Thread: [BUGS] BUG #14666: Question on money type as the key of partitioned table
[BUGS] BUG #14666: Question on money type as the key of partitioned table
From
tianbing@highgo.com
Date:
The following bug has been logged on the website: Bug reference: 14666 Logged by: tian bing Email address: tianbing@highgo.com PostgreSQL version: 10beta1 Operating system: CentOS 6.4(64bits) Description: Hi, When I use the money type as the key to create the partition table as follows: postgres=# create table test(m money) partition by list(m); CREATE TABLE postgres=# create table test_1 partition of test for values in (10); CREATE TABLE Partition bounds without apostrophe can be createed, but it store the null value,not '10' value. In fact, Correct grammar for creating partition is with apostrophe like this: postgres=# create table test_1 partition of test for values in ('10'); But the first creating partition without apostrophe should report an error like "ERROR: operator does not exist: money = integer" as adding a check constraint. Looking forward to your reply. -- Sent via pgsql-bugs mailing list (pgsql-bugs@postgresql.org) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-bugs
Re: [BUGS] BUG #14666: Question on money type as the key ofpartitioned table
From
Michael Paquier
Date:
On Tue, May 23, 2017 at 10:45 PM, <tianbing@highgo.com> wrote: > When I use the money type as the key to create the partition table as > follows: > > postgres=# create table test(m money) partition by list(m); > CREATE TABLE > postgres=# create table test_1 partition of test for values in (10); > CREATE TABLE > > Partition bounds without apostrophe can be created, but it store the null > value,not '10' value. > > In fact, Correct grammar for creating partition is with apostrophe like > this: > postgres=# create table test_1 partition of test for values in ('10'); > > But the first creating partition without apostrophe should report an error > like "ERROR: operator does not exist: money = integer" as adding a check > constraint. > > Looking forward to your reply. This looks like a justified complain to me, so added to the open item list. Those INSERTs should work: =# insert into test_1 values (10); ERROR: 23514: new row for relation "test_1" violates partition constraint DETAIL: Failing row contains ($10.00). LOCATION: ExecConstraints, execMain.c:2037 =# insert into test values (10); ERROR: 23514: no partition of relation "test" found for row DETAIL: Partition key of the failing row contains (m) = ($10.00). LOCATION: ExecFindPartition, execMain.c:3343 =# insert into test values ('10'); ERROR: 23514: no partition of relation "test" found for row DETAIL: Partition key of the failing row contains (m) = ($10.00). LOCATION: ExecFindPartition, execMain.c:3343 The shape of the partition definition looks broken. -- Michael -- Sent via pgsql-bugs mailing list (pgsql-bugs@postgresql.org) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-bugs
Re: [BUGS] BUG #14666: Question on money type as the key ofpartitioned table
From
Amit Langote
Date:
On Wed, May 24, 2017 at 10:54 AM, Michael Paquier <michael.paquier@gmail.com> wrote: > On Tue, May 23, 2017 at 10:45 PM, <tianbing@highgo.com> wrote: >> When I use the money type as the key to create the partition table as >> follows: >> >> postgres=# create table test(m money) partition by list(m); >> CREATE TABLE >> postgres=# create table test_1 partition of test for values in (10); >> CREATE TABLE >> >> Partition bounds without apostrophe can be created, but it store the null >> value,not '10' value. >> >> In fact, Correct grammar for creating partition is with apostrophe like >> this: >> postgres=# create table test_1 partition of test for values in ('10'); >> >> But the first creating partition without apostrophe should report an error >> like "ERROR: operator does not exist: money = integer" as adding a check >> constraint. >> >> Looking forward to your reply. > > This looks like a justified complain to me, so added to the open item > list. Those INSERTs should work: > =# insert into test_1 values (10); > ERROR: 23514: new row for relation "test_1" violates partition constraint > DETAIL: Failing row contains ($10.00). > LOCATION: ExecConstraints, execMain.c:2037 > =# insert into test values (10); > ERROR: 23514: no partition of relation "test" found for row > DETAIL: Partition key of the failing row contains (m) = ($10.00). > LOCATION: ExecFindPartition, execMain.c:3343 > =# insert into test values ('10'); > ERROR: 23514: no partition of relation "test" found for row > DETAIL: Partition key of the failing row contains (m) = ($10.00). > LOCATION: ExecFindPartition, execMain.c:3343 > The shape of the partition definition looks broken. Yep, looks like a bug; will look into it. Thanks for adding to the open items list. Regards, Amit -- Sent via pgsql-bugs mailing list (pgsql-bugs@postgresql.org) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-bugs
Re: [BUGS] BUG #14666: Question on money type as the key ofpartitioned table
From
Michael Paquier
Date:
On Wed, May 24, 2017 at 4:22 PM, Amit Langote <amitlangote09@gmail.com> wrote: > Yep, looks like a bug; will look into it. Cool. Just mentioning... I found surprising to see that failing with a partition error: =# create table test(m money) partition by list(m); CREATE TABLE =# create table test_1 partition of test for values in (10::money); ERROR: 42601: syntax error at or near "::" Likely the price to pay to get a minimal feature in 10. -- Michael -- Sent via pgsql-bugs mailing list (pgsql-bugs@postgresql.org) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-bugs
