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How to Replicate SQL LEAD and LAG Functionality in SQL Server 2008?

Patricia Arquette
Patricia ArquetteOriginal
2024-12-17 22:27:11909browse

How to Replicate SQL LEAD and LAG Functionality in SQL Server 2008?

Alternatives to SQL LEAD and LAG Functions in SQL Server 2008

In SQL Server 2008, LEAD and LAG functions are not available, which can pose challenges when comparing values between consecutive rows. This guide presents alternative approaches to achieve similar functionality.

Problem Statement

The goal is to identify rows where the current row has a StatusId of 1, and the subsequent row has a StatusId of 6, with the times being within the same minute. For example, in the provided table, we want to retrieve row Id 6 because row Id 6 has a StatusId of 1, and row Id 7 has a StatusId of 6, and their times are essentially the same: 2014-02-19 06:11.

Solution 1: Self-Join

One approach is a self-join:

select t.*
from table t join
     table tnext
     on t.id = tnext.id - 1 and
        t.StatusId = 1 and
        tnext.StatusId = 6 and
        datediff(second, t.MinStartTime, tnext.MinStartTime) < 60;

This query returns rows where the Id of the current row is one less than the Id of the next row, the StatusId is 1 for the current row, 6 for the next row, and the time difference between them is within 60 seconds.

Solution 2: Same Calendar Minute

If you require the times to be in the same calendar minute, you can modify the query as follows:

select t.*
from table t join
     table tnext
     on t.id = tnext.id - 1 and
        t.StatusId = 1 and
        tnext.StatusId = 6 and
        datediff(second, t.MinStartTime, tnext.MinStartTime) < 60 and
        datepart(minute, t.MinStartTime) = datepart(minute, tnext.MinStartTime);

This query additionally checks whether the minutes of the two rows are equal, ensuring that they belong to the same calendar minute.

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