How to find the minimum value for 1-many relationship? - openedge

Question

I want to find the minimum date value in a list of transactions that are associated with an investment. There are many transactions for one investment, clearly. How do I write this so that Progress will only give me the minimum transaction date? I get the minimum at the end of my list, but I do not want the list, just the minimum value.
FOR EACH ilinvest WHERE ilinvest.inv-num EQ 406885:
FOR EACH iltrans WHERE iltrans.reg-pin EQ ilinvest.reg-pin:
DISPLAY iltrans.tran-dt(MINIMUM).
END.
END.

Answer 1

If you have an index on the tran-dt field, you could do something like
FOR EACH ilinvest WHERE ilinvest.inv-num EQ 406885:
FOR EACH iltrans WHERE iltrans.reg-pin EQ ilinvest.reg-pin
BY iltrans.tran-dt ASCENDING:
// The iltrans.tran-dt value here is the lowest. Note that
// you may see the unknown value .
// Leave after getting the first record
LEAVE.
END.
END.

Answer 2

Thank you both for your help. Removing the ascending worked like a charm:
FOR EACH ilinvest:
FOR EACH iltrans WHERE iltrans.reg-pin EQ ilinvest.reg-pin
AND iltrans.acct-num EQ ilinvest.inv-num
BY iltrans.tran-dt:
iMin = iltrans.tran-dt.
LEAVE.
END.
END.

Answer 3

Beware if your date field has no value, ie the unknown value (?). If the unknown value sorts before or after other values depends on all sorts of black magic
Additionally, since I do not like leave, I prefer a while:
def var dt min as date no-undo.
for each ilinvest no-lock,
each iltrans
where iltrans.reg-pin = ilinvest.reg-pin
and iltrans.acct-num = ilinvest.inv-num
no-lock
by iltrans.tran-dt
while dtmin = ?:
if iltrans.trans-dt <> ? then
dtmin = iltrans.tran-dt.
end.