On Mon, 22 Sep 2008 12:48:16 -0500, Brain <[EMAIL PROTECTED]> wrote:
>Hello.
>
>Could some one please help me mt to get the mainframe id when i execute the
>cobol program, this id will hen be inserted in a table to see who has inserted
>the record.
>
>Thanks,
>RajeevV
If you want a "pure" COBOL program, then it is going to be a bit complicated
to code and might someday fail because IBM could change things. You'd need
to do a lot of "chain chasing". Instead, I would strongly suggest that you
use an assembler subroutine which uses the IAZXJSAB macro. This is the
correct, and documented, way to do it. The routine is very small:
USERID CSECT
USERID AMODE ANY
USERID RMODE ANY
SAVE (14,12),,USERID
LR 11,15
USING USERID,11
CNOP 0,4
BAS 2,*+76 BRANCH AROUND SAVE AREA
DC 18A(0) SAVE AREA
ST 13,4(,2) SAVE REGISTER 13
ST 2,8(,13) SAVE BACK CHAIN
LR 13,2 LOAD SAVE AREA POINTER
L 2,0(,1) POINT TO RETURN AREA FOR USERID
IAZXJSAB READ, X
USERID=(2)
L 13,4(,13) POINT TO OLD SAVE AREA
RETURN (14,12),T,RC=0 RETURN
LTORG *
IAZJSAB
IHAASCB
IHAASSB
IHAPSA
IKJTCB
IHASTCB
END USERID
You call it via:
...
77 RUNNING_USER_ID PIC X(8).
...
CALL "USERID" USING RUNNING_USER_ID.
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