Question

I have this query which groups the results by ORDER#.

SELECT ORDER#, MAX(SHIPDATE - ORDERDATE) DELAYDAYS FROM ORDERS GROUP BY ORDER#;

My goal is to return the single ORDER# that has the longest shipping delay. Not the entire set. How can I accomplish this?

Tried this too. Does not work. This gives me the single column, but not the ORDER#.

SELECT MAX(X.DELAYDAYS) FROM
   (SELECT ORDER#, MAX(SHIPDATE - ORDERDATE) DELAYDAYS 
    FROM ORDERS GROUP BY ORDER#) X;

Answer 1

To get a list of all orders that tie for the longest delay:

with order_delay as (
  select order#,
           max( shipdate - orderdate ) as delayDays
      from orders
     group by order#
),
ranked_order_delay as (
  select order#,
         delayDays,
         rank() over( order by delayDays desc ) as delayRank
    from order_delay
)
select order#, delayDays from ranked_order_delay
 where delayRank=1
;

Use the query below to get a single order, taking the lowest order# in case of a tie for the longest delay. (should always strive for a determinant result) I believe this is the optimal solution, requiring only one pass through the data.

select min(order#) keep (dense_rank last order by (shipdate-orderdate)) as order#,
       max(shipdate-orderdate) delaydays
  from orders

To take the highest order# in case of a tie for longest delay, then simply use max(order#) instead.

Edit – I knew there was a better way. It just took a wile. I was hung up thinking I had to determine the max delay for each order#, but then I realized it wasn’t necessary for this query.