Michael Paquier <michael.paquier@gmail.com> writes: > Just mentioning... I found surprising to see that failing with a > partition error: > =# create table test(m money) partition by list(m); > CREATE TABLE > =# create table test_1 partition of test for values in (10::money); > ERROR: 42601: syntax error at or near "::" Well, this is a misunderstanding of the syntax: the values in the values list have to be simple literals, not expressions. However, it's arguably not your fault, because the CREATE TABLE reference page says that the list elements are "bound_literal"s, a term which AFAICS is defined nowhere. I'm inclined to change that to something like IN ( { numeric_literal | string_literal | NULL } [, ...] ) which seems less likely to leave the reader wondering what is meant. regards, tom lane -- Sent via pgsql-bugs mailing list (pgsql-bugs@postgresql.org) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-bugs
tianbing@highgo.com writes: > When I use the money type as the key to create the partition table as > follows: > postgres=# create table test(m money) partition by list(m); > CREATE TABLE > postgres=# create table test_1 partition of test for values in (10); > CREATE TABLE > Partition bounds without apostrophe can be createed, but it store the null > value, not '10' value. That's not actually what it's doing. A look into pg_class shows that while, for an integer partitioning column, you'd get something like this for relpartbound: test1p | {PARTITIONBOUND :strategy l :listdatums ( {CONST :consttype 23 :consttypmod -1 :constcollid 0 :constlen 4 :constbyval true:constisnull false :location 54 :constvalue4 [ 10 0 0 0 0 0 0 0 ]}) :lowerdatu ms <> :upperdatums <>} in the case at hand you'd get test2p | {PARTITIONBOUND :strategy l :listdatums ( {FUNCEXPR :funcid 3811 :funcresulttype 790 :funcretset false :funcvariadic false:funcformat 2 :funccollid 0 :inputcollid0 :args ({CONST :consttype 23 :constty pmod -1 :constcollid 0 :constlen 4 :constbyval true :constisnull false :location54 :constvalue 4 [ 10 0 0 0 0 0 0 0 ]}) :location-1}) :lowerdatums <> :upperda tums <>} that is, what we have is a run-time coercion of integer to money. The partitioning code utterly fails to consider that what it might get from the partition list syntax is not a constant --- but since casts are not required to be immutable, it might not. This is exacerbated by the fact that subsequent code naively assumes that the elements of PartitionBoundSpec.listdatums are Consts, without any checking. It's a wonder you don't get runtime crashes. (You might if the partition column type is pass-by-ref, I suspect.) And I'm unimpressed by the fact that this assumption is nowhere documented, too. What we need to do here (at least in the short term) is throw an error if we don't get a simple Const out of const-simplification. I'm not sure if we need a separate error message for that case, or if we can get away with just re-using the existing text about "specified value cannot be cast to type ...". The point here would be that the cast exists but is not immutable. Maybe use the same primary message but explain that in an errdetail? regards, tom lane -- Sent via pgsql-bugs mailing list (pgsql-bugs@postgresql.org) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-bugs
Re: [BUGS] BUG #14666: Question on money type as the key ofpartitioned table
From
Amit Langote
Date:
Thanks a lot for getting to fixing this bug before I did. On 2017/05/29 6:53, Tom Lane wrote: > tianbing@highgo.com writes: >> When I use the money type as the key to create the partition table as >> follows: > >> postgres=# create table test(m money) partition by list(m); >> CREATE TABLE >> postgres=# create table test_1 partition of test for values in (10); >> CREATE TABLE > >> Partition bounds without apostrophe can be createed, but it store the null >> value, not '10' value. > > That's not actually what it's doing. A look into pg_class shows that > while, for an integer partitioning column, you'd get something like this > for relpartbound: > > test1p | {PARTITIONBOUND :strategy l :listdatums ( > {CONST :consttype 23 :consttypmod -1 :constcollid 0 :constlen 4 :constbyval true > :constisnull false :location 54 :constvalue 4 [ 10 0 0 0 0 0 0 0 ]}) :lowerdatu > ms <> :upperdatums <>} > > in the case at hand you'd get > > test2p | {PARTITIONBOUND :strategy l :listdatums ( > {FUNCEXPR :funcid 3811 :funcresulttype 790 :funcretset false :funcvariadic false > :funcformat 2 :funccollid 0 :inputcollid 0 :args ({CONST :consttype 23 :constty > pmod -1 :constcollid 0 :constlen 4 :constbyval true :constisnull false :location > 54 :constvalue 4 [ 10 0 0 0 0 0 0 0 ]}) :location -1}) :lowerdatums <> :upperda > tums <>} > > that is, what we have is a run-time coercion of integer to money. > The partitioning code utterly fails to consider that what it might > get from the partition list syntax is not a constant --- but since > casts are not required to be immutable, it might not. Oops, that's right. I had failed to consider that the cast step might introduce a coercion function call. > This is exacerbated by the fact that subsequent code naively assumes > that the elements of PartitionBoundSpec.listdatums are Consts, without > any checking. It's a wonder you don't get runtime crashes. (You might > if the partition column type is pass-by-ref, I suspect.) And I'm > unimpressed by the fact that this assumption is nowhere documented, too. > > What we need to do here (at least in the short term) is throw an error > if we don't get a simple Const out of const-simplification. I'm not > sure if we need a separate error message for that case, or if we can > get away with just re-using the existing text about "specified value > cannot be cast to type ...". The point here would be that the cast > exists but is not immutable. Maybe use the same primary message > but explain that in an errdetail? The message as committed in 76a3df6e5e621928fbf0cddf347e16a62e9433ec looks on point to me. Thanks again. Regards, Amit -- Sent via pgsql-bugs mailing list (pgsql-bugs@postgresql.org) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-bugs