let $x = datenow()

You should make sure that $x is a date variable, not a string variable. You do that in the setup block:

begin-setup
declare variable
date $x
end-declare
end-setup

Alternately, you could write:

let $x = datetostr(datenow(), ‘mm/dd/yyyy hh:mi:ss’)

Your note indicates that this is working properly. I assume you insert a row with the $x variable placed in a date, datetime, or string field. The insert should work with all three field types, but you would handle the fields differently when reading them back into an SQR program.

What type of field is &AUDIT_STAMP? Since you used to_char successfully, I assume it is a date or datetime field (which is a distinction in PeopleTools, not in the Oracle database). In that case, you can use to_char, edit, or datetostr to convert it to a string. You would not use strtodate() – that is a function that starts with a string, and you are starting with a date.

If &AUDIT_STAMP is a string, and you are storing it as a date, you may need datetostr() to format a string to output that date.

I’ve hit a problem inserting date values into SQL Server 2005 db – I’m using SQR and am reading values (Termination and Contract End Dt’s) from one table and inserting them into another (interface) table – when there is a value, I have no issue, but when the source table, and therefore the target variable, contains a NULL value, the data is inserted (or interpreted by SQL server) as 01/01/1900.

Any Ideas on how to get around this?

Thanks for the help!

If you are storing the date in a variable, like this:

let $now = datenow()

you need be declare the variable as a date variable, like this:

begin-setup
declare-variable
date $now
end-declare
end-setup

If you don’t declare $now as a date variable, SQR will make it a string variable. When you set $now equal to datenow(), SQR will convert the date to a string that looks like a date in your default format. Then, when you set a database date field to that value, SQR will convert the string to datetime that is 0 seconds after midnight.

You may find it easier to use the built-in database variable for system date and time. In Oracle, it is SYSDATE. You could write something like this:

begin-sql
update tablename
set datefield = sysdate
where keyfield = ‘X’
end-sql

I have a question about the use of the DATENOW()function in PeopleSoft SQR. In my pursuit of answers I came across this blog. I’m not sure asking this question is appropriate, but here goes.

I need to get the current datetime and update a field in a table. When I use the DATENOW() function I am getting the current date with time of 12:00:00 AM. Any ideas about what is occurring here

Thanks,
Kevin

The datediff function only works with date variables. You have to declare them in the setup section like this:

begin-setup
declare-variable
date $eff_dte $pay_end_dt
end-declare
end-setup

Then you need to set their values properly. If you are getting values from a database table, this is sufficient:

let $eff_dte = &EFFDT
let $pay_end_dt = &PAY_END_DT

But if you get them from any other source, you have to do something like this:

let $eff_dte = strtodate(’05/21/2009′, ‘mm/dd/yyyy’)
let $pay_end_dt = strtodate(’30-APR-2009′, ‘dd-mon-yyyy’)

This seems like a lot of work, but I think it’s worth it. Then your datediff will work and you can write:

if #date_diff < 6

I assume your “if ‘datediff’ < 6″ was a typo. You were comparing a string and a number.

Let #Date_diff = datediff ($Eff_Dte, $Pay_End_Dt, $MONTH)
If ‘datediff’ < 6

I am looking to compare employee start date vs. current pay period end date, to check whether a specific earn type was used within 6 months of their start date.

These are the errors I am getting:

Error on line 620:
(SQR 4045) Function or operator ‘datediff’ requires date argument.

Error on line 620:
(SQR 4045) Function or operator ‘datediff’ requires date argument.

Error on line 621:
(SQR 4048) Function or operator ‘<’ must be a string or date argument.

Thanks, Robert.