AboutGeoff Expertise I specialise in database analysis and design, SQL and database queries using QBE and VBA. In my work, I use MS Access together with MS SQL Server as ETL (Extraction - Transformation - Loading) tools for migrating data between business ERP systems and data stores. My forte is building bespoke functions and applications.
See my
website for example apps and downloads
Experience I am a chartered engineer with 30 years of engineering and business experience, member of the BCS and have been working specifically in database applications, including SQL Server (v7/8/2000) for the last 9 or so years. I previously taught a course in Database Analysis and Design, but am now a freelance consultant and systems analyst.
Commercial database design and development work undertaken.
Question Hi Geoff,
I have used the codes for calculating Median from Microsoft and it works fine. But I can use only for one list. In my report I have groupings (3layers). I want the median value to come out for a particular group and not for the complete field list. This is the function I used. I can supply tName with the table name and fldName with the field that I want median for and I will get the median. What should I do if I want median for the group. In the example below,
Type Amount
1 10
3 15
1 7
1 8
3 20
3 25
This function will get median for the whole list. How can I get median for Type 1 and median for Type 3 in the same report dynamically by using grouping/sorting.
Function Median(tName As String, fldName As String) As Single
Dim MedianDB As DAO.Database
Dim ssMedian As DAO.Recordset
Dim RCount As Integer, i As Integer, x As Double, y As Double, _
OffSet As Integer
Set MedianDB = CurrentDb()
Set ssMedian = MedianDB.OpenRecordset("SELECT [" & fldName & _
"] FROM [" & tName & "] WHERE [" & fldName & _
"] IS NOT NULL ORDER BY [" & fldName & "];")
'NOTE: To include nulls when calculating the median value, omit
'WHERE [" & fldName & "] IS NOT NULL from the example.
ssMedian.MoveLast
RCount% = ssMedian.RecordCount
x = RCount Mod 2
If x <> 0 Then
OffSet = ((RCount + 1) / 2) - 2
For i% = 0 To OffSet
ssMedian.MovePrevious
Next i
Median = ssMedian(fldName)
Else
OffSet = (RCount / 2) - 2
For i = 0 To OffSet
ssMedian.MovePrevious
Next i
x = ssMedian(fldName)
ssMedian.MovePrevious
y = ssMedian(fldName)
Median = (x + y) / 2
End If
ssMedian.Close
MedianDB.Close
End Function
Thanks.
Mark Kaphle
Answer Hello Mark
Modify the function as follows:
1. Pass the Type as a parameter
2. Modify the SQL to use the Type
See below:
Function Median(tName As String, fldName As String, nType as LONG) As Single
Dim MedianDB As DAO.Database
Dim ssMedian As DAO.Recordset
Dim RCount As Integer, i As Integer, x As Double, y As Double, _
OffSet As Integer
Set MedianDB = CurrentDb()
Set ssMedian = MedianDB.OpenRecordset("SELECT [" & fldName & _
"] FROM [" & tName & "] WHERE [" & fldName & _
"] IS NOT NULL AND [Type] = " & nType & " ORDER BY [" & fldName & "];")
'NOTE: To include nulls when calculating the median value, omit
'WHERE [" & fldName & "] IS NOT NULL from the example.
ssMedian.MoveLast
RCount% = ssMedian.RecordCount
x = RCount Mod 2
If x <> 0 Then
OffSet = ((RCount + 1) / 2) - 2
For i% = 0 To OffSet
ssMedian.MovePrevious
Next i
Median = ssMedian(fldName)
Else
OffSet = (RCount / 2) - 2
For i = 0 To OffSet
ssMedian.MovePrevious
Next i
x = ssMedian(fldName)
ssMedian.MovePrevious
y = ssMedian(fldName)
Median = (x + y) / 2
End If
ssMedian.Close
MedianDB.Close
End Function
Please check that the field name in the table is indeed [Type], and also beware that this is a reserved word so always use [] square brackets around it...
