IBM MAINFRAME HELP & SUPPORT FORUMS

annu zacson · Joined: 19 Aug 2008 Posts: 5 Location: Cochin

I have to create an output file in a semicolon delimited format. How to use STRING command to perform this:
For eg:
Assume the input data is given below:
customer id Account No First Name Last Name
001 2345 Mark Parker
002 4567 Any James
003 Missing field John Hanes

The output file should be:
001;2345;Mark;Parker;;
002;4567;Any;James;;
003;;John;Hanes;;

I need a semicolon at the end of each field plus at the end of each record.
If any fields are missing, it should be denoted a semicoln as specified in the third row.
please let me know how to do this using STRING command in cobol?

dbzTHEdinosauer · Joined: 20 Oct 2006 Posts: 1639 Location: germany

i assume a missing field is all spaces.
you can start with this:

Code:
string field-1 delimited by spaces
';'
field-2 delimited by spaces
';'
into receiving-field
end-string

Sambhaji · Joined: 16 Feb 2007 Posts: 267 Location: Pune, India

how you r seperating fields from input file?
is it fixed length input file?

or is there a space in between two fields of input file?

dick scherrer · Joined: 23 Nov 2006 Posts: 8733 Location: 221 B Baker St

Hello,

From the posted example, the input appears fixed-length. "Separation" is accomplished by using the input field names.

annu zacson · Joined: 19 Aug 2008 Posts: 5 Location: Cochin

Thanks for helping me out.
My input file is of fixed length and looks like this:
amex_cust_id A/N 12
cust_acct_ky N 11
national_id A/N 20
gender_cd A 01
last_nm A/N 30
cm_first_nm A/N 20
cm_mid_initl_nm A/N 20

my second doubt is , how will i remove the trailing spaces? for example, for the field last_nm, if the last_nm is only 5 chars, the actual length is 30 chars, so how ill i remove the last 25 chars so that i can concatenate that field in semicolon delimited format with the next field cm_first_nm?

Sambhaji · Joined: 16 Feb 2007 Posts: 267 Location: Pune, India

Can you please show how your output will look like?
I didn't get what exactly you want?

annu zacson · Joined: 19 Aug 2008 Posts: 5 Location: Cochin

Suppose my input file is like this (please see attachment) with the structure declared above:
Here Nat_id is declared as 20 CHARS. but the value is N12345. So i need to remove the trailing spaces ie 14 chars and concatenate with the next field.
so the output will look like this:
37222222222;111111111111;N12345;M;PARKER;A;MARK;;

The output file is also of the same record length as input file.

dbzTHEdinosauer · Joined: 20 Oct 2006 Posts: 1639 Location: germany

annu,

the example STRING instruction that I provided does just that.

I am afraid that you will just have to look at the documentation explaining the COBOL STRING instruction,

Sambhaji · Joined: 16 Feb 2007 Posts: 267 Location: Pune, India

hey.... :evil:
code given by Dick Brenholtz does exactly same you want....

annu zacson · Joined: 19 Aug 2008 Posts: 5 Location: Cochin

That code is giving an error as it is telling STRING function cannot be found. So I wrote like this:
string field-1 ';'
field-2 ';' �;�
into receiving-field DELIMITED BY SPACES
end-string

now its working..
Thanks.