Cobol Code to Count the total no of orphan records

mahesh pillai · New User Joined: 04 Oct 2011 Posts: 6 Location: india

I need to write a COBOL code. where i need to find out the total count of the records which do not have relation to the child.

for ex:- If there are 4 records A,B.C and D. where A is linked with B as
A-B, A is linked with C as A-C and A is linked with D as A-D. where B,C and D are child record of the A. I need to write a query where it counts the number of records available in A where it is not related with any of the child records B or C or D.

Can anyone help me out??

Robert Sample · Posted: Fri Oct 07, 2011 4:41 pm

I doubt you're going to get a lot of help based on what you've posted. You seem to be talking about a database -- IMS, perhaps? -- but did not make that clear. Records in files do not have parent / child relationships.

Why not write the code to go through all the A records, and set a flag for each unique value, if you find a child record then turn off the flag. If you hit the end of the unique value records and the flag is still set, you found one without a child.

mahesh pillai · New User Joined: 04 Oct 2011 Posts: 6 Location: india

am sorry.. am talking about IDMS database. In that we are having the child records for particular A record.

Escapa · Posted: Fri Oct 07, 2011 4:47 pm

Nic Clouston · Posted: Fri Oct 07, 2011 4:58 pm

obtain first in set p
check for child set(s)
if none exists increment counter
obtain next within p

Now, it is 15 years since I did IDMS but I think that should give you a clue as to how to proceed.

dick scherrer · Posted: Fri Oct 07, 2011 7:22 pm

Hello and welcome to the forum,

You will get more useful replies if you post your questions in the appropriate part(s) of the forum . . .

This has been moved to the IDMS part of the forum.

robert_wood · Posted: Fri Oct 07, 2011 7:24 pm

IF SET EMPTY condition.

Or OLQ (if you have it) for an online quick and dirty.

However... if you do not know how to do this, I would suggest that you should not be let loose on an IDMS database.