Thread: How to add days to date
Hi:
I have problem trying to add same days to a date.
I have this select :
SELECT '2006-08-01' + FP.carencia + ( FP.prazo * ( MFP.parcela - 1 )) AS
vencimento
FROM fi_mov_formas_pagamento MFP
LEFT OUTER JOIN fi_formas_pagamento FP ON ( MFP.idformapagamento =
FP.idformapagamento AND MFP.idempresa = FP.idempresa )
INNER JOIN hd_cabecalho HDC ON ( MFP.idmovimento = HDC.idhelpdesk AND
MFP.idempresa = HDC.idempresa )
WHERE MFP.idmovimento = 1
AND MFP.idempresa = 1
AND MFP.idtipomovimentacao = 1
'2006-08-01' + FP.carencia + ( FP.prazo * ( MFP.parcela - 1 )) --> This
results in
EX :
'2006-08-01' + 30 + ( 7 * ( 3 - 1 )) ==> '2006-08-01' + 44
All my trys fails.
Can you help me ?
Thanks in advance.
Alejandro Michelin Salmon
On Tue, Aug 15, 2006 at 10:10:27PM -0300, Alejandro Michelin Salomon ( Adinet ) wrote:
> EX :
> '2006-08-01' + 30 + ( 7 * ( 3 - 1 )) ==> '2006-08-01' + 44
>
> All my trys fails.
The error message hints at what's wrong:
test=> SELECT '2006-08-01' + 30 + (7 * (3 - 1));
ERROR: invalid input syntax for integer: "2006-08-01"
PostgreSQL doesn't know that the untyped string is supposed to be
interpreted as a date. Use a cast:
test=> SELECT '2006-08-01'::date + 30 + (7 * (3 - 1));
?column?
------------
2006-09-14
(1 row)
or
test=> SELECT CAST('2006-08-01' AS date) + 30 + (7 * (3 - 1));
?column?
------------
2006-09-14
(1 row)
--
Michael Fuhr
Alejandro Michelin Salomon ( Adinet ) wrote: > Hi: > > I have problem trying to add same days to a date. > > '2006-08-01' + FP.carencia + ( FP.prazo * ( MFP.parcela - 1 )) --> This > results in > > EX : > '2006-08-01' + 30 + ( 7 * ( 3 - 1 )) ==> '2006-08-01' + 44 Looks like you could use the interval type here too. I don't know what the above is supposed to do, but I'd prefer to write it like: '2006-08-01'::date + '1 month'::interval + (3-1) * '1 week'::interval Or: '2006-08-01'::date + INTERVAL '1 month' + (3-1) * INTERVAL '1 week' It at least saves you the trouble of determining how long what month takes, and it handles DST changes correctly. The drawback is that it's not a linear data type, which can cause some trouble if you need to convert values from your application to intervals. For details, have a look at the documentation: http://www.postgresql.org/docs/8.1/static/datatype-datetime.html -- Alban Hertroys alban@magproductions.nl magproductions b.v. T: ++31(0)534346874 F: ++31(0)534346876 M: I: www.magproductions.nl A: Postbus 416 7500 AK Enschede // Integrate Your World //
Michael :
I change my query to this :
SELECT CAST( '2006-08-01' AS DATE ) + FP.carencia + ( FP.prazo * (
MFP.parcela - 1 )) AS vencimento
FROM fi_mov_formas_pagamento MFP
LEFT OUTER JOIN fi_formas_pagamento FP ON ( MFP.idformapagamento =
FP.idformapagamento AND MFP.idempresa = FP.idempresa )
INNER JOIN hd_cabecalho HDC ON ( MFP.idmovimento = HDC.idhelpdesk AND
MFP.idempresa = HDC.idempresa )
WHERE MFP.idmovimento = 1
AND MFP.idempresa = 1
AND MFP.idtipomovimentacao = 1
And i run ok now.
Thanks for your help.
Alejandro
-->-----Mensagem original-----
-->De: Michael Fuhr [mailto:mike@fuhr.org]
-->Enviada em: terça-feira, 15 de agosto de 2006 22:17
-->Para: Alejandro Michelin Salomon ( Adinet )
-->Cc: Pgsql-General
-->Assunto: Re: [GENERAL] How to add days to date
-->
-->
-->On Tue, Aug 15, 2006 at 10:10:27PM -0300, Alejandro Michelin
-->Salomon ( Adinet ) wrote:
-->> EX :
-->> '2006-08-01' + 30 + ( 7 * ( 3 - 1 )) ==> '2006-08-01' + 44
-->>
-->> All my trys fails.
-->
-->The error message hints at what's wrong:
-->
-->test=> SELECT '2006-08-01' + 30 + (7 * (3 - 1));
-->ERROR: invalid input syntax for integer: "2006-08-01"
-->
-->PostgreSQL doesn't know that the untyped string is supposed
-->to be interpreted as a date. Use a cast:
-->
-->test=> SELECT '2006-08-01'::date + 30 + (7 * (3 - 1));
--> ?column?
-->------------
--> 2006-09-14
-->(1 row)
-->
-->or
-->
-->test=> SELECT CAST('2006-08-01' AS date) + 30 + (7 * (3 - 1));
--> ?column?
-->------------
--> 2006-09-14
-->(1 row)
-->
-->--
-->Michael Fuhr
-->