Counting number of orders depending on city - sql

Question

I have a temp table that is being created, we will say that column 1 is an order_id, and column 2 is user_id, column 3 is start_date, column 4 is end_date and column 5 is city.
order_id user_id Start_date end_date city
101 1 200001 200101 X
101 2 200101 200110 y
101 3 200110 200112 z
101 3 200112 200210 z
I want to count by city the number of order_ids that moved out of it to another city and in another column the number of order_ids that moved into it from another city.
I would like it to come out as a table, like this:
city moved_out_orders moved_into_orders
x 1 0
y 1 1
z 0 1

Answer 1

You can do:
with
x as (
select a.city as from_city, b.city as to_city
from t a
join t b on a.order_id = b.order_id
and a.city <> b.city
and a.end_date = b.start_date
),
o (city, cnt) as (
select from_city, count(*) from x group by from_city
),
i (city, cnt) as (
select to_city, count(*) from x group by to_city
)
select
coalesce(i.city, o.city) as city,
o.cnt as moved_out_orders,
i.cnt as moved_in_orders
from i
full join o on o.city = i.city

Answer 2

Hmmm . . . I think you just want to enumerate the rows for each order and then discard the highest and lowest for each count:
select city,
sum(case when seqnum_desc > 1 then 1 else 0 end) as moved_out,
sum(case when seqnum_asc > 1 then 1 else 0 end) as moved_in
from (select t.*,
row_number() over (partition by orderid order by startdate) as seqnum_asc,
row_number() over (partition by orderid order by startdate desc) as seqnum_desc
from t
) t
group by city;
EDIT:
You appear to have adjacent rows in the same city. Seems strange, but instead you can use lead() and lag():
select city,
sum(case when next_city <> city then 1 else 0 end) as moved_out,
sum(case when prev_city <> city then 1 else 0 end) as moved_in
from (select t.*,
lag(city) over (partition by orderid order by startdate) as prev_city,
lead(city) over (partition by orderid order by startdate) as next_city
from t
) t
group by city;