Sql-server – Get date based on interval days in sql

With partitioning you would have to do a partition per day, which puts the Pre-SQL 2012 limit of 1000 paritions in a new perspective as it would only allow for 3 years archive. With SQL Server 2012 you get 15000 partitions which is plenty for 1 partition per day.

Every day you would add a new partition. If you want to move the 61st past day partition you can do it efficiently, but is still an offline operation. See Move a Partition to a Different File Group Efficiently.

All your indexes would have to be aligned, see Special Guidelines for Partitioned Indexes.

Buying into partitioning is not an easy decission and it may be quite a big bite to chew... see How To Decide if You Should Use Table Partitioning. Specifically you should not expect performance improvements from partitioning. You should approach performance problems on time seriest by clustering by datetime.

As I read the question, the basic recursive algorithm required is:

1. Return the row with the earliest date in the set
2. Set that date as "current"
3. Find the row with the earliest date more than 90 days after the current date
4. Repeat from step 2 until no more rows are found

This is relatively easy to implement with a recursive common table expression.

For example, using the following sample data (based on the question):

DECLARE @T AS table (TheDate datetime PRIMARY KEY);

INSERT @T (TheDate)
VALUES
    ('2014-01-01 11:00'),
    ('2014-01-03 10:00'),
    ('2014-01-04 09:30'),
    ('2014-04-01 10:00'),
    ('2014-05-01 11:00'),
    ('2014-07-01 09:00'),
    ('2014-07-31 08:00');

The recursive code is:

WITH CTE AS
(
    -- Anchor:
    -- Start with the earliest date in the table
    SELECT TOP (1)
        T.TheDate
    FROM @T AS T
    ORDER BY
        T.TheDate

    UNION ALL

    -- Recursive part   
    SELECT
        SQ1.TheDate
    FROM 
    (
        -- Recursively find the earliest date that is 
        -- more than 90 days after the "current" date
        -- and set the new date as "current".
        -- ROW_NUMBER + rn = 1 is a trick to get
        -- TOP in the recursive part of the CTE
        SELECT
            T.TheDate,
            rn = ROW_NUMBER() OVER (
                ORDER BY T.TheDate)
        FROM CTE
        JOIN @T AS T
            ON T.TheDate > DATEADD(DAY, 90, CTE.TheDate)
    ) AS SQ1
    WHERE
        SQ1.rn = 1
)
SELECT 
    CTE.TheDate 
FROM CTE
OPTION (MAXRECURSION 0);

The results are:

╔═════════════════════════╗
║         TheDate         ║
╠═════════════════════════╣
║ 2014-01-01 11:00:00.000 ║
║ 2014-05-01 11:00:00.000 ║
║ 2014-07-31 08:00:00.000 ║
╚═════════════════════════╝

With an index having TheDate as a leading key, the execution plan is very efficient:

You could choose to wrap this in a function and execute it directly against the view mentioned in the question, but my instincts are against it. Usually, performance is better when you select rows from a view into a temporary table, provide the appropriate index on the temporary table, then apply the logic above. The details depend on the details of the view, but this is my general experience.

For completeness (and prompted by ypercube's answer) I should mention that my other go-to solution for this type of problem (until T-SQL gets proper ordered set functions) is a SQLCLR cursor (see my answer here for an example of the technique). This performs much better than a T-SQL cursor, and is convenient for those with skills in .NET languages and the ability to run SQLCLR in their production environment. It may not offer much in this scenario over the recursive solution because the majority of the cost is the sort, but it is worth mentioning.