Thank you, Razzak, unfortunately I'm not using a date variable I'm using a
DATETIME variable which the help indicates is valid:
Where arg is a value that has either a DATE or DATETIME data type, IDAY
returns the integer day of the month for a particular date.
Returns the integer month for a particular date, where arg is a value that
has a DATE or DATETIME data type.
Where arg is a value that has a DATE or DATETIME data type, IYR4 always
returns a four-digit integer year of date,
I'm not actually using the old IYR - I found that because I left off the 4
when I was first checking.
This used to work in v7.6:
SET VAR vNowDT = .#NOW
SET VAR vTxtDay = (CTXT( (IDAY(.vNowDT)) ))
SET VAR vTxtMon = (CTXT( (IMON(.vNowDT)) ))
SET VAR vTxtYear = (CTXT( (IYR4(.vNowDT)) ))
IF (SLEN(.vTxtDay)) < 2 THEN
SET VAR vTxtDay = ('0' + .vTxtDay)
ENDIF
IF (SLEN(.vTxtMon)) < 2 THEN
SET VAR vTxtMon = ('0' + .vTxtMon)
ENDIF
SET VAR vNowText = +
(.vTxtYear + '/' + .vTxtMon + '/' + .vTxtDay & (SGET( (CTXT(
(TEXTRACT(.vNowDT)) )), 5, 1)) )
The day & month fail in v9.0 so I get, as an example: 2010/00/00 23:11
This then gets added to some more text...
Regards,
Alastair.
--------------------------------------------------
From: "A. Razzak Memon" <[email protected]>
Sent: Wednesday, June 09, 2010 9:03 PM
To: "RBASE-L Mailing List" <[email protected]>
Subject: [RBASE-L] - Re: v9.0 INT day/month/year
On 6/9/10, Alastair Burr <[email protected]> wrote:
----------------------------Am I imagining this or just going nuts:
R>SET VAR vIntYear INTEGER = (IYR4(.#Now)) ; SHOW VAR vIntYear
2010
R>SET VAR vIntYear INTEGER = (IYR(.#Now)) ; SHOW VAR vIntYear
0
R>SET VAR vIntMon INTEGER = (IMON(.#Now)) ; SHOW VAR vIntMon
0
R>SET VAR vIntDay INTEGER = (IDAY(.#Now)) ; SHOW VAR vIntDay
0
Alastair,
Here's how:
SET VAR vIntYear INTEGER = (IYR(.#DATE))
SET VAR vIntMon INTEGER = (IMON(.#DATE))
SET VAR vIntDay INTEGER = (IDAY(.#DATE))
Very Best R:egards,
Razzak.
