Access Maximum Function: Return the second highest value across 10 fields

[liamthequietman]

The code works great except if the first field is a zero or all the fields in the row contain zeros (see pic).

Many thanks!

Function Maximum(ParamArray FieldArray() As Variant)
' Declare the two local variables.
Dim I As Integer
Dim currentVal As Variant
Dim secondHighest As Variant
 
' Set the variable currentVal equal to the array of values.
  currentVal = FieldArray(0)
 
' Cycle through each value from the row to find the largest.
 
    Dim tmpArray
    For I = 0 To UBound(FieldArray)
        If FieldArray(I) > currentVal Then
        currentVal = FieldArray(I)
        End If
    Next
 
tmpArray = Filter(FieldArray, currentVal, False, vbTextCompare)
'This will fill the tmpArray with all your array values EXCEPT the highest one.
 
    secondHighest = tmpArray(0)
        For I = 0 To UBound(tmpArray)
        If tmpArray(I) > secondHighest Then
        secondHighest = tmpArray(I)
    End If
    Next
 
' Return the maximum value found.
Maximum = secondHighest
 
' Expr1: Maximum ([nPP1CSF],[nPP2CSF],[nPP3CSF],[nPP4CSF],[nPP5CSF],[nPP6CSF],[nPP7CSF],[nPP8CSF],[nPP9CSF],[nPP0CSF])
 
End Function

[gnawoncents]

Liam,
Welcome to Bytes! There are a couple issues keeping you from the answer you want. First, anytime all the numbers are the same (not just zeroes), the code won't be able to handle it because you're filtering out all the variables. You can solve this by counting how many times the numbers are the same--if they are always the same, just force insert what you want into the field (since there is no second-highest number, I'm not sure what this would be). Also, you'll run into some interesting problems if you leave the values as variables. I changed them to integers in the code below.

Function Maximum(ParamArray FieldArray() As Variant)
' Declare the two local variables.
Dim I As Integer
Dim currentVal As Integer
Dim secondHighest As Integer
Dim intCount As Integer
 
' Set the variable currentVal equal to the array of values.
currentVal = FieldArray(0)
 
' Cycle through each value from the row to find the largest.
For I = 0 To UBound(FieldArray)
    If FieldArray(I) > currentVal Then
        currentVal = FieldArray(I)
      ElseIf FieldArray(I) = currentVal Then
        intCount = intCount + 1
    End If
Next
 
If intCount - 1 = UBound(FieldArray) Then
    ' Return the starting value since they were all the same
    Maximum = FieldArray(0) 'MODIFY THIS TO WHATEVER YOU WANT WHEN THEY ARE ALL THE SAME
  Else
    Dim tmpArray As Variant
    tmpArray = Filter(FieldArray, currentVal, False, vbTextCompare)
    ' This will fill the tmpArray with all your array values EXCEPT the highest one.
 
    secondHighest = tmpArray(0)
 
    For I = 0 To UBound(tmpArray)
        If tmpArray(I) > secondHighest Then
            secondHighest = tmpArray(I)
        End If
    Next
 
    ' Return the maximum value found.
    Maximum = secondHighest
End If
 
' Expr1: Maximum ([nPP1CSF],[nPP2CSF],[nPP3CSF],[nPP4CSF],[nPP5CSF],[nPP6CSF],[nPP7CSF],[nPP8CSF],[nPP9CSF],[nPP0CSF])
 
End Function
 

[twinnyfo]

liamthequietman,

If you will indulge me the opportunity to modify gnawoncents's excellent suggestion....

Function Maximum(ParamArray FieldArray() As Variant)
On Error GoTo EH
    ' Declare the two local variables.
    Dim I As Integer
    Dim currentVal As Integer
    Dim secondHighest As Integer
    Dim intCount As Integer
 
    ' Set the variable currentVal equal to the array of values.
    currentVal = FieldArray(0)
 
    ' Cycle through each value from the row to find the largest.
    For I = 0 To UBound(FieldArray)
        If FieldArray(I) = currentVal Then intCount = intCount + 1
        If FieldArray(I) > currentVal Then
            secondHighest = currentVal
            currentVal = FieldArray(I)
        Else
            If FieldArray(I) > secondHighest _
                And FieldArray(I) < currentVal Then
                secondHighest = FieldArray(I)
            End If
        End If
    Next
 
    If intCount - 1 = UBound(FieldArray) Then
        ' Return the starting value since they were all the same
        Maximum = FieldArray(0) 'MODIFY THIS TO WHATEVER YOU WANT WHEN THEY ARE ALL THE SAME
    Else
        ' Return the maximum value found.
        Maximum = secondHighest
    End If
 
    Exit Function
EH:
    MsgBox "There was an error finding the second highest value!  " & _
        "Please contact your Database Administrator.", vbCritical, "WARNING!"
    Exit Function
End Function

If you look closely above, I've eliminated the need to go through your array twice. Assuming your beginning highest value is in currentVal (which will begin at 0), then if there is any value greater than that, the higher value becomes currentVal. But, before we do that, we asign the value of currentVal to secondHighest (which initially will also be 0). Then if any future values are greater than currentVal, then secondHighest takes on the value of the previous highest value. There is an Else added to the code to handle cases in which the new value is less than currentVal but greater than secondHighest.

