2018 at 5:36 PM, amul sul wrote:
>> On Wed, Jul 11, 2018 at 5:10 PM Dilip Kumar wrote:
>
>>>
>> I am not sure that I have understand the following comments
>> 11 +
On Wed, Jul 11, 2018 at 5:36 PM, amul sul wrote:
> On Wed, Jul 11, 2018 at 5:10 PM Dilip Kumar wrote:
>>
> I am not sure that I have understand the following comments
> 11 +* Generate one prune step for the information derived from IS NULL,
> 12 +* if any. To prune hash partitions, we
On Wed, Jul 11, 2018 at 5:10 PM Dilip Kumar wrote:
>
> On Wed, Jul 11, 2018 at 4:20 PM, Dilip Kumar wrote:
> > On Wed, Jul 11, 2018 at 3:06 PM, Ashutosh Bapat
> > wrote:
> >> Hi,
> >> Consider following test case.
> >> create table prt (a int, b int, c int) partition by range(a, b);
> >> create
On Wed, Jul 11, 2018 at 4:20 PM, Dilip Kumar wrote:
> On Wed, Jul 11, 2018 at 3:06 PM, Ashutosh Bapat
> wrote:
>> Hi,
>> Consider following test case.
>> create table prt (a int, b int, c int) partition by range(a, b);
>> create table prt_p1 partition of prt for values (0, 0) to (100, 100);
>> cr
On Wed, Jul 11, 2018 at 3:06 PM, Ashutosh Bapat
wrote:
> Hi,
> Consider following test case.
> create table prt (a int, b int, c int) partition by range(a, b);
> create table prt_p1 partition of prt for values (0, 0) to (100, 100);
> create table prt_p1 partition of prt for values from (0, 0) to (
Hi,
Consider following test case.
create table prt (a int, b int, c int) partition by range(a, b);
create table prt_p1 partition of prt for values (0, 0) to (100, 100);
create table prt_p1 partition of prt for values from (0, 0) to (100, 100);
create table prt_p2 partition of prt for values from (1
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