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 -- Sent via pgsql-hackers mailing list (pgsql-hackers@postgresql.org) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-hackers
