function with unknown params

giozh wrote:

i've declared and implemented e function like:

CREATE OR REPLACE FUNCTION acquistoLotto(m_g INTEGER, grossista VARCHAR(20),
produttore BIGINT, costo INTEGER, dat DATE, descr VARCHAR(120), num_prod
INTEGER) RETURNS VOID AS $$

but when i'm trying to call it like

select acquistoLotto(0, 'grossista', 52187073424, 10, 22/1/2013, 'ciao ciao
ciao', 10);

You need to enclose the date in quotes, too.

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On Tue, Jul 9, 2013 at 10:00 PM, giozh <giozh@yahoo.it> wrote:

i've declared and implemented e function like:

CREATE OR REPLACE FUNCTION acquistoLotto(m_g INTEGER, grossista
VARCHAR(20),
produttore BIGINT, costo INTEGER, dat DATE, descr VARCHAR(120), num_prod
INTEGER) RETURNS VOID AS $$

but when i'm trying to call it like

select acquistoLotto(0, 'grossista', 52187073424, 10, 22/1/2013, 'ciao ciao
ciao', 10);

Seems you missing on how to pass date field. Try this it will work.

postgres=# select acquistoLotto(0, 'grossista', 52187073424, 10,
'1-22-2013', 'ciao ciaociao', 10);
acquistolotto
---------------

(1 row)

or

postgres=# select acquistoLotto(0, 'grossista', 52187073424, 10,
'1-22-2013'::date, 'ciao ciaociao', 10);
acquistolotto
---------------

(1 row)

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Regards,
Raghavendra
EnterpriseDB Corporation
Blog: http://raghavt.blogspot.com/

ok, it works. But why on error message i had that two unknown data type? if
was an error on date type, why it don't signal that?

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You passed in:

22/1/2013

Which is 22 divided by 1, divided by 2013 - which is an integer..

On Tue, Jul 9, 2013 at 10:17 AM, giozh <giozh@yahoo.it> wrote:

giozh <giozh@yahoo.it> wrote:

ok, it works. But why on error message i had that two unknown
data type? if was an error on date type, why it don't signal
that?

Because PostgreSQL treats a quoted literal as being of unknown type
-- each of these could have matched a parameter of any type, and
would have been treated as a literal of the needed type.  On the
other hand, if you match up the parameters of *known* types against
the function's signature, you will see that the fifth parameter is
declared to be a date, but the value you supplied is an integer --
because 22 divided by 1 divided by 2013 evaluates to an integer of
zero.  Since there can be many different functions with any given
name, with different parameter types, it doesn't try to match up
parameters and guess which one might be the wrong type -- for all
it knows the problem is that there's a missing function with a
signature compatible with the values you specified.  It tells you
the types it sees.

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Kevin Grittner
EnterpriseDB: http://www.enterprisedb.com
The Enterprise PostgreSQL Company

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