This should give you a more streamlined code (and faster, at that).

Hope this hepps!

[gnawoncents]

twinnyfo,

Very nice--much more elegant. *initiates slow clap*

[twinnyfo]

gnawoncents,

Aww shucks.... I just stole what you posted!

:-P

[liamthequietman]

BRILLIANT!

What an excellent collaboration between the two of you!

I hope others can benefit from this excellent code.

I'm very grateful!

[twinnyfo]

Glad the two of us could hepp!

[liamthequietman]

I've made changes to the code for "Minimum" values "Second Lowest" The values are Single and contain -1

It works perfect except if a -1 appears in any of the columns.

Greatly appreciated!

(see pic)

Function Minimum(ParamArray FieldArray() As Variant)
    On Error GoTo EH
        ' Declare the two local variables.
        Dim I As Integer
        Dim currentVal As Single
        Dim secondLowest As Single
        Dim intCount As Integer
 
        ' Set the variable currentVal equal to the array of values.
        currentVal = FieldArray(0)
 
        ' Cycle through each value from the row to find the lowest.
        For I = 0 To UBound(FieldArray)
            If FieldArray(I) = currentVal Then intCount = intCount + 1
            If FieldArray(I) < currentVal Then
                secondLowest = currentVal
                currentVal = FieldArray(I)
            Else
                If FieldArray(I) < secondLowest _
                    And FieldArray(I) > currentVal Then
                    secondLowest = FieldArray(I)
                End If
            End If
        Next
 
        If intCount - 1 = UBound(FieldArray) Then
            ' Return the starting value since they were all the same
            Minimum = FieldArray(0) 'Modify this to whatever you want when they are all the same
        Else
            ' Return the Minimum value found.
            Minimum = secondLowest
        End If
 
        Exit Function
EH:
        MsgBox "There was an error finding the second Lowest value!  " & _
            "Please contact your Database Administrator.", vbCritical, "WARNING!"
        Exit Function
 
' Expr1: Minimum([nPP1SHF],[nPP2SHF],[nPP3SHF],[nPP4SHF],[nPP5SHF],[nPP6SHF],[nPP7SHF],[nPP8SHF],[nPP9SHF],[nPP0SHF])
 
End Function

[twinnyfo]

All you need to do is reverse you ">" and "<" operators.

[liamthequietman]

I edited my post. Please read again I almost have the solution!

Thanks!

[twinnyfo]

liam - for future posts, please use Code Tags when posting your code. The moderators are able to edit your content, but we shouldn't have to do that every time.

[twinnyfo]

What happens when there is a -1 in any of the values. It should work with positive or negative numbers.

[liamthequietman]

I'm a novice and will now Google "Code Tags" to understand what they are.

My apologies!

[liamthequietman]

My changes to the code work perfect except if a -1 appears in any column.

see pic in post I edited

[twinnyfo]

Code tags are what you place your code in so that they are properly formatted on this forum. There is a button in the text entry display "[CODE/] which will insert code tags around any of your selected text.

Your pic does not tell me what the code is doing wrong. It looks to me to be doing exactly what the code is telling it to do. What do you want the result to be when it is all -1? Are these by chance text values?

[liamthequietman]

If there is a column(s) with a -1 it is returning the lowest number, not the second lowest.

Columns without -1 are returning the correct second lowest number

Columns with all -1 returns -1 which is perfect

[twinnyfo]

Have you stepped through the code while it is running to see how the evaluations of the values are working? Particularly, when there is a -1 value and the code gets here:

If FieldArray(I) < currentVal Then
    secondLowest = currentVal
    currentVal = FieldArray(I)
Else
    If FieldArray(I) < secondLowest _
        And FieldArray(I) > currentVal Then
        secondLowest = FieldArray(I)
    End If
End If

How are the evaluations responding?

Not sure if it has anything to do with it, but you may need to declare your array as a Double?

Function Minimum(ParamArray FieldArray() As Double)

[liamthequietman]

Being a complete VBA novice, I'm not familiar on how to step through the code. I placed my cursor in the code and pressed F8 and received a ping which I associate with an error.

Declaring the array as Double resulted in a error

[twinnyfo]

No worries, we've all been novices at one point.

Before you run your code, click your mouse to the left of the first line of your function in the VBA editor. It should leave a dark brown circle in the left margin. Run your code and the code should stop at that point (called a break point). Then, hit F8 repeatedly to watch the code work through itself. Any time the code is paused, you can hover your mouse over any of the variables and you will be able to see their current values.

Hope that hepps!

[liamthequietman]

Thanks for not giving up on this project!

My step through attempt did not work, a Macro box appears (see pic).

I found another issue with the results. In review:

CORRECT

Columns without -1 are returning the correct second lowest number

Columns with all -1 returns -1 which is perfect

INCORRECT

If there is a column(s) with a -1 it is returning the lowest number, not the second lowest.

