SQL sum over partition for preceding period - sql

Question

I have the following table, which represent Customers for each day:
+----------+-----------+
| Date | Customers |
+----------+-----------+
| 1/1/2014 | 4 |
| 1/2/2014 | 7 |
| 1/3/2014 | 5 |
| 1/4/2014 | 5 |
| 1/5/2014 | 10 |
| 2/1/2014 | 7 |
| 2/2/2014 | 4 |
| 2/3/2014 | 1 |
| 2/4/2014 | 5 |
+----------+-----------+
I would like to add 2 additional columns:
Summary of the customers for the current month
Summary of the customers for the preceding month
here's the desired outcome:
+----------+-----------+----------------------+------------------------+
| Date | Customers | Sum_of_Current_month | Sum_of_Preceding_month |
+----------+-----------+----------------------+------------------------+
| 1/1/2014 | 4 | 31 | 0 |
| 1/2/2014 | 7 | 31 | 0 |
| 1/3/2014 | 5 | 31 | 0 |
| 1/4/2014 | 5 | 31 | 0 |
| 1/5/2014 | 10 | 31 | 0 |
| 2/1/2014 | 7 | 17 | 31 |
| 2/2/2014 | 4 | 17 | 31 |
| 2/3/2014 | 1 | 17 | 31 |
| 2/4/2014 | 5 | 17 | 31 |
+----------+-----------+----------------------+------------------------+
I have managed to calculate the 3rd column by a simple sum over partition function:
Select
Date,
Customers,
Sum(Customers) over (Partition by (Month(Date)||year(Date) Order by 1) as Sum_of_Current_month
From table
However, I can't find a way to calculate the Sum_of_preceding_month column.
Appreciate your support.
Asaf

Answer 1

The previous month is a bit tricky. What's your Teradata release, TD14.10 supports LAST_VALUE:
SELECT
dt,
customers,
Sum_of_Current_month,
-- return the previous sum
COALESCE(LAST_VALUE(x ignore NULLS)
OVER (ORDER BY dt
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING)
,0) AS Sum_of_Preceding_month
FROM
(
SELECT
dt,
Customers,
SUM(Customers) OVER (PARTITION BY TRUNC(dt,'mon')) AS Sum_of_Current_month,
CASE -- keep the number only for the last day in month
WHEN ROW_NUMBER()
OVER (PARTITION BY TRUNC(dt,'mon')
ORDER BY dt)
= COUNT(*)
OVER (PARTITION BY TRUNC(dt,'mon'))
THEN Sum_of_Current_month
END AS x
FROM tab
) AS dt

Answer 2

I think this might be easier by using lag() and an aggregation sub-query. The ANSI Standard syntax is:
Select t.*, tt.sumCustomers, tt.prev_sumCustomers
From table t join
(select extract(year from date) as yyyy, extract(month from date) as mm,
sum(Customers) as sumCustomers,
lag(sum(Customers)) over (order by extract(year from date), extract(month from date)
) as prev_sumCustomers
from table t
group by extract(year from date), extract(month from date)
) tt
on extract(year from date) = tt.yyyy and extract(month from date) = t.mm;
In Teradata, this would be written as:
Select t.*, tt.sumCustomers, tt.prev_sumCustomers
From table t join
(select extract(year from date) as yyyy, extract(month from date) as mm,
sum(Customers) as sumCustomers,
min(sum(Customers)) over (order by extract(year from date), extract(month from date)
rows between 1 preceding and 1 preceding
) as prev_sumCustomers
from table t
group by extract(year from date), extract(month from date)
) tt
on extract(year from date) = tt.yyyy and extract(month from date) = t.mm;

Answer 3

Try this:
SELECT
[Date],
[Customers],
(SELECT SUM(customers) FROM table WHERE MONTH(dte) = MONTH(tbl.dte)),
ISNULL((SELECT SUM(customers) FROM table WHERE MONTH(dte) = MONTH(DATEADD(MONTH, -1, tbl.dte))), 0)
FROM table tbl