Here's a nice equivalent to Oracle
"count(distinct fieldname || fieldname)"
for when you need to know how many unique combinations of two fields are in a dataset.

It concatenates the values of the two fields in a DEFINE, then uses SUM CNT.DST to get the count of distinct combinations. Note that the result is defined as I/9, else it returns ***** as a value.


DEFINE FILE EXAMPLETABLE
CONCAT_FIELDS/A70 = FIELD1 || FIELD2;
END
TABLE FILE EXAMPLETABLE
SUM CNT.DST.CONCAT_FIELDS/I9
END

Sorry - I didn't figure this out for CAR, but am sure it could be done there. I'm putting this entry in for next time I need it and can't remember/find what I did!

Thanks to an entry in the IBI TechSupport for the basic idea, which I extended to use a concatenated DEFINE.

Cheers!

But use weak catenation:
CONCAT_FIELDS/A70 = FIELD1 | FIELD2;

lest rows like

field1  field2
======  ======
A       BC 
AB      C

be confounded.