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Okay, I changed my program to use UDate instead of *Date and it works fine, thanks everyone. But, as someone stated, UDate is the job date, not the system date. Users have been known to use // DATE mmddyy (A S/36 emulation way to change to job date) so I think the system date would be better. I think I'll try this TIME, which I've never used, and Peter, I can call you Peter, can't I?, are you saying that TIME gives the system time and not the job time? Regards, Jim Langston phcoffin@us.ibm.com wrote: > > So, without going into a whole rigamarole of createing a data structure, > breaking it > > down into MM DD CC YY and moving the parts, how do I initialize this > field to > > todays date? And I should be creating it as Zones 6,0 right? > > DTODAY S L > * ... > C TIME TODAY > C *MDY MOVE TODAY ParmAsOf > > I love date fields. > > Peter H. Coffin > Database Application Management Tools Developer > phcoffin@us.ibm.com > T/L 665-6298 (Outside: 414-223-6298) > http://w3.wwmkt.ibm.com/americasmi +--- | This is the RPG/400 Mailing List! | To submit a new message, send your mail to RPG400-L@midrange.com. | To subscribe to this list send email to RPG400-L-SUB@midrange.com. | To unsubscribe from this list send email to RPG400-L-UNSUB@midrange.com. | Questions should be directed to the list owner/operator: david@midrange.com +---
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