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Re: Strange error message when reference non-existent column foo."count"

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Patrick Krecker wrote
> I encountered this today and it was quite surprising:
> 
> select version();
>                                                version
> 
> ------------------------------------------------------------------------------------------------------
>  PostgreSQL 9.3.5 on x86_64-unknown-linux-gnu, compiled by gcc (Ubuntu
> 4.8.2-19ubuntu1) 4.8.2, 64-bit
> 
> create table foo as (select generate_series(1,3));
> 
> As expected, the following fails:
> 
> select count from foo;
> ERROR:  column "count" does not exist
> LINE 1: select count from foo;
>                ^
> But if I change the syntax to something I thought was equivalent:
> 
> select foo."count" from foo;
>  count
> -------
>      3
> (1 row)
> 
> It works! This was quite surprising to me. Is this expected behavior, that
> you can call an aggregate function without any parentheses (I can't find
> any other syntax that works for count() sans parentheses, and this
> behavior
> doesn't occur for any other aggregate)?

That fact that this is an aggregate function is beside the point - the
syntax works for any function.

The following two expressions are equivalent:

count(foo) = foo.count

I do not immediately recall where this is documented but it is.  It should
probably be documented or cross-referenced at:

http://www.postgresql.org/docs/9.3/static/sql-syntax-calling-funcs.html#SQL-SYNTAX-CALLING-FUNCS-NAMED

but alas that is not so.

The basic idea is to hide the function invocation and allow for
syntactically similar derived columns to be described.

(goes looking)

4.2.6 - the note therein:
http://www.postgresql.org/docs/9.3/static/sql-expressions.html#FIELD-SELECTION
pointing to 35.4.3
http://www.postgresql.org/docs/9.3/static/xfunc-sql.html#XFUNC-SQL-COMPOSITE-FUNCTIONS

This relies on the rule that every table automatically has an implicit type
created and so a "composite function" can act on that type.  The "foo."
reference in your example is technically referring to the type "foo" and not
the table "foo".

David J.




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