Calculate and find the second day of the week in SQL - sql

Question

Let's say I have data like this
CustomerID
Trans_date
C001
01-sep-22
C001
04-sep-22
C001
14-sep-22
C002
03-sep-22
C002
01-sep-22
C002
18-sep-22
C002
20-sep-22
C003
02-sep-22
C003
28-sep-22
C004
08-sep-22
C004
18-sep-22
But I'm unable to find the first and second transaction based on Trans_date.
I wish for the result to look like this:
CustomerID
Trans_week
first
second
C001
35
35
37
C001
35
35
37
C001
37
35
37
C002
35
35
37
C002
35
35
37
C002
37
35
37
C002
38
35
37
C003
35
35
39
C003
39
35
39
C004
36
36
37
C004
37
36
37
And for the last result will show like this:
CustomerID
first
second
C001
35
37
C002
35
37
C003
35
39
C004 didnt include because i would need who cust id who come first in their 1st week.

Answer 1

You may use ROW_NUMBER() function -inside a subquery- to get the first and second transaction dates for a customer, then use conditional MAX window function on the results of that subquery.
SELECT CustomerID, DATEPART(week,CustTrans) AS Trans_week,
DATEPART(week, MAX(CASE rn WHEN 1 THEN CustTrans END) OVER (PARTITION BY CustomerID)) first,
DATEPART(week, MAX(CASE rn WHEN 2 THEN CustTrans END) OVER (PARTITION BY CustomerID)) second
FROM
(
SELECT *,
ROW_NUMBER() OVER (PARTITION BY CustomerID ORDER BY CustTrans) rn
FROM trydata
) T
ORDER BY CustomerID, Trans_week
See a demo on SQL Server.
As you requested in the comments, if you want to select only one row per customer that showing the first and second weeks, use the following query:
SELECT CustomerID,
DATEPART(week, MAX(CASE rn WHEN 1 THEN CustTrans END)) first,
DATEPART(week, MAX(CASE rn WHEN 2 THEN CustTrans END)) second
FROM
(
SELECT *,
ROW_NUMBER() OVER (PARTITION BY CustomerID ORDER BY CustTrans) rn
FROM trydata
) T
WHERE rn <= 2
GROUP BY CustomerID
ORDER BY CustomerID
See a demo.

Answer 2

with cte (RN,CustomerID, FirstWeek,SecondWeek ) as
( SELECT ROW_NUMBER() over(partition by CustomerID ORDER BY CustomerID ) RN, CustomerID,FirstWeek, isnull((select TOP 1 (DATEPART(week,CustTrans))
from trydata c
where c.CustomerID = SRC.CustomerID AND DATEPART(week,C.CustTrans) > SRC.FirstWeek
ORDER BY DATEPART(week,C.CustTrans) ),'0') AS SecondWeek
FROM (
SELECT CustomerID,DATEPART(week,CustTrans) TransWeek,
(select MIN(DATEPART(week,CustTrans)) from trydata c where c.CustomerID = trydata.CustomerID) AS FirstWeek
FROM trydata
) SRC )
select CustomerID,FirstWeek,SecondWeek from cte where RN = 1
Output:
Example 2 :
WITH CTE (CustomerID,FIrstWeek,RN) AS (
SELECT CustomerID,MIN(DATEPART(week,CustTrans)) TransWeek,
ROW_NUMBER() over(partition by CustomerID ORDER BY DATEPART(week,CustTrans) asc ) FROM TryData
GROUP BY CustomerID,DATEPART(week,CustTrans)
)
SELECT CTE.CustomerID, CTE.FIrstWeek,
(select TOP 1 (DATEPART(week,c.CustTrans))
from trydata c
where c.CustomerID = CTE.CustomerID AND DATEPART(week,C.CustTrans) > CTE.FIrstWeek
) SecondWeek
FROM CTE
WHERE RN = 1
FIddle Demo
Edit: This can be done on easier way and less complex.