yes,,, that is what the printout shows,, I setup a printer file in the program,, and printed the contents of the paramters, right before I seton LR. The dates are originally entered as 010196,, and 060103,, when the julian routine runs,, it says the first 3 bytes are the century,year,, ie 096 for 1996,, and 103,, for 2003, the last 3 bytes are the julian day of the year..and in the program the paramters are defined as 6/0. They come from the LDA, loaded by the cl,, and converted,, and put into the parameters. And NO,, I do not have any idea how to debug a CLP.
The acutal clp looks like this..
DCL VAR(&RFDAT) TYPE(*CHAR) LEN(6)
DCL VAR(&TFDAT) TYPE(*CHAR) LEN(6)
SNDUSRMSG MSG('ENTER MONTH BEGIN DATE - MMDDYY') +
TOMSGQ(*) MSGTYPE(*INQ) MSGRPY(&RFDAT)
SNDUSRMSG MSG('ENTER MONTH ENDING DATE - MMDDYY') +
TOMSGQ(*) MSGTYPE(*INQ) MSGRPY(&TFDAT)
CHGDTAARA DTAARA(*LDA (1 6)) VALUE(&RFDAT)
CHGDTAARA DTAARA(*LDA (7 6)) VALUE(&TFDAT)
CALL PGM(LABDATE) PARM(&RFDAT &TFDAT)
CLRPFM FILE(HCCCUSTOM/LABEQP)
OVRDBF FILE(F0911) TOFILE(*LIBL/F0911) SHARE(*YES)
OPNQRYF FILE((F0911)) OPTION(*INP) +
QRYSLT('GLOBJ *EQ "6323 " *AND GLDGJ *EQ +