split i/p to o/p based on the input

anishkannath · New User Joined: 23 Oct 2007 Posts: 5 Location: Hyderabad

Hi,

I have a requirement to split the records into various o/p files based on the input value. We dont know how many o/p files will be required as this purely depends on a indicator in input record

Sample:- ( here splitting records based on 4 th byte. So here there will be 3 o/p files) Some cases we may have 5 or even 10 o/p files depneding on input.

SORTIN
AAA1
AAA1
AAA2
AAA3

O/P
File - 1
AAA1
AAA1

File - 2
AAA2

File - 3
AAA3

Can we allocate these o/p files based on the input value ?

Thanks,
Anish kannath

Skolusu · Posted: Tue Mar 30, 2010 2:14 am

anishkannath,

If the indicator is just 1 byte numeric then you can only have a max of 10 different value (0 thru 9). Allocate 10 files and delete all the empty files.

Did you look at the smart dfsort tricks

# Split a file to n output files dynamically
# Five ways to split a data set

www.ibm.com/support/docview.wss?rs=114&uid=isg3T7000094

anishkannath · New User Joined: 23 Oct 2007 Posts: 5 Location: Hyderabad

Hi,

My indicator is actually 8 bytes.. I have just indicates as one byte in the example. So we can have more number of records.

Thanks,
Anish Kannath

gcicchet · Posted: Tue Mar 30, 2010 3:06 am

Hi,

are you saying there could be 100 million different values ?

Good luck and I hope the 8 bytes is not a PD field.

Gerry

anishkannath · New User Joined: 23 Oct 2007 Posts: 5 Location: Hyderabad

I never said we can have that many records , only thing is that i am not sure abt how many o/p files needs to be created as this depends on the input. BTB that field is an alpha numeric one.

I saw an example using internal reader and am trying to use that. Just want to know if there is any other way rather than submitting a JCL thru internal reader.

Skolusu · Posted: Tue Mar 30, 2010 3:31 am

gcicchet · Posted: Tue Mar 30, 2010 5:19 am

Hi Kolusu,

senjay · Active User Joined: 10 May 2007 Posts: 147 Location: India

With all respect to Skolusu and though my answer is not directly related to OP, here is the answer for Gcicchet's question.

Since the OP can choose from 26 alphabets (A-Z) and 10 numerals (0-9), taking the number of digits as X (which is here 26 + 10 = 36), and the number of bytes the OP has to form is Y (which is here 8), the maximum possible number of different values for the 8 byte field, with repeated characters allowed, is X^Y = 36^8 = 2821109907456

gcicchet · Posted: Tue Mar 30, 2010 2:41 pm

Hi Senthil,

that was how I arrived at 100 million, mine was based on numeric only

Skolusu · Posted: Tue Mar 30, 2010 10:01 pm