step3 wants to run only once in a month
Posted: Tue May 22, 2012 12:16 pmIf i have 3 steps in a job and i want to run step1, step2 daily and step3 wants to run only once in a month how?
IBM Mainframe Forum
Mainframe Technical Support Forums
https://www.ibmmainframeforum.com/
step3 wants to run only once in a month
Page 1 of 1
Posted: Tue May 22, 2012 12:16 pmRe: jcl
Posted: Tue May 22, 2012 12:23 pmPresumably your site has a Schduler? Split into two jobs. Talk to your Scheduler people explaining what you want.
Re: jcl
Posted: Tue May 22, 2012 1:02 pmgood posting habits suggest the use "descriptive" titles.
using a stupid title ( same as the forum section ) will lower the chances of getting good answers
if You do not want spend time on thinking and writing a proper title
there is no reason for us to waste our time helping You
using a stupid title ( same as the forum section ) will lower the chances of getting good answers
if You do not want spend time on thinking and writing a proper title
there is no reason for us to waste our time helping You
Re: jcl
Posted: Tue May 22, 2012 5:54 pmAs BillyBoyo suggests, this is an issue better resolved by using your shop's production scheduler. If you are going to run your JCL once a day, you need to determine the day of month using a utility like the program that follows, which returns the day of the month as a return code. Then it is a simple matter to arrange your JCL to run a step on the first day of the month.
MDAY CSECT
USING *,4
SAVE (14,4),,*
LR 4,15
TIME DEC
* TRANSLATE PL4'0CYYDDD' TO BINARY
XC 64(4,13),64(13)
ST 1,68(,13)
CVB 3,64(,13)
* ISOLATE CYY AND DDD
SR 2,2
D 2,=F'1000'
* CONVERT CYY TO A TRUE YEAR
LA 3,1900(,3)
* DETERMINE IF LEAP YEAR. THIS IS THE
* FULL GREGORIAN CALENDAR LEAP YEAR TEST
LR 14,3
LR 15,3
N 14,=A(B'11')
JNZ MD0100
* YEAR IS DIVISIBLE BY 4, TEST IF CENTURY YEAR
D 14,=F'100'
LTR 14,14
JNZ MD0200
* TEST IF CENTURY YEAR DIVISIBLE BY 400
N 15,=A(B'11')
JZ MD0200
* IT'S NOT A LEAP YEAR, TEST IF DAY OF YEAR
* IS IN JANUARY OR FEBRUARY
MD0100 C 2,=A(31+28)
JNH MD0200
* CHANGE NON-LEAP YEAR DAY OF YEAR TO
* LEAP YEAR DAY OF YEAR
LA 2,1(,2)
* LOAD DAY OF MONTH INTO REG 15
MD0200 SR 15,15
IC 15,T1(2)
RETURN (14,4),T,RC=(15)
* TABLE TO TRANSLATE DAY OF YEAR IN A LEAP
* YEAR TO DAY OF MONTH
T1 DC 32AL1(*-T1)
T2 DC 29AL1(*-T2+1)
T3 DC 31AL1(*-T3+1)
T4 DC 30AL1(*-T4+1)
T5 DC 31AL1(*-T5+1)
T6 DC 30AL1(*-T6+1)
T7 DC 31AL1(*-T7+1)
T8 DC 31AL1(*-T8+1)
T9 DC 30AL1(*-T9+1)
T10 DC 30AL1(*-T10+1)
T11 DC 30AL1(*-T11+1)
T12 DC 30AL1(*-T12+1),0D'0'
LTORG ,
DC 0D'0'
END MDAY
USING *,4
SAVE (14,4),,*
LR 4,15
TIME DEC
* TRANSLATE PL4'0CYYDDD' TO BINARY
XC 64(4,13),64(13)
ST 1,68(,13)
CVB 3,64(,13)
* ISOLATE CYY AND DDD
SR 2,2
D 2,=F'1000'
* CONVERT CYY TO A TRUE YEAR
LA 3,1900(,3)
* DETERMINE IF LEAP YEAR. THIS IS THE
* FULL GREGORIAN CALENDAR LEAP YEAR TEST
LR 14,3
LR 15,3
N 14,=A(B'11')
JNZ MD0100
* YEAR IS DIVISIBLE BY 4, TEST IF CENTURY YEAR
D 14,=F'100'
LTR 14,14
JNZ MD0200
* TEST IF CENTURY YEAR DIVISIBLE BY 400
N 15,=A(B'11')
JZ MD0200
* IT'S NOT A LEAP YEAR, TEST IF DAY OF YEAR
* IS IN JANUARY OR FEBRUARY
MD0100 C 2,=A(31+28)
JNH MD0200
* CHANGE NON-LEAP YEAR DAY OF YEAR TO
* LEAP YEAR DAY OF YEAR
LA 2,1(,2)
* LOAD DAY OF MONTH INTO REG 15
MD0200 SR 15,15
IC 15,T1(2)
RETURN (14,4),T,RC=(15)
* TABLE TO TRANSLATE DAY OF YEAR IN A LEAP
* YEAR TO DAY OF MONTH
T1 DC 32AL1(*-T1)
T2 DC 29AL1(*-T2+1)
T3 DC 31AL1(*-T3+1)
T4 DC 30AL1(*-T4+1)
T5 DC 31AL1(*-T5+1)
T6 DC 30AL1(*-T6+1)
T7 DC 31AL1(*-T7+1)
T8 DC 31AL1(*-T8+1)
T9 DC 30AL1(*-T9+1)
T10 DC 30AL1(*-T10+1)
T11 DC 30AL1(*-T11+1)
T12 DC 30AL1(*-T12+1),0D'0'
LTORG ,
DC 0D'0'
END MDAY
Re: jcl
Posted: Tue May 22, 2012 9:13 pmWhat about a simple COBOL-program !?
Identification Division.
Program-Id. DAYPGM.
Procedure Division.
Move Function Current-date(7:2) to Return-Code
GoBack
.
Program-Id. DAYPGM.
Procedure Division.
Move Function Current-date(7:2) to Return-Code
GoBack
.
Re: step3 wants to run only once in a month
Posted: Wed May 23, 2012 7:03 pmI hate to be t"that guy" but what about midnight? When is this going to run?