Nope. Civilian contractor currently deployed in Afghanistan. I work 7 days a week, 12+ hours a day....

I would...

Tammy-274861 (7/11/2014)

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Hi Lynn, J and autoexcrement!

Lynn is correct, and that is why I had clarified by saying that as long as the sum of the gaps for the measurement year...

Looking back it looks like the OP may have been on the forums when I posted my solution. Will have to wait and see.

I'm still lost. Based on everything you just posted, what is the expected results. What should you see from the processing of a query that does everything correctly...

J Livingston SQL (7/11/2014)

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Lynn Pettis (7/11/2014)

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First, to clarify. A person is considered continuously enrolled as long as the total of any gaps in enrollment are less than 45 days...

I made an assumption you are using at least SQL Server 2000.

DECLARE @Temperature TABLE (CalendarDate datetime, Temperature decimal(10,2),UOM char(1))

INSERT INTO @Temperature

SELECT GETDATE(), 55, 'F'

UNION ALL

SELECT GETDATE(), 15, 'C'

UNION ALL

SELECT GETDATE()...

Second question, since you posted this in a SQL Server 7/2000 forum, are you using SQL Server 2000 or SQL Server 7? It matters in helping you develop a...

To see what is happening, run the following. If you look at the second query I have removed the MAX and GROUP BY. The result set you see...

Glad to see it worked. Just for clarity, what was...

Greg Edwards-268690 (7/11/2014)

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Lynn Pettis (7/10/2014)

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Okay. Curious if the OP on anther thread is ever going to say anything. Asks for help and doesn't quite provide enough to help....

First, to clarify. A person is considered continuously enrolled as long as the total of any gaps in enrollment are less than 45 days in the given calendar year.

Based...

Arno Ho (7/11/2014)

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Hi,

Your algorithm is faster than mine but it does not fully meet my expectations

We need to calculate the hours scheduled and that is really the hours worked...

Vague, can you elaborate as to what you are looking for here?

To the OP, any updates?