Horton

Thread

Commits

  1. Test partition functions with legacy inheritance children, too

  2. Consider only relations part of partition trees in partition functions

  3. Make pg_partition_tree return no rows on unsupported and undefined objects

  4. Tweak pg_partition_tree for undefined relations and unsupported relkinds

  1. pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2018-12-07T01:04:06Z

    Hi all,

While testing another patch, I have bumped into the issue of
$subject...  I should have put some more negative testing from the start
on this stuff, here is a culprit query when passing directly an OID:
select pg_partition_tree(0);

I think that we should make the function return NULL if the relation
defined does not exist, as we usually do for system-facing functions.
It is also easier for the caller to know that the relation does not
exist instead of having a plpgsql try/catch wrapper or such.

Thoughts?
--
Michael

  2. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2018-12-07T23:05:40Z

    On Fri, Dec 07, 2018 at 10:04:06AM +0900, Michael Paquier wrote:
> While testing another patch, I have bumped into the issue of
> $subject...  I should have put some more negative testing from the start
> on this stuff, here is a culprit query when passing directly an OID:
> select pg_partition_tree(0);
> 
> I think that we should make the function return NULL if the relation
> defined does not exist, as we usually do for system-facing functions.
> It is also easier for the caller to know that the relation does not
> exist instead of having a plpgsql try/catch wrapper or such.
> 
> Thoughts?

Are there any objections about fixing this issue?  I would rather fix it
sonner than later.
--
Michael

  3. Re: pg_partition_tree crashes for a non-defined relation

    Amit Langote <amitlangote09@gmail.com> — 2018-12-08T03:28:53Z

    Hi,

Sorry for not replying sooner.

On Sat, Dec 8, 2018 at 8:06 Michael Paquier <michael@paquier.xyz> wrote:

> On Fri, Dec 07, 2018 at 10:04:06AM +0900, Michael Paquier wrote:
> > While testing another patch, I have bumped into the issue of
> > $subject...  I should have put some more negative testing from the start
> > on this stuff, here is a culprit query when passing directly an OID:
> > select pg_partition_tree(0);
> >
> > I think that we should make the function return NULL if the relation
> > defined does not exist, as we usually do for system-facing functions.
> > It is also easier for the caller to know that the relation does not
> > exist instead of having a plpgsql try/catch wrapper or such.
> >
> > Thoughts?
>
> Are there any objections about fixing this issue?  I would rather fix it
> sonner than later.


Thanks for noticing it and creating the patch.  The fix makes sense.

Regards,
Amit

  4. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2018-12-08T03:45:49Z

    On Sat, Dec 08, 2018 at 12:28:53PM +0900, Amit Langote wrote:
> Thanks for noticing it and creating the patch.  The fix makes sense.

Thanks a lot for looking at it!
--
Michael

  5. Re: pg_partition_tree crashes for a non-defined relation

    Tom Lane <tgl@sss.pgh.pa.us> — 2018-12-08T04:33:32Z

    Michael Paquier <michael@paquier.xyz> writes:
> On Fri, Dec 07, 2018 at 10:04:06AM +0900, Michael Paquier wrote:
>> I think that we should make the function return NULL if the relation
>> defined does not exist, as we usually do for system-facing functions.
>> It is also easier for the caller to know that the relation does not
>> exist instead of having a plpgsql try/catch wrapper or such.

> Are there any objections about fixing this issue?  I would rather fix it
> sonner than later.

Return NULL seems a reasonable behavior.

How about cases where the relation OID exists but it's the wrong
kind of relation?

			regards, tom lane

  6. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2018-12-08T06:54:36Z

    On Fri, Dec 07, 2018 at 11:33:32PM -0500, Tom Lane wrote:
> How about cases where the relation OID exists but it's the wrong
> kind of relation?

Such cases already return an error:
=# create sequence popo;
CREATE SEQUENCE
=# select pg_partition_tree('popo');
ERROR:  42809: "popo" is not a table, a foreign table, or an index
LOCATION:  pg_partition_tree, partitionfuncs.c:54

I think that's a sensible choice, because it makes no sense to look for
the inheritors of unsupported relkinds.  Perhaps you have a different
view on the matter?
--
Michael

  7. Re: pg_partition_tree crashes for a non-defined relation

    Stephen Frost <sfrost@snowman.net> — 2018-12-08T13:46:08Z

    Greetings,

* Michael Paquier (michael@paquier.xyz) wrote:
> On Fri, Dec 07, 2018 at 11:33:32PM -0500, Tom Lane wrote:
> > How about cases where the relation OID exists but it's the wrong
> > kind of relation?
> 
> Such cases already return an error:
> =# create sequence popo;
> CREATE SEQUENCE
> =# select pg_partition_tree('popo');
> ERROR:  42809: "popo" is not a table, a foreign table, or an index
> LOCATION:  pg_partition_tree, partitionfuncs.c:54
> 
> I think that's a sensible choice, because it makes no sense to look for
> the inheritors of unsupported relkinds.  Perhaps you have a different
> view on the matter?

