Group By / Having correlated subquery - sql

Question

I am trying to solve this query, I'm really lost
least one order in which the quantity they ordered was greater than
the average quantity of all other orders.
My tables are:
Customer
Cust_ID | CustName | Region | Phone
Orders
Ordernum | Cust_ID | Item_ID | Quantity
Vendor
Vendor_ID | Item_ID | Costs | Region
stock
Item_ID | Description | Price | On_hand
Database schema is:
Customer (Cust_ID, CustName, Region, Phone)
Orders (Ordernum, Cust_ID, Item_ID, Quantity) Foreign key Cust_ID
references Customer Not Null, On Delete Restrict
Foreign key Item_ID references Stock Not Null, On Delete Restrict
Stock (Item_ID, Description, Price, On_hand)
Vendor (Vendor_ID, Item_ID, Cost, Region)
Foreign key Item_ID references Stock Not null, On Delete Restrict
Here is what I have tried so far. What am I doing wrong?
select custname, phone, count(distinct(item_id)), sum(quantity)
from customer c, orders o
Where c.cust_id= o.cust_id and count (o.ordernum) >= 2
group by cust_id
having sum (quantity) >
(select avg(quantity)
from orders o2
where o.item_id != o2.item_id)
order by custname;
I re-wrote my code and this is what I came up with. I'm really lost:
select c1.custname, phone, count(distinct(o.item_id)), sum(quantity) as quantity
from customer c1, orders o
Where c1.cust_id= o.cust_id
group by c1.cust_id, custname, phone, quantity
Having 2>=
(select count(o1.item_id)
From orders o1
where c1.cust_ID = o1.cust_ID)
AND sum(quantity) >
(select AVG(quantity)
from orders o2
Where c1.cust_ID != o2.cust_ID AND o.Item_ID = o2.Item_ID)
order by custname;

Answer 1

Try this query
select a.cust_id, count(distinct Item_ID) as itemOrderd,
sum(b.quantity) as sum
from
customer a
inner join
orders b
on a.cust_id=b.cust_id
group by a.cust_id
having count(distinct ordernum) > 1
and max(b.quantity) > avg(b.quantity)
Fiddle
| CUST_ID | ITEMORDERD | SUM |
|---------|------------|-----|
| 1 | 2 | 61 |

Answer 2

I think it will be:
SELECT MAX(c.CustName) as CustName,
MAX(c.Phone) as Phone,
COUNT(DISTINCT o.Item_ID) as CountOfDistinctItems,
SUM(o.Quantity) as SumOfQuantity
FROM Customer as c
JOIN Orders as o on c.Cust_ID=o.Cust_ID
JOIN (SELECT Ordernum, SUM(Quantity) as Order_q_sum
FROM Orders
GROUP BY Ordernum) as oq
ON o.Ordernum=oq.Ordernum
GROUP BY c.Cust_ID
HAVING COUNT(DISTINCT o.Ordernum)>=2
AND
MAX(oq.Order_q_sum)>
(
SELECT AVG(SumQuantity) FROM
(SELECT SUM(Quantity) SumQuantity FROM Orders GROUP BY OrderNum) as t1
)

Answer 3

HOPE THIS CLOSER TO YOUR NEEDS
SELECT x.custname,
x.phone,
Sum(x.itemcnt),
Sum(x.quantity)
FROM (SELECT c.custname,
c.phone,
1 AS itemCnt,
Sum(o.quantity),
Isnull((SELECT Avg(y.quantity)
FROM orders y
WHERE y.cust_id = o.cust_id), 0.00) AvgQty
FROM customer c
LEFT OUTER JOIN orders o
ON c.cust_id = o.cust_id
GROUP BY c.custname,
c.phone,
o.item_id) x
GROUP BY x.custname,
x.phone
HAVING Sum(x.itemcnt) > 1
AND Sum(x.quantity) > x.avgqty