Hello,
I do not understand why CREATE SUBSCRIPTION does not pick up .pgpass
(when psql does):
root@pg.newtest:/# psql 'host=pg.oldtest port=5432 user=pg_replication
dbname=oldtest'
oldtest=# \q
root@pg.newtest:/# psql newtest postgres
newtest=# CREATE SUBSCRIPTION sub_pgupgrade CONNECTION
'host=
On Wed, Aug 31, 2022 at 04:03:31PM +0300, Kristjan Mustkivi wrote:
> Hello,
>
> I do not understand why CREATE SUBSCRIPTION does not pick up .pgpass
> (when psql does):
>
> root@pg.newtest:/# psql 'host=pg.oldtest port=5432 user=pg_replication
> dbname=oldtest'
> oldtest=# \q
>
> root@pg.newtest
On Wed, Aug 31, 2022 at 4:07 PM hubert depesz lubaczewski
wrote:
> How/where you provide it?
>
> Why would you assume that postgres (running from user postgres
> presumably) would look for pgpass in /root/.pgpass?
>
> postgres should have it in ~postgres/.pgpass
> with proper ownership.
Hi, depe
On Wed, Aug 31, 2022 at 04:26:22PM +0300, Kristjan Mustkivi wrote:
> And as said, the psql utility has no problems finding the .pgass where
> it is. If I lie to it about the pgpass location i.e by giving
> passfile=/root/.pgpassx it will ask for password.
of course it doesn't have problem, because
On Wed, Aug 31, 2022 at 4:27 PM hubert depesz lubaczewski
wrote:
>
> On Wed, Aug 31, 2022 at 04:26:22PM +0300, Kristjan Mustkivi wrote:
> > And as said, the psql utility has no problems finding the .pgass where
> > it is. If I lie to it about the pgpass location i.e by giving
> > passfile=/root/.p
Dear Postgres Devs,
I use Postgres everywhere I can and for me it is the by far best
database system out there. Great job, thank you!
Now I would like to humbly propose a feature that gives an easy way to
get a quick count estimate for any condition - index based or not -
based on a random s
Torge writes:
> Now I would like to humbly propose a feature that gives an easy way to
> get a quick count estimate for any condition - index based or not -
> based on a random sample of rows, that does not require a custom
> function creation or complex SQL statement
Can't you do that already
I am not aware of this, sounds like what I want. Will look into it. Thanks!
Am 01.09.22 um 03:47 schrieb Tom Lane:
Torge writes:
Now I would like to humbly propose a feature that gives an easy way to
get a quick count estimate for any condition - index based or not -
based on a random sample o