We should really have a more clearly defined policy around this, but my
recollection is that we often prefer to return NULL rather than throwing
an error for the convenience of people doing things like querying
pg_class using similar functions.

I wonder if we maybe should have a regression test for every such
function which just queries the catalog in a way to force the function
to be called for every relation defined in the regression tests, to
ensure that it doesn't segfault or throw an error..

Thanks!

Stephen

  8. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2018-12-08T23:15:07Z

    On Sat, Dec 08, 2018 at 08:46:08AM -0500, Stephen Frost wrote:
> We should really have a more clearly defined policy around this, but my
> recollection is that we often prefer to return NULL rather than throwing
> an error for the convenience of people doing things like querying
> pg_class using similar functions.

Yes, that's visibly right.  At least that's what I can see from the
various pg_get_*def and pg_*_is_visible.  Returning NULL would indeed
be more consistent.

> I wonder if we maybe should have a regression test for every such
> function which just queries the catalog in a way to force the function
> to be called for every relation defined in the regression tests, to
> ensure that it doesn't segfault or throw an error..

Like sqlsmith?  It looks hard to me to make something like that part of
the main regression test suite, as that's going to be costly and hard to
scale with.
--
Michael

  9. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2018-12-08T23:59:26Z

    On Sun, Dec 09, 2018 at 08:15:07AM +0900, Michael Paquier wrote:
> On Sat, Dec 08, 2018 at 08:46:08AM -0500, Stephen Frost wrote:
>> We should really have a more clearly defined policy around this, but my
>> recollection is that we often prefer to return NULL rather than throwing
>> an error for the convenience of people doing things like querying
>> pg_class using similar functions.
> 
> Yes, that's visibly right.  At least that's what I can see from the
> various pg_get_*def and pg_*_is_visible.  Returning NULL would indeed
> be more consistent.

Thinking more about your argument, scanning fully pg_class is quite
sensible as well because there is no need to apply an extra qual on
relkind, so let's change the function as you suggest, by returning NULL
on invalid relation type.  Any opinions about the attached then which
does the switch?
--
Michael

  10. Re: pg_partition_tree crashes for a non-defined relation

    Stephen Frost <sfrost@snowman.net> — 2018-12-09T17:48:15Z

    Greetings,

* Michael Paquier (michael@paquier.xyz) wrote:
> On Sat, Dec 08, 2018 at 08:46:08AM -0500, Stephen Frost wrote:
> > I wonder if we maybe should have a regression test for every such
> > function which just queries the catalog in a way to force the function
> > to be called for every relation defined in the regression tests, to
> > ensure that it doesn't segfault or throw an error..
> 
> Like sqlsmith?  It looks hard to me to make something like that part of
> the main regression test suite, as that's going to be costly and hard to
> scale with.

No, I mean something like:

with x as (select pg_partition_tree(relname) from pg_class)
select 1 from x limit 1;

or whatever it takes to make sure that the function is run against every
entry in pg_class (or whatever is appropriate) while not returning the
results (since we don't actually care about the output, we just want to
make sure it doesn't ERROR or crash).

sqlsmith, as I recall, doesn't care about ERROR cases, it's just looking
for crashes, so it's not quite the same thing.

Thanks!

Stephen

  11. Re: pg_partition_tree crashes for a non-defined relation

    Stephen Frost <sfrost@snowman.net> — 2018-12-09T17:49:18Z

    Greetings,

* Michael Paquier (michael@paquier.xyz) wrote:
> On Sun, Dec 09, 2018 at 08:15:07AM +0900, Michael Paquier wrote:
> > On Sat, Dec 08, 2018 at 08:46:08AM -0500, Stephen Frost wrote:
> >> We should really have a more clearly defined policy around this, but my
> >> recollection is that we often prefer to return NULL rather than throwing
> >> an error for the convenience of people doing things like querying
> >> pg_class using similar functions.
> > 
> > Yes, that's visibly right.  At least that's what I can see from the
> > various pg_get_*def and pg_*_is_visible.  Returning NULL would indeed
> > be more consistent.
> 
> Thinking more about your argument, scanning fully pg_class is quite
> sensible as well because there is no need to apply an extra qual on
> relkind, so let's change the function as you suggest, by returning NULL
> on invalid relation type.  Any opinions about the attached then which
> does the switch?

