IBM MainFrame

First need to sort both files

READ FILE1

READ FILE2

PERFORM UNTIL EOF = 'Y'

IF FILE1-ID > FILE2-ID THEN

READ FILE2 AT END MOVE 'Y' TO EOF

ELSE IF FILE1-ID < FILE2-ID

READ FILE1 AT END MOVE 'Y' TO EOF

ELSE IF FILE1-ID = FILE2-ID

WRITE FILE3

READ FILE1 AT END MOVE 'Y' TO EOF

READ FILE2 AT END MOVE 'Y' TO EOF

END-PERFORM

Linkage section is used to accept data from outside the program.either its parm part of Jcl or Call from any other programs,they pass data into called program thru linkage section only.

//STEP1 EXEC PGM=SORT

//SYSPRINT DD SYSOUT=*

//SYSOUT DD SYSOUT=*

//SORTIN DD DSN=INPUT.DATA,DISP=SHR

//OUTDATA DD DSN=OUTPUT.DATA,DISP=(NEW,CATLG,DELETE)

.

.

.

//SYSIN DD *

OPTION COPY

OUTFIL FNAMES=OUTDATA,SAMPLE=2

THIS WILL COPY ALL EVEN RECORDS

-------------------------------------------

FOR ODD RECORDS

OUTFIL FNAMES=OUTDATA,STARTREC=1,SAMPLE=2

By using DFSORT we can eliminate the duplicates

//Sysin dd *

sort fields=(1,5,ch,A)

sum fields=

/*

//

in the above dfsort we use sum fields= empty then it

eliminate the duplicates

if u want all dupli.in one file u can use

sum fields = xsum........

File-aid tool is online tool (application) used for

processing files using online screens. File-aid utility is

used in jcl for processing files in batch.

Move highvalues to the key of the VSAM file and the issue Readnext record command. Then give Readprev command. This will read the last record. In VSAM, there is one command HURBA, Using that we can read the last record, If you want to know more about go to IBM RED BOOKS. You will get the solution.

by using the HURBA we can get the max records by using the CKD count key data we will get the last record

In IBM Cobol

1) In IBM's Language Environment you can dynamically

acquire storage (look up HEAP in the manual). The storage

you acquire you can use as you like. Can be an array,

therefore a dynamic array. It is a "work-around" as dynamic

arrays are not directly supported.

2) 01 SINGLE-DIM-ARRAY

05 SDA-ITEM PIC X OCCURS 30.

A single-dimension array of thirty one-byte characters.

3) 01 MULTI-DIM-ARRAY.

05 MDA-FIRST-DIM OCCURS 20.

07 MDA-SECOND-DIM OCCURS 30 PIC X.

A multi-dimension array of 20 times 30 one-byte characters.

Comp is binary.its the type how system stores ur data.

pic s9(1)-->pics9(4) --occupies 2bytes

pic s9(5)---pic s9(8) ----4 bytes

pic s9(9)---pics9(18)-----8 bytes

comp-3 is for packed decimal data where each character occupies half byte and sign is stored in last nibble.

As picture clause for variable b is x(2), which means that

it can accept both numeric and characters of length 2, but

the care we should take here is that we should enclose them

in the single quotes, therefore to my expectation option b

is write.

//PROC1 PROC

//STEP1 EXEC PGM=IEFBR14

//DD1 DD DSN=XYZ, DISP=SHR

//SYSPRINT DD SYSOUT=*

//SYSIN DD DUMMY

// PEND

//JJJ JCLLIB ORDER=PROCLIB PATH

//PROC2 PROC

//STEP2 EXEC PROC1

//JOBNAME JOB PARAMETERS

//JJJ JCLLIB ORDER=PROCLIB PATH

//PROC3 PROC

//STEP2.DD1 DSN=NEW NAME,DISP=SHR

THIS IS D WAY....

THIS PROG. WILL DO NOTHING IT'S JUST AN EXAMPLE