SQL Server: Average counts by hour and day of week

Question

Background

I have a table set up in a SQL Server environment that contains a log of various activity that I\'m tracking. Particular log items use unique codes to categ

Answer 1

; WITH a AS (
    SELECT CAST([time] AS date) AS ForDate
       , DATEPART(hour, [time]) AS OnHour
       , txtW=DATENAME(WEEKDAY,[time])
       , intW=DATEPART(WEEKDAY,[time])
       , Totals=COUNT(*)
    FROM [log] WHERE [code] = 'tib_imp.8'
    GROUP BY CAST(time AS date)
    , DATENAME(WEEKDAY,[time])
    , DATEPART(WEEKDAY,[time])
    , DATEPART(hour,[time])
)
SELECT [Day]=txtW
, [Hour]=OnHour
, [Average Count]=AVG(Totals)
FROM a
GROUP BY txtW, intW, OnHour
ORDER BY intW, OnHour

Answer 2

Guessing here:

SELECT [Day], [Hour], [DayN], AVG(Totals) AS [Avg]
FROM
  (
        SELECT 
          [Day]  = DATENAME(WEEKDAY, [time]),
          [DayN] = DATEPART(WEEKDAY, [time]),
          [Hour] = DATEPART(HOUR,    [time]),
          Totals = COUNT(*)
        FROM dbo.[log] 
            WHERE [code] = 'tib_imp.8'
        GROUP BY 
          DATENAME(WEEKDAY, [time]),
          DATEPART(WEEKDAY, [time]),
          DATEPART(HOUR,    [time])
  ) AS q
GROUP BY [Day], [Hour], [DayN]
ORDER BY DayN; 

Again, without data, I might once again be throwing a handful of mud at the wall and hoping it sticks, but perhaps what you need is:

SELECT [Day], [Hour], [DayN], AVG(Totals) AS [Avg]
FROM
(
    SELECT 
  w = DATEDIFF(WEEK, 0, [time]),
      [Day]  = DATENAME(WEEKDAY, [time]),
      [DayN] = DATEPART(WEEKDAY, [time]),
      [Hour] = DATEPART(HOUR,    [time]),
      Totals = COUNT(*)
    FROM dbo.[log] 
      WHERE [code] = 'tib_imp.8'
    GROUP BY 
  DATEDIFF(WEEK, 0, [time]),
      DATENAME(WEEKDAY, [time]),
      DATEPART(WEEKDAY, [time]),
      DATEPART(HOUR,    [time])
  ) AS q
GROUP BY [Day], [Hour], [DayN]
ORDER BY DayN; 

This is also going to produce integer-based averages, so you may want to cast the Totals alias on the inner query to DECIMAL(something,something).