Looks alright on a quick glance, but shouldn't you also update the
comment..?

Thanks!

Stephen

  12. Re: pg_partition_tree crashes for a non-defined relation

    Tom Lane <tgl@sss.pgh.pa.us> — 2018-12-09T18:05:43Z

    Stephen Frost <sfrost@snowman.net> writes:
> * Michael Paquier (michael@paquier.xyz) wrote:
>> On Sat, Dec 08, 2018 at 08:46:08AM -0500, Stephen Frost wrote:
>>> I wonder if we maybe should have a regression test for every such
>>> function which just queries the catalog in a way to force the function
>>> to be called for every relation defined in the regression tests, to
>>> ensure that it doesn't segfault or throw an error..

> No, I mean something like:
> with x as (select pg_partition_tree(relname) from pg_class)
> select 1 from x limit 1;
> or whatever it takes to make sure that the function is run against every
> entry in pg_class (or whatever is appropriate) while not returning the
> results (since we don't actually care about the output, we just want to
> make sure it doesn't ERROR or crash).

I'm going to object to that on cost grounds.  It is not reasonable to
run moderately-expensive functions like this on more than a thousand
pg_class entries in order to test what are just a few distinct cases,
especially in code that's highly unlikely to break once written.
Or at least, it's not reasonable to do that every time anybody runs
the regression tests for the rest of our lives.

Moreover, this would only help check a specific new function if the
author or reviewer remembered to add a test case that does it.
Since the whole problem here is "I forgot to consider this", I don't
have much faith in that happening.

Maybe we should have some sort of checklist of things to think about
when adding new SQL-visible functions, rather than trying to
memorialize every such consideration as a regression test no matter
how expensive.  Admittedly, "I forgot to go over the checklist" is
still a problem; but at least it's not penalizing every developer and
every buildfarm run till kingdom come.

			regards, tom lane

  13. Re: pg_partition_tree crashes for a non-defined relation

    Stephen Frost <sfrost@snowman.net> — 2018-12-09T18:54:52Z

    Greetings,

* Tom Lane (tgl@sss.pgh.pa.us) wrote:
> Stephen Frost <sfrost@snowman.net> writes:
> > * Michael Paquier (michael@paquier.xyz) wrote:
> >> On Sat, Dec 08, 2018 at 08:46:08AM -0500, Stephen Frost wrote:
> >>> I wonder if we maybe should have a regression test for every such
> >>> function which just queries the catalog in a way to force the function
> >>> to be called for every relation defined in the regression tests, to
> >>> ensure that it doesn't segfault or throw an error..
> 
> > No, I mean something like:
> > with x as (select pg_partition_tree(relname) from pg_class)
> > select 1 from x limit 1;
> > or whatever it takes to make sure that the function is run against every
> > entry in pg_class (or whatever is appropriate) while not returning the
> > results (since we don't actually care about the output, we just want to
> > make sure it doesn't ERROR or crash).
> 
> I'm going to object to that on cost grounds.  It is not reasonable to
> run moderately-expensive functions like this on more than a thousand
> pg_class entries in order to test what are just a few distinct cases,
> especially in code that's highly unlikely to break once written.

Yeah, I had been wondering if it would be too expensive also.

I don't entirely buy off on the argument that it's code that's 'highly
unlikely to break once written' though- we do add new relkinds from time
to time, for example.  Perhaps we could have these functions run just
once per relkind.

> Moreover, this would only help check a specific new function if the
> author or reviewer remembered to add a test case that does it.

We could possibly write a test which would run every function that
accepts the typical data types (oid/regclass/name/etc) instead of
depending on the author to remember to add it.

> Since the whole problem here is "I forgot to consider this", I don't
> have much faith in that happening.

Yeah, I agree that we'd want something more automated as, otherwise, it
would definitely be likely to be forgotten.

> Maybe we should have some sort of checklist of things to think about
> when adding new SQL-visible functions, rather than trying to
> memorialize every such consideration as a regression test no matter
> how expensive.  Admittedly, "I forgot to go over the checklist" is
> still a problem; but at least it's not penalizing every developer and
> every buildfarm run till kingdom come.

This seems like something we should do regardless.  Moreover, I'd
suggest that we start documenting some of these policies in the way that
we have a style guide for errors and such- we need a "system function
style guide" that could start with something like "prefer to return NULL
instead of ERROR on unexpected but otherwise valid inputs, and test that
the code doesn't segfault when given a variety of inputs".

