Example 8: Renumber and Insert Records into a Library Member

In this example, two blocks (three logical records per block) are inserted into an existing member, and the member is placed in the existing partitioned data set. A portion of the output member is also renumbered.

  //UPDATE   JOB  ...
  //STEP1    EXEC PGM=IEBUPDTE,PARM=MOD
  //SYSPRINT DD  SYSOUT=A
  //SYSUT1   DD  DSNAME=PDS,UNIT=disk,DISP=(OLD,KEEP),
  //             VOLUME=SER=111112
  //SYSUT2   DD  DSNAME=PDS,UNIT=disk,DISP=(OLD,KEEP),
  //             VOLUME=SER=111112
  //SYSIN    DD  *
  ./  CHANGE NAME=RENUM,LIST=ALL,LEVEL=01,SOURCE=0,SEQFLD=(765,783)
  ./  NUMBER SEQ1=AA015,NEW1=AA020,INCR=5,INSERT=YES
 
  (Data statement 1)
  (Data statement 2)
  (Data statement 3)
 
  ./  NUMBER SEQ1=AA030,INSERT=YES
 
  (Data statement 4)
  (Data statement 5)
  (Data statement 6)
  (Data statement 7, sequence number AA035)
 
  /*

The control statements are as follows:

SYSUT1 and SYSUT2 DD define the partitioned data set PDS.
SYSIN DD defines the control data set, which follows in the input stream.
The CHANGE function statement identifies the input member RENUM. The entire member is listed in the message data set.
The NUMBER detail statements specify the insert operations (INSERT=YES) and control the renumbering operation.
Data statements 1, 2, 3, and 4, 5, 6 are the blocks of logical records to be inserted. Because they contain blank sequence numbers, sequence numbers are assigned when the data statements are inserted.
Data statement 7, because it contains a sequence number, stops the insert operation. The sequence number is identical to the number on the next record in the old master data set; consequently, data statement 7 will replace the equally numbered old master record in the output data set.

The existing logical records in this example have sequence numbers AA010, AA015, AA020, AA025, AA030, AA035, AA040, AA045, AA050, BB010, and BB015. The insertion and renumbering operations are performed as follows:

Data statement 1 is assigned sequence number AA020 (NEW1=AA020) and inserted after existing logical record AA015 (SEQ1=AA015).
Data statements 2 and 3 are assigned sequence numbers AA025 and AA030 (INCR=5) and are inserted after data statement 1.
Existing logical records AA020, AA025, and AA030 are assigned sequence numbers AA035, AA040, and AA045, respectively.
Data statement 4 is assigned sequence number AA050 and inserted. (The SEQ1=AA030 specification in the second NUMBER statement places this data statement after existing logical record AA030, which has become logical record AA045.)
Data statements 5 and 6 are assigned sequence numbers AA055 and AA060 and are inserted after data statement 4.
Existing logical record AA035 is replaced by data statement 7, which is assigned sequence number AA065.
The remaining logical records in the member are renumbered until logical record BB010 is encountered. Because this record has a sequence number higher than the next number to be assigned, the renumbering operation is ended.

Table 1 shows existing sequence numbers, data statements inserted, and the new sequence numbers.

Table 1. Reordered sequence numbers
Sequence Numbers and Data Statements Inserted	New Sequence Numbers
AA010	AA010
AA015	AA015
Data statement 1	AA020
Data statement 2	AA025
Data statement 3	AA030
AA020	AA035
AA025	AA040
AA030	AA045
Data statement 4	AA050
Data statement 5	AA055
Data statement 6	AA060
AA035 (Data statement 7)	AA065
AA040	AA070
AA045	AA070
AA050	AA075
BB010	BB010
BB015	BB015