Regression in Postgres 17?

Στις 22/10/24 18:54, ο/η Colin 't Hart έγραψε:

Hi,

This works in Postgres 15:

pg15> create function json_test(out value text, out json jsonb)
returns record
language sql
as
$$
  select null::text, null::jsonb;
$$
;
CREATE FUNCTION
pg15> select * from json_test();
┌───────┬──────┐
│ value │ json │
├───────┼──────┤
│       │      │
└───────┴──────┘
(1 row)

In Postgres 17 trying to create the function yields an error:

pg17> create function json_test(out value text, out json jsonb)
returns record
language sql
as
$$
  select null::text, null::jsonb;
$$
;
ERROR:  syntax error at or near "jsonb"
LINE 1: create function json_test(out value text, out json jsonb)

Am I doing something wrong? Or is this a regression?

Do this instead :

create function json_test(out value text, out jsonparam jsonb)
returns record
language sql
as
$$
 select null::text, null::jsonb;
$$
;
CREATE FUNCTION

apparently json is a reserved word (now) and won't be accepted as
function parameter name.

On 10/22/24 08:54, Colin 't Hart wrote:

Hi,

This works in Postgres 15:

pg15> create function json_test(out value text, out json jsonb)
returns record
language sql
as
$$
  select null::text, null::jsonb;
$$
;
CREATE FUNCTION
pg15> select * from json_test();
┌───────┬──────┐
│ value │ json │
├───────┼──────┤
│       │      │
└───────┴──────┘
(1 row)

In Postgres 17 trying to create the function yields an error:

pg17> create function json_test(out value text, out json jsonb)
returns record
language sql
as
$$
  select null::text, null::jsonb;
$$
;
ERROR:  syntax error at or near "jsonb"
LINE 1: create function json_test(out value text, out json jsonb)

Am I doing something wrong? Or is this a regression?

Yes you are doing something wrong, naming an argument with a type
name(json) is not a good idea. Guessing you got caught by this:

https://www.postgresql.org/docs/16/sql-keywords-appendix.html

JSON non-reserved reserved

https://www.postgresql.org/docs/17/sql-keywords-appendix.html

JSON non-reserved (cannot be function or type) reserved

Thanks,

Colin

--
Adrian Klaver
adrian.klaver@aklaver.com

Adrian Klaver <adrian.klaver@aklaver.com> writes:

In Postgres 17 trying to create the function yields an error:

pg17> create function json_test(out value text, out json jsonb)
returns record
...
ERROR:  syntax error at or near "jsonb"
LINE 1: create function json_test(out value text, out json jsonb)

Am I doing something wrong? Or is this a regression?

Yes you are doing something wrong, naming an argument with a type
name(json) is not a good idea.

The actual problem is that the SQL standards committee invented
some bizarre syntax that we couldn't parse without making JSON
a partially-reserved word. It still works as a type name, but
in this particular syntax where it's not initially clear which
names are type names, you lose. Double-quote the argument name,
or name it something other than "json".

regards, tom lane

On Tue, Oct 22, 2024 at 6:03 PM Achilleas Mantzios
<a.mantzios@cloud.gatewaynet.com> wrote:

Στις 22/10/24 18:54, ο/η Colin 't Hart έγραψε:
This works in Postgres 15:
Do this instead :
create function json_test(out value text, out jsonparam jsonb)

...

apparently json is a reserved word (now) and won't be accepted as function parameter name.

Or properly double-quote the param name:

```
ddevienne=> show server_version;
server_version
----------------
17.0
(1 row)
^
ddevienne=> create function json_test(out value text, out "json"
jsonb) returns record language sql as 'select null::text,
null::jsonb;';
CREATE FUNCTION
ddevienne=> select * from json_test();
value | json
-------+------
|
(1 row)
```