IBM Mainframe Forum

HI

I need to create a file which takes 3 volser 's to fit the data in.But i need only 2000 records in each vol ser.
I now that if the i/p records are more then the o/p file will take more than one volser to hold the data.Each volser will hold more than 10k records .
But now i want each vol ser to take only 2k records ...

So Do we have any option to set volser to take these many no. of records from i/p n then take another volser for the next set of 2k records?

Hello and welcome to the forum,

If you are willing to create 3 different datasets on 3 separate volumes, you can do this easily using your sort product.

Define the control statements to copy 2000 records to the first 2 output files and the remainder to the 3rd output file.

I think i cannot use SORT b/c my JCL uses some other program to create the o/p file from the given 3 i/p files.
And I want the i/p name to be the same.

Hello,

And I want the i/p name to be the same.

You have to use the file name that is already associated with the input file(s).

I think i cannot use SORT b/c my JCL uses some other program to create the o/p file from the given 3 i/p files.

Sure you can. . . Run the existing process as-is. Then split the output file into 3 using the sort. . .

If writing to disk, size the primary allocation for 2000 records with 0 secondary and UNIT=(xxxxx,3) in the JCL.

If writing to tape, you can use Assembler to invoke DCB exits and force EOV after 2000 records.

The whole thing sounds ridiculous, though.

Hi dick

I have tried with SORT ..n created 3 VOLSER with same o/p file name.

But here we want the VOLSER shud have dependency ..like the second VOLSER shud have the dependency with 1st n so on.
If we split using SORT there wil not b any dependency rt? each volser will b seperate.

The O/p tape shud have 3 VOLSER n those 3 shud have dependency like wen we mount on system it shu show like

1st volser seq.no(=1)--------1 st VOLSER
1st volser seq.no(=2)--------2 nd VOLSER
1st volser seq.no(=3)--------3rd VOLSER


This is how it shud show.
This can be achieved only wen we create a file with 3 VOLSER...I have created like dat but wat we want is each VOLSER shud contain only abt 2k records to ease testing.


Hi Bill

The solution you have given may work.bt I dnt have aces to all those things.

Thank u

Hello,

Without doing something that is highly customized (like using the assembler that Bill mentioned), you won't create vol1, vol2, vol3 of the same dataset with 2000 records . . .

What you can do is create 3 dataset names on 3 separate volumes and put 2000 records on the first 2 and the remainder on the 3rd volume.

If you explain why this is even being considered, someone may have an alternative suggestion.

B/c we have been asked for testing accordingly.
But now i have created tapes with 3VOLSER bt not with 2k records in it n explained ,They have approved
But pls tel me how many records a VOLSER can hold n also the no.of records a file can hold on SYSDA.

But pls tel me how many records a VOLSER can hold n also the no.of records a file can hold on SYSDA.

Things that are needed to be known to answer this question:
1. What kind of VOLSER -- tape or disk?
2. If tape, what kind of tape -- 3490? 3590E? 3590J? Something else? is it compressed or not?
3. If disk, is is 3390 mod 3? mod 9? "mod 27"? "mod 54"? If not 3390, what is the device geometry?
4. Are the records fixed or variable length? How many bytes per record? How many records per block?
5. What are you calling "SYSDA"? Is this a physical device or VIO? If a physical device, see question number 3?

For example, unblocked fixed length 80-byte records on a 3390 allows 78 records per track which is 1170 records per cylinder and 3,901,950 on a mod 3 device (3335 cylinders). Blocked at 27920 bytes per block, you get 698 records per track, 10470 per cylinder and 34,917,450 records on a mod 3.

Hello,

Suggest you work to convince whoever wants this kind of info that this is not time well spent. . .

A better "thing" to calculate is for a given amount of data, how much media will it require (blocked optimally).