Thanks!

Stephen

  14. Re: pg_partition_tree crashes for a non-defined relation

    Tom Lane <tgl@sss.pgh.pa.us> — 2018-12-09T19:07:29Z

    Stephen Frost <sfrost@snowman.net> writes:
> * Tom Lane (tgl@sss.pgh.pa.us) wrote:
>> ... especially in code that's highly unlikely to break once written.

> I don't entirely buy off on the argument that it's code that's 'highly
> unlikely to break once written' though- we do add new relkinds from time
> to time, for example.  Perhaps we could have these functions run just
> once per relkind.

Well, the relevant code is likely to be "if relkind is not x, y, or z,
then PG_RETURN_NULL".  If we add a new relkind and forget to consider the
function, the outcome is a NULL result that perhaps should not have been
NULL ... but a test like this won't help us notice that.

			regards, tom lane

  15. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2018-12-09T22:39:41Z

    On Sun, Dec 09, 2018 at 02:07:29PM -0500, Tom Lane wrote:
> Stephen Frost <sfrost@snowman.net> writes:
>> I don't entirely buy off on the argument that it's code that's 'highly
>> unlikely to break once written' though- we do add new relkinds from time
>> to time, for example.  Perhaps we could have these functions run just
>> once per relkind.
> 
> Well, the relevant code is likely to be "if relkind is not x, y, or z,
> then PG_RETURN_NULL".  If we add a new relkind and forget to consider the
> function, the outcome is a NULL result that perhaps should not have been
> NULL ... but a test like this won't help us notice that.

Yes, in order to prevent problems with newly-introduced relkinds, I
think that the checks within functions should be careful to check only
for relkinds that they support, and not list those they do not support.
--
Michael

  16. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2018-12-09T22:42:21Z

    On Sun, Dec 09, 2018 at 12:49:18PM -0500, Stephen Frost wrote:
> Looks alright on a quick glance, but shouldn't you also update the
> comment..?

The comment on HEAD or with the patch is that:
    /* Only allow relation types that can appear in partition trees. */

This still looks adapted to me.  Or would you reword it because "allow"
rather implies that an ERROR is returned?  Would you prefer changing it
something like that perhaps:
"Return NULL for relation types that do not appear in partition trees."
--
Michael

  17. Re: pg_partition_tree crashes for a non-defined relation

    Stephen Frost <sfrost@snowman.net> — 2018-12-10T13:21:22Z

    Greetings,

* Michael Paquier (michael@paquier.xyz) wrote:
> On Sun, Dec 09, 2018 at 02:07:29PM -0500, Tom Lane wrote:
> > Stephen Frost <sfrost@snowman.net> writes:
> >> I don't entirely buy off on the argument that it's code that's 'highly
> >> unlikely to break once written' though- we do add new relkinds from time
> >> to time, for example.  Perhaps we could have these functions run just
> >> once per relkind.
> > 
> > Well, the relevant code is likely to be "if relkind is not x, y, or z,
> > then PG_RETURN_NULL".  If we add a new relkind and forget to consider the
> > function, the outcome is a NULL result that perhaps should not have been
> > NULL ... but a test like this won't help us notice that.
> 
> Yes, in order to prevent problems with newly-introduced relkinds, I
> think that the checks within functions should be careful to check only
> for relkinds that they support, and not list those they do not support.

Yes, but it's certainly possible for someone to add another relkind to
that list without looking through the rest of the function carefully
enough.  Perhaps not the best example.  I still don't like the
assumption that the code won't be broken once it's written correctly the
first time though, and I continue to feel that it would be good to have
regression tests which run these functions with interesting arguments.

I also agree that we don't want to make the regression tests take a lot
of additional time.

Thanks!

Stephen

  18. Re: pg_partition_tree crashes for a non-defined relation

    Stephen Frost <sfrost@snowman.net> — 2018-12-10T13:21:43Z

    Greetings,

* Michael Paquier (michael@paquier.xyz) wrote:
> On Sun, Dec 09, 2018 at 12:49:18PM -0500, Stephen Frost wrote:
> > Looks alright on a quick glance, but shouldn't you also update the
> > comment..?
> 
> The comment on HEAD or with the patch is that:
>     /* Only allow relation types that can appear in partition trees. */
> 
> This still looks adapted to me.  Or would you reword it because "allow"
> rather implies that an ERROR is returned?  Would you prefer changing it
> something like that perhaps:
> "Return NULL for relation types that do not appear in partition trees."

