how to List the top ten customers form the given start date with greater cost/price of orders using progress 4gl - progress-4gl

Question

a single customer could have multiple orders and corresponding cost.
now the cost/price has to be added and according to the max price the customer's should be displayed.
i have just written the skeleton of the prog , please help me.
define temp-table ttorder
field ocust-num like order.cust-num
field oorder-num like order.order-num.
define variable i as int.
define variable a as int.
define temp-table ttorderln
field oprice like order-line.price.
for each order where order-date > 01/08/93 /*by cust-num*/ .
create ttorder .
assign
ttorder.ocust-num =cust-num
ttorder.oorder-num = order-num.
for each order-line where order-line.order-num = ttorder.oorder-num break by ocust-num.
i = order-line.price.
a = i + a.
/* display cust-num order-num a. */
if last-of (ttorder.ocust-num) then
do:
create ttorderln.
ttorderln.oprice=a.
end.
display ttorderln.
end.
end.

Answer 1

If you're looking for the top customers, you should use a temp table based on customers, not orders. Go through the orders and order lines, accumulating the amounts by customer.
define temp-table ttcust
field cust-num like order.cust-num
field order-tot as decimal
index idx1 cust-num
index idx2 order-tot.
define variable i as integer.
for each order where order-date > 01/08/93:
/* Find or create a customer temp record. */
find first ttcust where ttcust.cust-num = order.cust-num no-error.
if not available(ttcust) then
do:
create ttcust.
ttcust.cust-num = order.cust-num.
end.
/* Add up the order lines. */
for each order-line where order-line.order-num = order.order-num no-lock:
ttcust.order-tot = ttcust.order-tot + order-line.extended-price.
end.
end.
/* Display the top 10. */
for each ttcust by order-tot descending:
i = i + 1.
message "Cust: " ttcust.cust-num skip
"Total: " ttcust.order-tot view-as alert-box.
if i >= 10 then leave.
end.