Thanks Robert. I did some online finding and found that this job will work:
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//STEP01 EXEC PGM=IEBCOPY
//SYSUT1 DD DISP=SHR,DSN=input.pds
//SYSUT2 DD DISP=(,CATLG),DSN=output.pds,SPACE=(TRK,(50,50))
// UNIT=3390
//SYSPRINT DD SYSOUT=*
//SYSIN DD *
COPY INDD=SYSUT1,OUTDD=SYSUT2
//
But then I looked at the manual for IEBCOPY,
http://publibz.boulder.ibm.com/cgi-bin/ ... 0113105507 and the first example is confusing:
"In this example, a partitioned data set (DATASET5) is copied from one disk volume to another. Figure 17 shows the input and output data sets before and after processing.
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//COPY JOB ...
//JOBSTEP EXEC PGM=IEBCOPY
//SYSPRINT DD SYSOUT=A
//SYSUT1 DD DSNAME=DATASET5,UNIT=disk,VOL=SER=111113,
// DISP=SHR
//SYSUT2 DD DSNAME=DATASET4,UNIT=disk,VOL=SER=111112,
// DISP=(NEW,KEEP),SPACE=(TRK,(5,1,2))
The control statements are discussed below:
SYSUT1 DD defines a partitioned data set, DATASET5, that contains two members (A and C).
SYSUT2 DD defines a new partitioned data set, DATASET4, that is to be kept after the copy operation. Five tracks are allocated for the data set; two blocks are allocated for directory entries.
Because the partitioned data set has only two members, SYSUT3 and SYSUT4 DD are not needed.
Because the input and output data sets are identified as SYSUT1 and SYSUT2, the SYSIN data set is not needed. The SYSUT1 data set will be copied in full to the SYSUT2 data set. After the copy operation is finished, DATASET4 will contain the same members that are in DATASET5. However, there will be no embedded, unused space in DATASET4. If you are copying a PDSE, the processing is the same, except that there is no embedded, unused space in a PDSE."
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I could not understand why they say like this,"
Because the partitioned data set has only two members, SYSUT3 and SYSUT4 DD are not needed."? Can you please help in knowing this?