Function Help
Hello all,
I am working on a function to determine the date of the first saturday of
the month.
Currently I have:
CREATE FUNCTION first_saturday(date)
RETURNS date
AS '
Select CASE WHEN date_part(\'dow\',\'$1\'::DATE)=0 THEN date(\'$1\')+6
WHEN date_part(\'dow\',\'$1\'::DATE)=1 THEN date(\'$1\')+5
WHEN date_part(\'dow\',\'$1\'::DATE)=2 THEN date(\'$1\')+4
WHEN date_part(\'dow\',\'$1\'::DATE)=3 THEN date(\'$1\')+3
WHEN date_part(\'dow\',\'$1\'::DATE)=4 THEN date(\'$1\')+2
WHEN date_part(\'dow\',\'$1\'::DATE)=5 THEN date(\'$1\')+1
WHEN date_part(\'dow\',\'$1\'::DATE)=6 THEN date(\'$1\')+0
END'LANGUAGE 'sql'
I get an error that $1 is not a valid date. But I want that to be the
variable I enter...
what Am I doing wrong???
Brian C. Doyle writes:
CREATE FUNCTION first_saturday(date)
RETURNS date
AS '
Select CASE WHEN date_part(\'dow\',\'$1\'::DATE)=0 THEN date(\'$1\')+6
WHEN date_part(\'dow\',\'$1\'::DATE)=1 THEN date(\'$1\')+5
WHEN date_part(\'dow\',\'$1\'::DATE)=2 THEN date(\'$1\')+4
WHEN date_part(\'dow\',\'$1\'::DATE)=3 THEN date(\'$1\')+3
WHEN date_part(\'dow\',\'$1\'::DATE)=4 THEN date(\'$1\')+2
WHEN date_part(\'dow\',\'$1\'::DATE)=5 THEN date(\'$1\')+1
WHEN date_part(\'dow\',\'$1\'::DATE)=6 THEN date(\'$1\')+0
END'LANGUAGE 'sql'I get an error that $1 is not a valid date.
Don't quote the $1. E.g.,
WHEN date_part(\'dow\',$1)=6 THEN date($1)+0
The $1 etc. are not macros, they are identifiers representing a typed
expression.
--
Peter Eisentraut peter_e@gmx.net http://funkturm.homeip.net/~peter
That was it... I knew it was something simple.. Thanks Peter!!!
At 11:06 PM 9/24/01 +0200, Peter Eisentraut wrote:
Show quoted text
Brian C. Doyle writes:
CREATE FUNCTION first_saturday(date)
RETURNS date
AS '
Select CASE WHEN date_part(\'dow\',\'$1\'::DATE)=0 THEN date(\'$1\')+6
WHEN date_part(\'dow\',\'$1\'::DATE)=1 THEN date(\'$1\')+5
WHEN date_part(\'dow\',\'$1\'::DATE)=2 THEN date(\'$1\')+4
WHEN date_part(\'dow\',\'$1\'::DATE)=3 THEN date(\'$1\')+3
WHEN date_part(\'dow\',\'$1\'::DATE)=4 THEN date(\'$1\')+2
WHEN date_part(\'dow\',\'$1\'::DATE)=5 THEN date(\'$1\')+1
WHEN date_part(\'dow\',\'$1\'::DATE)=6 THEN date(\'$1\')+0
END'LANGUAGE 'sql'I get an error that $1 is not a valid date.
Don't quote the $1. E.g.,
WHEN date_part(\'dow\',$1)=6 THEN date($1)+0
The $1 etc. are not macros, they are identifiers representing a typed
expression.--
Peter Eisentraut peter_e@gmx.net http://funkturm.homeip.net/~peter---------------------------(end of broadcast)---------------------------
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"Brian" == Brian C Doyle <bcdoyle@mindspring.com> writes:
Brian> Hello all,
Brian> I am working on a function to determine the date of the first saturday
Brian> of the month.
Brian> Currently I have:
Brian> CREATE FUNCTION first_saturday(date)
Brian> RETURNS date
Brian> AS '
Brian> Select CASE WHEN date_part(\'dow\',\'$1\'::DATE)=0 THEN date(\'$1\')+6
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=1 THEN date(\'$1\')+5
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=2 THEN date(\'$1\')+4
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=3 THEN date(\'$1\')+3
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=4 THEN date(\'$1\')+2
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=5 THEN date(\'$1\')+1
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=6 THEN date(\'$1\')+0
Brian> END'LANGUAGE 'sql'
Brian> I get an error that $1 is not a valid date. But I want that to be the
Brian> variable I enter...
Brian> what Am I doing wrong???
Working too hard? :)
why not just:
My_Col + 6 - date_part('dow', My_Col)
Don't even need a function for that. :)
--
Randal L. Schwartz - Stonehenge Consulting Services, Inc. - +1 503 777 0095
<merlyn@stonehenge.com> <URL:http://www.stonehenge.com/merlyn/>
Perl/Unix/security consulting, Technical writing, Comedy, etc. etc.
See PerlTraining.Stonehenge.com for onsite and open-enrollment Perl training!
The main reason is that I am not quering a table with it as of yet...
Ultimately it will before query that is a Month do date query that does not
use a calender month
At 05:29 PM 9/24/01 -0700, Randal L. Schwartz wrote:
Show quoted text
"Brian" == Brian C Doyle <bcdoyle@mindspring.com> writes:
Brian> Hello all,
Brian> I am working on a function to determine the date of the first saturday
Brian> of the month.Brian> Currently I have:
Brian> CREATE FUNCTION first_saturday(date)
Brian> RETURNS date
Brian> AS '
Brian> Select CASE WHEN date_part(\'dow\',\'$1\'::DATE)=0 THEN date(\'$1\')+6
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=1 THEN date(\'$1\')+5
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=2 THEN date(\'$1\')+4
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=3 THEN date(\'$1\')+3
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=4 THEN date(\'$1\')+2
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=5 THEN date(\'$1\')+1
Brian> WHEN date_part(\'dow\',\'$1\'::DATE)=6 THEN date(\'$1\')+0
Brian> END'LANGUAGE 'sql'Brian> I get an error that $1 is not a valid date. But I want that to be the
Brian> variable I enter...
Brian> what Am I doing wrong???Working too hard? :)
why not just:
My_Col + 6 - date_part('dow', My_Col)
Don't even need a function for that. :)
--
Randal L. Schwartz - Stonehenge Consulting Services, Inc. - +1 503 777 0095
<merlyn@stonehenge.com> <URL:http://www.stonehenge.com/merlyn/>
Perl/Unix/security consulting, Technical writing, Comedy, etc. etc.
See PerlTraining.Stonehenge.com for onsite and open-enrollment Perl training!