SQL Server日期之间的差异

Question

I need to make a query that will compare the dates with the same 'ID_2'

If the difference between dates in column "ProdDateTime" is more than 7 and difference between dates in column "Sold" is more than 1 than it should show this rows.
here's an example: the difference between row #12 and #13 row with ID_2=444 is more than 7 in column ProdDateTime and is more than 1 in column Sold, that's why it should show this row as an error and all the following (14,15) because they have the same error.

PS I used row_number and dense_rank to make ID_1 and ID_2 columns accordingly.

Do you have any tips on how to do so? Thank you all in advance!

Answer 1

I assumed that date in "ProdDateTime" column is always in increasing order. So, here we need to compare dates from previous row value and next row value of ProdDateTime column. I have used LAG and LEAD analytical functions to solve this problem. It's just a hint for you, I have implemented for only single condition below, Maybe it works for you:-

    ;WITH ResultData 
    AS
    (
        SELECT *
            ,LEAD(ProdDateTime)  OVER (PARTITION BY ID_2 ORDER BY ID_2 ) AS Lead
            ,LAG(ProdDateTime)  OVER (PARTITION BY ID_2 ORDER BY ID_2 ) AS Lag
            ,DATEDIFF(DAY, ProdDateTime, LEAD(ProdDateTime)  OVER (PARTITION BY ID_2 ORDER BY ID_2 )) AS LeadDiff
            ,DATEDIFF(DAY, LAG(ProdDateTime)  OVER (PARTITION BY ID_2 ORDER BY ID_2 ), ProdDateTime) AS LagDiff
        FROM tb
    )

    SELECT * 
    FROM ResultData
    WHERE LeadDiff > 1 OR LagDiff > 1

If you are unaware from above functions, Please do search for same. Best of luck!

Answer 2

You can select next and previous rows as the below.

SELECT
    * -- Your DATEDIFF operations
FROM
    Tbl CurrentRow LEFT JOIN 
    Tbl PreviousRow ON CurrentRow.ID_1 = (PreviousRow.ID_1 + 1) 
                       CurrentRow.ID_2 = PreviousRow.ID_2 
                   LEFT JOIN
    Tbl NextRow     ON CurrentRow.ID_1 = (NextRow.ID_1 - 1) 
                       CurrentRow.ID_2 = NextRow.ID_2