Answer / shriram supalwar, dharmabad

//jobcard
//step1 exec pgm=iebgener
//sysprint dd sysout=*
//sysut1 dd dsn=proj.test.file1,disp=shr
// dd dsn=proj.test.file2,disp=shr
//sysut2 dd dsn=proj.output,disp=shr
//sysin dd dummy
//if (step1.rc eq 0) then
//step2 exec pgm=sort
//sysprint dd sysout=*
//sortin dd dsn=proj.output,disp=shr
//sortout dd dsn=proj.sortout,disp=shr
//sortxsum dd dsn=proj.duplicate,disp=shr
//sysin dd *
sort fields=(1,5,ch,a)
sum fields=none,xsum
/*

Now, all these records are copied to proj.output
and out of those records the duplicate records are copied to
proj.duplicate

Correct me if i am wrong

Answer / aditi

hi,

I have also faced the same question.

as we have the two seperate input files.
Nitika, U have taken only one input file so how we sort it
using only one file?

Answer / nitika

Use SORTXSUM utility in JCL to get the output.
//JOB1 JOB ,, CLASS = A
//STEP1 EXEC PGM=SORT
//SORTIN DD DSN=TEST.PS.INPUT
//SORTOUT DD DSN=TEST.PS.OUTPUT
//SORTXSUM DD DSN=TEST.PS.OUTPUT
//SYSIN DD *
SORT FIELDS = (1,1,CH,A)
SUM FIELDS = (NONE,XSUM)
/*

Now SUM FIELDS = NONE means no duplicates in the output file
(TEST.PS.OUTPUT)and XSUM means duplicates, in the two files
being matched, should be sent to the file mentioned in
SORTXSUM ddname.

So, output file will have 1,4.

Answer / karthik

@aditi n nikita

i think we can use ibm utility iebcompare

Answer / mdv

@Shriram Supalwar, Dharmabad

You don't need the IEBGENER STEP.

Your second step sort can have both the files concatenated
as one and given as Input.


Thanks

Answer / abhusha

Use join keys
//STEP01 EXEC PGM=SORT
//SYSPRINT DD SYSOUT=*
//SYSOUT DD SYSOUT=*
//SORTJNF1 DD DSN=INPUT.FILE.ONE,DISP=SHR
//SORTJNF2 DD DSN=INPUT.FILE.TWO,DISP=SHR
//SORTOUT DD DSN=OUTPUT.FILE,
// DISP=(NEW,CATLG,DELETE),
// SPACE=(CYL,(100,200),RLSE),
// DCB=(RECFM=FB,BLKSIZE=0,LRECL=80)
//SYSIN DD *
SORT FIELDS=COPY
JOINKEYS FILES=F1,FIELDS=(1,10,A)
JOINKEYS FILES=F2,FIELDS=(1,10,A)
REFORMAT FIELDS=(F1:1,71,F2:1,9)
/*

In SPACE,what is directory,RLSE

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