3.8 Complicated judgment on each group
One example is to count customers who place an order for at least three days consecutively in each month.
SQL
WITH a AS(
SELECT count (1) AS num, day (OrderDate) days,
CustomerID,year(OrderDate) Years,month(OrderDate) Months
FROM Orders
WHERE OrderDate>='2022-01-01'
GROUP BY CustomerID, day (OrderDate), year(OrderDate),month(OrderDate)),
b AS (
SELECT
lag(days,1) OVER(PARTITION BY CustomerID, Years,Months ORDER BY days) pd,
days, CustomerID,Years,Months
FROM a),
c AS (
SELECT *,
CASE WHEN days-pd=1 THEN 0 ELSE 1 END AS tmp
FROM b),
d AS (
SELECT CustomerID,Years,Months,sum (tmp) OVER (
PARTITION BY CustomerID,Years,Months ORDER BY days) NoOrderDays
FROM c),
e AS (
SELECT CustomerID, Years,Months,NoOrderDays,count (*) ContinuousDays
FROM d
GROUP BY CustomerID,Years,Months,NoOrderDays),
f AS (
SELECT CustomerID,Years,Months,max (ContinuousDays) mCon
FROM e
GROUP BY CustomerID,Years,Months
HAVING max (ContinuousDays)>=3)
SELECT Years,Months,count (*)
FROM f
GROUP BY Years,Months
SPL
A | |
---|---|
1 | >st=date(“2022-01-01”),start=days@o(st) |
2 | =file(“Orders_Account.ctx”).open().cursor@x(CustomerID,OrderDate;OrderDate>=start) |
3 | =A2.group(CustomerID,month@y(OrderDate)) |
4 | =A3.(~.select@1((a=case(OrderDate-OrderDate[-1],0:a,1:a+1;1))==3)) |
5 | =A4.groups(year(OrderDate):Years,month(OrderDate):Months; count(1)) |
A3 Fetch order records of each customer in each month in order as a set.
A4 On each grouped subset, find whether there are customers who place an order for three days continuously.
A5 As A4 returns data that is already ordered by CustomerID, we just perform count to achieve the count distinct.
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