Yes.

Thanks!

Stephen

  19. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2018-12-10T13:52:37Z

    On Mon, Dec 10, 2018 at 08:21:43AM -0500, Stephen Frost wrote:
> * Michael Paquier (michael@paquier.xyz) wrote:
>> This still looks adapted to me.  Or would you reword it because "allow"
>> rather implies that an ERROR is returned?  Would you prefer changing it
>> something like that perhaps:
>> "Return NULL for relation types that do not appear in partition trees."
> 
> Yes.

OK.  Sure, let's do as you suggest then.  I'll wait a couple of days
before committing the patch so as if someone has extra comments they
have the time to post.  So please feel free to comment!
--
Michael

  20. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2018-12-12T00:52:25Z

    On Mon, Dec 10, 2018 at 10:52:37PM +0900, Michael Paquier wrote:
> OK.  Sure, let's do as you suggest then.  I'll wait a couple of days
> before committing the patch so as if someone has extra comments they
> have the time to post.  So please feel free to comment!

And done this way.  Thanks for the input, Stephen and others!
--
Michael

  21. Re: pg_partition_tree crashes for a non-defined relation

    Amit Langote <langote_amit_f8@lab.ntt.co.jp> — 2018-12-13T06:34:56Z

    On 2018/12/12 9:52, Michael Paquier wrote:
> On Mon, Dec 10, 2018 at 10:52:37PM +0900, Michael Paquier wrote:
>> OK.  Sure, let's do as you suggest then.  I'll wait a couple of days
>> before committing the patch so as if someone has extra comments they
>> have the time to post.  So please feel free to comment!
> 
> And done this way.  Thanks for the input, Stephen and others!

Thank you Michael for taking care of this.  I agree with the consensus
that instead of throwing an error for unsupported relkinds,
pg_partition_tree should rather return single row containing NULL for all
columns.

Regards,
Amit

  22. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2018-12-14T00:27:24Z

    On Thu, Dec 13, 2018 at 03:34:56PM +0900, Amit Langote wrote:
> Thank you Michael for taking care of this.  I agree with the consensus
> that instead of throwing an error for unsupported relkinds,
> pg_partition_tree should rather return single row containing NULL for all
> columns.

Cool, thanks for confirming!  If there are complains on the matter,
let's of course tune it again if necessary before the release.  There is
still time for that.
--
Michael

  23. Re: pg_partition_tree crashes for a non-defined relation

    Alvaro Herrera <alvherre@2ndquadrant.com> — 2019-02-27T18:48:08Z

    On 2018-Dec-09, Tom Lane wrote:

> Stephen Frost <sfrost@snowman.net> writes:
> > * Tom Lane (tgl@sss.pgh.pa.us) wrote:
> >> ... especially in code that's highly unlikely to break once written.
> 
> > I don't entirely buy off on the argument that it's code that's 'highly
> > unlikely to break once written' though- we do add new relkinds from time
> > to time, for example.  Perhaps we could have these functions run just
> > once per relkind.
> 
> Well, the relevant code is likely to be "if relkind is not x, y, or z,
> then PG_RETURN_NULL".  If we add a new relkind and forget to consider the
> function, the outcome is a NULL result that perhaps should not have been
> NULL ... but a test like this won't help us notice that.

I just happened to come across the result of this rationale in
pg_partition_tree() (an SRF) while developing a new related function,
pg_partition_ancestors(), and find the resulting behavior rather absurd
-- it returns one row with all NULL columns, rather than no rows.  I
think the sensible behavior would be to do SRF_RETURN_DONE() before
stashing any rows to the output, so that we get an empty result set
instead.

alvherre=# select * from pg_partition_tree('information_schema.sequences');
 relid | parentrelid | isleaf | level 
-------+-------------+--------+-------
       |             |        |      
(1 fila)

-- 
Álvaro Herrera                https://www.2ndQuadrant.com/
PostgreSQL Development, 24x7 Support, Remote DBA, Training & Services

  24. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2019-02-28T01:45:49Z

    On Wed, Feb 27, 2019 at 03:48:08PM -0300, Alvaro Herrera wrote:
> I just happened to come across the result of this rationale in
> pg_partition_tree() (an SRF) while developing a new related function,
> pg_partition_ancestors(), and find the resulting behavior rather absurd
> -- it returns one row with all NULL columns, rather than no rows.  I
> think the sensible behavior would be to do SRF_RETURN_DONE() before
> stashing any rows to the output, so that we get an empty result set
> instead.

