Remove all non contiguous records with identical fields - sql

Question

I got a table with some columns like
ID RecordID DateInserted
1 10 now + 1
2 10 now + 2
3 4 now + 3
4 10 now + 4
5 10 now + 5
I would like to remove all non contiguous duplicates of the RecordID Column when they are sorted by DateInserted
In my example I would like to remove record 4 and 5 because between 2 and 4 there is a record with different id.
Is there a way to do it with 1 query ?

Answer 1

You can use window functions. One method is to count the changes in value that occur up to each row and just take the rows with one change:
select t.*
from (select t.*,
sum(case when prev_recordid = recordid then 0 else 1 end) over (order by dateinserted) as grp_num
from (select t.*,
lag(recordid) over (order by dateinserted) as prev_recordid
from t
) t
) t
where grp_num = 1;

Answer 2

One way would be to "flag" all the rows where it is not the first time this RecordID appeared and the prior row contained a different RecordID. Then you just exclude any row beyond that point for that RecordID.
;WITH cte AS
(
SELECT ID, RecordID, DateInserted,
dr = DENSE_RANK() OVER (PARTITION BY RecordID ORDER BY DateInserted),
prior = COALESCE(LAG(RecordID,1) OVER (ORDER BY DateInserted), RecordID)
FROM dbo.table_name
),
FlaggedRows AS
(
SELECT RecordID, dr
FROM cte
WHERE dr > 1 AND prior <> RecordID
)
SELECT cte.ID, cte.RecordID, cte.DateInserted
FROM cte
LEFT OUTER JOIN FlaggedRows AS f
ON cte.RecordID = f.RecordID
WHERE cte.dr < COALESCE(f.dr, cte.dr + 1)
ORDER BY cte.DateInserted;
If you want to actually delete the rows from the source (remove will typically be inferred as removing from the result), then change the SELECT at the end to:
DELETE cte
FROM cte
INNER JOIN FlaggedRows f
ON cte.RecordID = f.RecordID
WHERE cte.dr >= f.dr;