Re: Dateadd function error

[Bob Helm <Bob.Helm@MEDIWARE.COM>]

I believe you have to declare $sunday as a date variable, like:

declare-variable
  date $sunday
end-declare

That ought to do it, hopefully.

B
The Programmer formerly known as Bob
Mediware Information Systems Inc.
(913) 307-1045
Bob.Helm@Mediware.com <mailto:Bob.Helm@Mediware.com>








-----Original Message-----
From: jim nowlin [mailto:jim_nowlin@NAVIGATORS.ORG]
Sent: Monday, July 02, 2001 1:58 PM
To: SQR-USERS@list.iex.net
Subject: Dateadd function error


Brio product and version number:  SQR V4.3.4
Operating system and version number: True64 (Compaq) 5.0A
Database name and version number: Oracle 7.3.4.0.0

This is the first time I tried to use "dateadd". The code I'm using
follows:

  let $sunday = dateadd($parm-end-dt,'DAY',6)

I get the following error in the sqrlog:

Error on line 118:
   (SQR 4045) Function or operator 'dateadd' requires date argument.

Any ideas on what I'm doing wrong?

If I can get this to work, can I use a negitive number to subtract six
days?

Thanks for the help,
Jim Nowlin
jim_nowlin@navigators.org