Hmm.  Going through the thread again NULL was decided to make the
whole experience consistent, now by returning nothing we would get
a behavior as consistent as when NULL is used in input, so point taken
to tune the behavior for unsupported relkinds and undefined objects.

Does the attached look fine to you?
--
Michael

  25. Re: pg_partition_tree crashes for a non-defined relation

    Amit Langote <langote_amit_f8@lab.ntt.co.jp> — 2019-02-28T02:21:59Z

    Hi,

On 2019/02/28 10:45, Michael Paquier wrote:
> On Wed, Feb 27, 2019 at 03:48:08PM -0300, Alvaro Herrera wrote:
>> I just happened to come across the result of this rationale in
>> pg_partition_tree() (an SRF) while developing a new related function,
>> pg_partition_ancestors(), and find the resulting behavior rather absurd
>> -- it returns one row with all NULL columns, rather than no rows.  I
>> think the sensible behavior would be to do SRF_RETURN_DONE() before
>> stashing any rows to the output, so that we get an empty result set
>> instead.
> 
> Hmm.  Going through the thread again NULL was decided to make the
> whole experience consistent, now by returning nothing we would get
> a behavior as consistent as when NULL is used in input, so point taken
> to tune the behavior for unsupported relkinds and undefined objects.
> 
> Does the attached look fine to you?

Reading the discussion, we just don't want to throw an "ERROR: unsupported
relkind" error, to avoid, for example, aborting a query that's iteratively
calling pg_partition_tree on all relations in a given set.  Returning no
rows at all seems like a better way of ignoring such relations.

with rels as (select relname from pg_class where relnamespace =
'public'::regnamespace) select pg_partition_tree(relname::regclass) from
rels;
           pg_partition_tree
────────────────────────────────────────
 (pk1,pk,t,0)
 (pk,,f,0)
 (pk1,pk,t,1)
 (pk1_pkey,pk_pkey,t,0)
 (pk_pkey,,f,0)
 (pk1_pkey,pk_pkey,t,1)

 (another_pk1,another_pk,t,0)
 (another_pk,,f,0)
 (another_pk1,another_pk,t,1)
 (another_pk1_pkey,another_pk_pkey,t,0)
 (another_pk_pkey,,f,0)
 (another_pk1_pkey,another_pk_pkey,t,1)
(13 rows)

Now that Alvaro mentions it, I too think that returning no rows at all is
better than 1 row filled with all NULLs, so +1.

Thanks,
Amit

  26. Re: pg_partition_tree crashes for a non-defined relation

    Alvaro Herrera <alvherre@2ndquadrant.com> — 2019-02-28T19:32:03Z

    On 2019-Feb-28, Michael Paquier wrote:

> On Wed, Feb 27, 2019 at 03:48:08PM -0300, Alvaro Herrera wrote:
> > I just happened to come across the result of this rationale in
> > pg_partition_tree() (an SRF) while developing a new related function,
> > pg_partition_ancestors(), and find the resulting behavior rather absurd
> > -- it returns one row with all NULL columns, rather than no rows.  I
> > think the sensible behavior would be to do SRF_RETURN_DONE() before
> > stashing any rows to the output, so that we get an empty result set
> > instead.
> 
> Hmm.  Going through the thread again NULL was decided to make the
> whole experience consistent, now by returning nothing we would get
> a behavior as consistent as when NULL is used in input, so point taken
> to tune the behavior for unsupported relkinds and undefined objects.

Right, thanks.

> Does the attached look fine to you?

Yeah, looks good, please push.

What about legacy inheritance?  I see that pg_partition_tree handles
that case perfectly well -- it seems to return the complete hierarchy
rooted at the given relation.  However, it seems odd that it works at
all, don't you think?  Consider this:

create table t1 (a int);
create table t11 () inherits (t1);
create table t2 (b int);
create table t111() inherits (t1, t2);

alvherre=# select * from pg_partition_tree('t1');
 relid | parentrelid | isleaf | level 
-------+-------------+--------+-------
 t1    | t           | t      |     0
 t11   | t1          | t      |     1
 t111  | t1          | t      |     1
(3 filas)

OK so far... but look at t2's tree:

alvherre=# select * from pg_partition_tree('t2');
 relid | parentrelid | isleaf | level 
-------+-------------+--------+-------
 t2    | t           | t      |     0
 t111  | t1          | t      |     2

I think this one is just weird -- t1 is not listed as "relid" anywhere,
and why does t111 has level=2?

I would opt for returning the empty set for legacy inheritance too.

More generally, I think we should return empty for anything that's
either not RELKIND_PARTITIONED_TABLE or has relispartition set.

