LAG within CASE giving false negative offset - sql

Question

TL;DR: scroll down to TASK 2.
I am dealing with the following data set:
email,createdby,createdon
a#b.c,jsmith,2016-10-10
a#b.c,nsmythe,2016-09-09
a#b.c,vstark,2016-11-11
b#x.y,ajohnson,2015-02-03
b#x.y,elear,2015-01-01
...
and so on. Each email is guaranteed to have at least one duplicate in the data set.
Now, there are two tasks to resolve; I resolved one of them but am struggling with the other one. I will now present both tasks for completeness.
TASK 1 (resolved):
For each row, for each email, return an additional column with the name of the user that created the first record with this email.
Expected result for the above sample data set:
email,createdby,createdon,original_createdby
a#b.c,jsmith,2016-10-10,nsmythe
a#b.c,nsmythe,2016-09-09,nsmythe
a#b.c,vstark,2016-11-11,nsmythe
b#x.y,ajohnson,2015-02-03,elear
b#x.y,elear,2015-01-01,elear
Code to get the above:
;WITH q0 -- this is just a security measure in case there are unique emails in the data set
AS ( SELECT t.email
FROM t
GROUP BY t.email
HAVING COUNT(*) > 1) ,
q1
AS ( SELECT q0.email
, createdon
, createdby
, ROW_NUMBER() OVER ( PARTITION BY q0.email ORDER BY createdon ) rn
FROM t
JOIN q0
ON t.email = q0.email)
SELECT q1.email
, q1.createdon
, q1.createdby
, LAG(q1.createdby, q1.rn - 1) OVER ( ORDER BY q1.email, q1.createdon ) original_createdby
FROM q1
ORDER BY q1.email
, q1.rn
Brief explanation: I partition data set by email, then I number rows in each partition ordered by creation date, finally I return [createdby] value from (rn-1)th record. Works exactly as expected.
Now, similar to the above, there is TASK 2:
TASK 2:
For each row, for each email, return name of the user that created the first duplicate. I.e. name of a user where rn=2.
Expected result:
email,createdby,createdon,first_dupl_createdby
a#b.c,jsmith,2016-10-10,jsmith
a#b.c,nsmythe,2016-09-09,jsmith
a#b.c,vstark,2016-11-11,jsmith
b#x.y,ajohnson,2015-02-03,ajohnson
b#x.y,elear,2015-01-01,ajohnson
I want to keep things performant so trying to employ LEAD-LAG functions:
WITH q0
AS ( SELECT t.email
FROM t
GROUP BY t.email
HAVING COUNT(*) > 1) ,
q1
AS ( SELECT q0.email
, createdon
, createdby
, ROW_NUMBER() OVER ( PARTITION BY q0.email ORDER BY createdon ) rn
FROM t
JOIN q0
ON t.email = q0.email)
SELECT q1.email
, q1.createdon
, q1.createdby
, q1.rn
, CASE q1.rn
WHEN 1 THEN LEAD(q1.createdby, 1) OVER ( ORDER BY q1.email, q1.createdon )
ELSE LAG(q1.createdby, q1.rn - 2) OVER ( ORDER BY q1.email, q1.createdon )
END AS first_dupl_createdby
FROM q1
ORDER BY q1.email
, q1.rn
Explanation: for the first record in each partition, return [createdby] from the following record (i.e. from the record containing the first duplicate). For all other records in the same partition return [createdby] from (rn-2) records ago (i.e. for rn = 2 we're staying on the same record, for rn = 3 we're going 1 record back, for rn = 4 - 2 records back and so on).
An issue comes up on the
ELSE LAG(q1.createdby, q1.rn - 2)
operation. Apparently, against any logic, despite the existence of the preceding line (WHEN 1 THEN...), the ELSE block is also evaluated for rn = 1, resulting in a negative offset value passed to the LAG function:
Msg 8730, Level 16, State 2, Line 37
Offset parameter for Lag and Lead functions cannot be a negative value.
When I comment out that ELSE line, the whole thing works fine but obviously I am not getting any results in the first_dupl_createdby column for rn > 1.
QUESTION:
Is there any way of re-writing the above CASE statement (in TASK #2) so that it always returns the value from a record where rn = 2 within each partition but - and this is important bit - without doing a self-JOIN operation (I know I could prepare rows where rn = 2 in a separate sub-query but this would mean extra scans on the whole table and also running an unnecessary self-JOIN).

Answer 1

I think you can simply use the max window function as you are trying to get the value from rownumber = 2 for each partition.
SELECT q1.email
, q1.createdon
, q1.createdby
, q1.rn
, max(case when rn=2 then q1.createdby end) over(partition by q1.email) first_dup_created_by
FROM q1
ORDER BY q1.email, q1.rn
You can use a similar query to get the results for rownumber=1 for the 1st scenario as well.

Answer 2

You can get the information for each email using row_number() and conditional aggregation:
select email,
max(case when seqnum = 1 then createdby end) as createdby_first,
max(case when seqnum = 2 then createdby end) as createdby_second
from (select t.*,
row_number() over (partition by email order by createdon) as seqnum
from t
) t
group by email;
You can join this information back to the original data to get the information you want. I don't see how lag() naturally would be used to solve this problem.

Answer 3

/shrug
; WITH duplicate_email_addresses AS (
SELECT email
FROM t
GROUP
BY email
HAVING Count(*) > 1
)
, records_with_duplicate_email_addresses AS (
SELECT email
, createdon
, createdby
, Row_Number() OVER (PARTITION BY email ORDER BY createdon) AS sequencer
FROM t
WHERE EXISTS (
SELECT *
FROM duplicate_email_addresses
WHERE email = t.email
)
)
, second_duplicate_record AS ( -- Why do you need any more than this?
SELECT email
, createdon
, createdby
FROM records_with_duplicate_email_addresses
WHERE sequencer = 2
)
SELECT records_with_duplicate_email_addresses.email
, records_with_duplicate_email_addresses.createdon
, records_with_duplicate_email_addresses.createdby
, second_duplicate_record.createdby AS first_duplicate_createdby
FROM records_with_duplicate_email_addresses
INNER
JOIN second_duplicate_record
ON second_duplicate_record.email = records_with_duplicate_email_addresses.email
;