Re: Dateadd function error

[Thyleen Tenney <ttenney@TALBOTCORP.COM>]

Usually fixed by declaring your variable as a date.  I find that if I need a
date variable in SQR, I declare it so in the beginning.

T

-----Original Message-----
From: Discussion of SQR, Brio Technology's database reporting language
[mailto:SQR-USERS@list.iex.net]On Behalf Of jim nowlin
Sent: Monday, July 02, 2001 12:58 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