-- 
Álvaro Herrera                https://www.2ndQuadrant.com/
PostgreSQL Development, 24x7 Support, Remote DBA, Training & Services

  27. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2019-03-01T00:22:04Z

    On Thu, Feb 28, 2019 at 04:32:03PM -0300, Alvaro Herrera wrote:
> Yeah, looks good, please push.

Done for this part.

> I would opt for returning the empty set for legacy inheritance too.
> 
> More generally, I think we should return empty for anything that's
> either not RELKIND_PARTITIONED_TABLE or has relispartition set.

I think that one option is to make the function return only the table
itself if it is not a partitioned table, which would be more
consistent with what pg_partition_root() does.

What I am writing next sounds perhaps a bit fancy, but in my opinion a
normal table is itself a partition tree, made of one single member:
itself.
--
Michael

  28. Re: pg_partition_tree crashes for a non-defined relation

    Amit Langote <langote_amit_f8@lab.ntt.co.jp> — 2019-03-01T04:38:28Z

    Hi,

On 2019/03/01 9:22, Michael Paquier wrote:
> On Thu, Feb 28, 2019 at 04:32:03PM -0300, Alvaro Herrera wrote:
>> Yeah, looks good, please push.
> 
> Done for this part.
> 
>> I would opt for returning the empty set for legacy inheritance too.
>>
>> More generally, I think we should return empty for anything that's
>> either not RELKIND_PARTITIONED_TABLE or has relispartition set.
> 
> I think that one option is to make the function return only the table
> itself if it is not a partitioned table, which would be more
> consistent with what pg_partition_root() does.
> 
> What I am writing next sounds perhaps a bit fancy, but in my opinion a
> normal table is itself a partition tree, made of one single member:
> itself.

That's what we discussed, but it seems that we ended up allowing regular
standalone tables (possibly in inheritance trees) only because it
*appeared* to work.  Alvaro already pointed out what appears to be a bug
in how we compute the value of level.  Instead of trying to fix it, I
agree we should just disallow tables that are not a partitioned
table/index or a partition (relispartition).  Maybe there won't be any use
cases, so we should change that while we still can.

So, maybe change the check in check_rel_can_be_partition() as follows:

    relkind = get_rel_relkind(relid);
    relispartition = get_rel_relispartition(relid);

    /* Only allow relation types that can appear in partition trees. */
    if (!relispartition &&
        relkind != RELKIND_PARTITIONED_TABLE &&
        relkind != RELKIND_PARTITIONED_INDEX)
        return false;

Thanks,
Amit

  29. Re: pg_partition_tree crashes for a non-defined relation

    Tom Lane <tgl@sss.pgh.pa.us> — 2019-03-01T04:50:16Z

    Amit Langote <Langote_Amit_f8@lab.ntt.co.jp> writes:
> On 2019/03/01 9:22, Michael Paquier wrote:
>> What I am writing next sounds perhaps a bit fancy, but in my opinion a
>> normal table is itself a partition tree, made of one single member:
>> itself.

> That's what we discussed, but it seems that we ended up allowing regular
> standalone tables (possibly in inheritance trees) only because it
> *appeared* to work.  Alvaro already pointed out what appears to be a bug
> in how we compute the value of level.  Instead of trying to fix it, I
> agree we should just disallow tables that are not a partitioned
> table/index or a partition (relispartition).

FWIW, I don't agree with Michael's suggestion above.  A plain table is
significantly different from a partitioned table with no children:
you can store rows in the former but not the latter, and you can add
partitions to the latter but not the former.  So conflating the two
doesn't seem likely to lead to any good outcome.

But, having said that, we've learned that it's generally bad for
catalog-query functions to fail outright just because they're pointed
at the wrong kind of catalog object.  So I think that what we want here
is for pg_partition_tree to return NULL or an empty set or some such
for a plain table, while its output for a childless partitioned table
should be visibly different from that.  I'm not wedded to details
beyond that idea.

			regards, tom lane

  30. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2019-03-01T07:58:29Z

    On Thu, Feb 28, 2019 at 11:50:16PM -0500, Tom Lane wrote:
> FWIW, I don't agree with Michael's suggestion above.  A plain table is
> significantly different from a partitioned table with no children:
> you can store rows in the former but not the latter, and you can add
> partitions to the latter but not the former.  So conflating the two
> doesn't seem likely to lead to any good outcome.

Okay, so the opinion moves into this direction.  What would be needed
is just something like the attached patch then based on Amit's
suggestion?  The routine names and comments looked fine to me so I
have not touched the surroundings, and comments are welcome.

