Combining max, average and group by

Question

Hi!
I have a simple table called Rates consisting of RateTime (datetime), Ask (decimal) and OrderSymbol (char)
I'm trying to create a select statement that will give me one row for each unique value in OrderSymbol consisting of columns for "CurrentAsk", "AVG(Ask)", and "OrderSymbol".
CurrentAsk has to be the latest one registered, so MAX(RateTime) for each OrderSymbol.
Is this possible?
TIA!
Dennis

Answer 1

Sure, using OVER() clause added in SQL Server 2005, e.g.
;with cte as (select RateTime, Ask,
AVG(ask) OVER (partition by OrderSymbol) as AverageAsk,
OrderSymbol, ROW_NUMBER() OVER (partition by OrderSymbol ORDER BY RateTime DESC) as RN
FROM dbo.Rates)
select Ask as CurrentAsk, AverageAsk, OrderSymbol
FROM cte where Rn = 1
In SQL 2012 and up you can use 
LAST_VALUE function, but be careful http://technet.microsoft.com/en-us/library/hh231517.aspx
E.g.
SELECT DISTINCT LAST_VALUE(Ask) OVER (partition by OrderSymbol ORDER BY RateTime
ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS CurrentAsk,
AVG(ask) OVER (partition by OrderSymbol) AS AverageAsk, OrderSymbol
FROM Rates
Although since we need to use DISTINCT in this query I don't know if it's going to perform better than ROW_NUMBER() version of it.
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