On 2017/03/08 18:27, Ashutosh Bapat wrote:
>>
>> About the other statement you changed, I just realized that we should
>> perhaps do one more thing. Show the Number of partitions, even if it's 0.
>> In case of inheritance, the parent table stands on its own when there are
>> no child tables, but a partitioned table doesn't in the same sense. I
>> tried to implement that in attached patch 0002. Example below:
>>
>> create table p (a int) partition by list (a);
>> \d p
>> <snip>
>> Partition key: LIST (a)
>> Number of partitions: 0
>>
>> \d+ p
>> <snip>
>> Partition key: LIST (a)
>> Number of partitions: 0
>>
>> create table p1 partition of p for values in (1);
>> \d p
>> <snip>
>> Partition key: LIST (a)
>> Number of partitions: 1 (Use \d+ to list them.)
>>
>> \d+ p
>> <snip>
>> Partition key: LIST (a)
>> Partitions: p1 FOR VALUES IN (1)
>
> I liked that. PFA 0002 updated. I changed one of \d output to \d+ to
> better test partitioned tables without partitions in verbose and
> non-verbose mode. Also, refactored the your code to have less number
> of conditions. Please let me know if it looks good.
Thanks, looks good.
Regards,
Amit