SQL query for latest Order Status in MySQL - mysql

Question

We are tracking order statuses sent by our shipping partners via a webhook fired by them. The webhook adds a row everytime its fired and therefore each order has multiple rows associated with it.
Structure of the table
enter image description here
We are trying to create a sql query to get the following
Find the last received row for a 'awb'. Get the current_status in that row. If the current_status is any of 'PICKUP EXCEPTION', 'OUT FOR PICKUP', 'PICKUP RESCHEDULED' then find the row with the first occurrence of these statuses for that specific 'awb'
Check number of days between first occurrence and last occurrence of those statuses for the awb
and output the awbs that have more than 2 days difference.
Here is the query i have been able to create.
WITH ranked_order_status AS (
SELECT os.*,
datediff(
now() ,
first_value(recived_at) over (partition by awb order by recived_at asc)
) as diff,
ROW_NUMBER() OVER (PARTITION BY awb ORDER BY recived_at desc) AS rn
FROM order_status AS os where current_status in ('PICKUP EXCEPTION', 'OUT FOR PICKUP', 'PICKUP RESCHEDULED')
)
SELECT * FROM ranked_order_status WHERE rn = 1 and diff > 2
This unfortunately shows me all awbs with rows having those statuses and not only the ones with the last received current status of 'PICKUP EXCEPTION', 'OUT FOR PICKUP', 'PICKUP RESCHEDULED'
Any idea how i can edit this?

Answer 1

So if I understood it correctly, this should be a clear case of an analytical function using RANK().
This would be my approach regarding the constraints you mentioned:
WITH t1 AS (
SELECT os.*,
FIRST_VALUE(os.received_at) OVER(PARTITION BY os.awb
ORDER BY os.received_at) AS first_received_at
FROM order_status AS os
WHERE os.current_status IN ('PICKUP EXCEPTION', 'OUT FOR PICKUP', 'PICKUP RESCHEDULED')
),
t2 AS (SELECT RANK() OVER (PARTITION BY t1.awb
ORDER BY t1.received_at DESC) AS reverse_event_sequence,
DATE_DIFF(t1.received_at, t1.first_received_at, DAY) AS day_diff
t1.*
FROM t1
),
final AS (
SELECT *
FROM t2
WHERE t2.day_diff > 2
AND t2.reverse_event_sequence = 1
)
SELECT *
FROM final
Basically, you want to first grab the first value of the received_at for every row, then you want to rank all the events for every awb and order it on descending order to always have the last event as rank=1, and then you apply the desired constraints on the date difference :)
I have to mention that not having a data sample doesn't help, though. And would appreciate any feedback on my approach :)

Answer 2

You can enumerate the rows from the most recent rows in two ways:
Partitioned by awb and ordered by received at descending
Partitioned by awb and status and ordered by received at descending
For the last status for each awb, the difference between these will be zero. You can select this and then aggregate:
select awb, current_status,
min(received_at), max(received_at)
from (select os.*,
row_number() over (partition by awb order by received_at desc) as seqnum,
row_number() over (partition by awb, current_status order by received_at desc) as seqnum_2
from order_status os
) os
where seqnum = seqnum_2 and
current_status in ('PICKUP EXCEPTION', 'OUT FOR PICKUP', 'PICKUP RESCHEDULED')
group by awb, current_status
having max(received_at) > min(received_at) + interval 2 day;