> But, having said that, we've learned that it's generally bad for
> catalog-query functions to fail outright just because they're pointed
> at the wrong kind of catalog object.  So I think that what we want here
> is for pg_partition_tree to return NULL or an empty set or some such
> for a plain table, while its output for a childless partitioned table
> should be visibly different from that.  I'm not wedded to details
> beyond that idea.

Yep, that's the intention since cc53123.  I won't come back to return
an ERROR in any case.  Here is what the patch gives for childless
partitions FWIW:
=# CREATE TABLE ptif_test (a int, b int) PARTITION BY range (a);
CREATE TABLE
=# SELECT * FROM pg_partition_tree('ptif_test');
   relid   | parentrelid | isleaf | level
-----------+-------------+--------+-------
 ptif_test | null        | f      |     0
(1 row)    
--
Michael

  31. Re: pg_partition_tree crashes for a non-defined relation

    Tom Lane <tgl@sss.pgh.pa.us> — 2019-03-01T16:38:20Z

    Michael Paquier <michael@paquier.xyz> writes:
> On Thu, Feb 28, 2019 at 11:50:16PM -0500, Tom Lane wrote:
>> But, having said that, we've learned that it's generally bad for
>> catalog-query functions to fail outright just because they're pointed
>> at the wrong kind of catalog object.  So I think that what we want here
>> is for pg_partition_tree to return NULL or an empty set or some such
>> for a plain table, while its output for a childless partitioned table
>> should be visibly different from that.  I'm not wedded to details
>> beyond that idea.

> Yep, that's the intention since cc53123.  I won't come back to return
> an ERROR in any case.  Here is what the patch gives for childless
> partitions FWIW:
> =# CREATE TABLE ptif_test (a int, b int) PARTITION BY range (a);
> CREATE TABLE
> =# SELECT * FROM pg_partition_tree('ptif_test');
>    relid   | parentrelid | isleaf | level
> -----------+-------------+--------+-------
>  ptif_test | null        | f      |     0
> (1 row)    

Right, while you'd get zero rows out for a non-partitioned table.
WFM.

			regards, tom lane

  32. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2019-03-02T09:21:54Z

    On Fri, Mar 01, 2019 at 11:38:20AM -0500, Tom Lane wrote:
> Right, while you'd get zero rows out for a non-partitioned table.
> WFM.

Exactly.  I have committed a patch doing exactly that, and I have
added test cases with a partitioned table and a partitioned index
which have no partitions. 
--
Michael

  33. Re: pg_partition_tree crashes for a non-defined relation

    Amit Langote <langote_amit_f8@lab.ntt.co.jp> — 2019-03-04T01:44:10Z

    On 2019/03/02 18:21, Michael Paquier wrote:
> On Fri, Mar 01, 2019 at 11:38:20AM -0500, Tom Lane wrote:
>> Right, while you'd get zero rows out for a non-partitioned table.
>> WFM.
> 
> Exactly.  I have committed a patch doing exactly that, and I have
> added test cases with a partitioned table and a partitioned index
> which have no partitions.

Thanks for committing and adding me as an author.

Regards,
Amit

  34. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2019-03-04T03:01:35Z

    On Mon, Mar 04, 2019 at 10:44:10AM +0900, Amit Langote wrote:
> Thanks for committing and adding me as an author.

Sure.  A good portion of the changes suggested on the backend was
mainly yours, so that looked right to me. 
--
Michael

  35. Re: pg_partition_tree crashes for a non-defined relation

    Alvaro Herrera <alvherre@2ndquadrant.com> — 2019-03-04T18:56:00Z

    On 2019-Feb-28, Alvaro Herrera wrote:

> What about legacy inheritance?  I see that pg_partition_tree handles
> that case perfectly well -- it seems to return the complete hierarchy
> rooted at the given relation.  However, it seems odd that it works at
> all, don't you think?  Consider this:
> [...]
> I would opt for returning the empty set for legacy inheritance too.

I added tests to immortalize the current behavior for legacy inheritance
relations too.  Thanks!

-- 
Álvaro Herrera                https://www.2ndQuadrant.com/
PostgreSQL Development, 24x7 Support, Remote DBA, Training & Services

  36. Re: pg_partition_tree crashes for a non-defined relation

    Michael Paquier <michael@paquier.xyz> — 2019-03-04T22:41:22Z

    On Mon, Mar 04, 2019 at 03:56:00PM -0300, Alvaro Herrera wrote:
> I added tests to immortalize the current behavior for legacy inheritance
> relations too.  Thanks!

Makes sense, thanks.
--
Michael