If there are columns with identical numbers it's returning zero

[twinnyfo]

By "running your code", I mean you have to execute it the way you would normally execute it. Then, when the code hits your break point, then you can step through it using F8. Hope that hepps.

[twinnyfo]

Or, maybe try using the values -2 and -1.2 to see if that will have any effect on the logic the code returns. At this point, I am grasping at straws, because based upon what I am seeing int he code, it "should" work, but it just doesn't. I must be missing something obvious, which would not be the first time something like that happened.

[twinnyfo]

Was able to do some testing and proofing. Note lines 23-31. We forgot about the situation when the first number is the lowest. We never checked to see if the second was still the lowest value. This should solve it.

Public Function Minimum(FieldArray() As Variant) As Single
On Error GoTo EH
    ' Declare the two local variables.
    Dim I               As Integer
    Dim currentVal      As Single
    Dim secondLowest    As Single
    Dim intCount        As Integer
 
    ' Set the variable currentVal equal to the array of values.
    currentVal = FieldArray(0)
    secondLowest = FieldArray(0)
 
    ' Cycle through each value from the row to find the lowest.
    For I = 0 To UBound(FieldArray)
        If FieldArray(I) = currentVal Then intCount = intCount + 1
        If FieldArray(I) < currentVal Then
            secondLowest = currentVal
            currentVal = FieldArray(I)
        ElseIf secondLowest = currentVal _
            And FieldArray(I) > secondLowest Then
                secondLowest = FieldArray(I)
        Else
            If secondLowest = currentVal _
                And FieldArray(I) > secondLowest Then
                secondLowest = FieldArray(I)
            Else
                If FieldArray(I) < secondLowest _
                    And FieldArray(I) > currentVal Then
                    secondLowest = FieldArray(I)
                End If
            End If
        End If
    Next
 
    If intCount - 1 = UBound(FieldArray) Then
        ' Return the starting value since they were all the same
        Minimum = FieldArray(0) 'Modify this to whatever you want when they are all the same
    Else
        ' Return the Minimum value found.
        Minimum = secondLowest
    End If
 
    Exit Function
EH:
    MsgBox "There was an error finding the second Lowest value!  " & _
        "Please contact your Database Administrator.", vbCritical, "WARNING!"
    Exit Function
End Function

[liamthequietman]

Can I attach a test version of my database?

[twinnyfo]

Check my last post and see if that code works. It ran fine for me. I knew I was overlooking something obvious. It wasn't until I was able to step through the code that I found it.

[liamthequietman]

I applaud you for your tenacity! The expression I enter into the query has worked in all your codes except now. Error wrong number of arguments:

Expr1: Minimum([nPP1SHF],[nPP2SHF],[nPP3SHF],[nPP4SHF],[nPP5SHF],[nPP6SHF],[nPP7SHF],[nPP8SHF],[nPP9SHF],[nPP0SHF])

[twinnyfo]

I changed Line 1 from:

Public Function Minimum(ParamArray FieldArray() As Variant) As Single

to:

Public Function Minimum(FieldArray() As Variant) As Single

Not sure if that makes a difference. I couldn't get it to work any other way.

I was using my code from within VBA, you are calling it as an expression in a Query, which is causing the disconnect.

[liamthequietman]

I found a code which works perfect EXCEPT ...

If there is a column(s) with a -1 it is returning the lowest number, not the second lowest.

Otherwise all the result are correct.

Function SecondMinimum(ParamArray FieldArray() As Variant) As Variant
    Dim I As Integer
    Dim LowestVal As Variant
    Dim secondVal As Variant
 
'   Default values
    LowestVal = Null
    secondVal = Null
 
'   Populate the LowestVal and secondVal temporary variables
    For I = 0 To UBound(FieldArray)
        If IsNull(FieldArray(I)) = False Then
            If IsNull(LowestVal) Then
                LowestVal = FieldArray(I)
            ElseIf IsNull(secondVal) Then
'               Before populating the secondVal, make sure LowestVal will be
'               less than or equal to secondVal.
                If FieldArray(I) > LowestVal Then
                    secondVal = FieldArray(I)
                Else
                    secondVal = LowestVal
                    LowestVal = FieldArray(I)
                End If
                Exit For
            End If
        End If
    Next I
 
'   Make sure there are at least two non-Null values in the temporary variables
    If IsNull(LowestVal) = False And IsNull(secondVal) = False Then
 
'       Loop through the array values, comparing them to the lowest and second-lowest
        For I = 0 To UBound(FieldArray)
'           Skip duplicate values.
            If FieldArray(I) <> LowestVal Then
'               New lowest value?
                If FieldArray(I) < LowestVal Then
'                   New lowest value.  Push the Lowest up to second, then make this
'                   array value the new lowest value.
                    secondVal = LowestVal
                    LowestVal = FieldArray(I)
'               Skip duplicate values.
                ElseIf FieldArray(I) <> secondVal Then
'                   New second-lowest value?
                    If FieldArray(I) < secondVal Then
'                       Replace the second-lowest value with the current array value.
                        secondVal = FieldArray(I)
                    End If
                End If
            End If
        Next I
    End If
 
'   This function will return Null if less than 2 non-Null values were passed.
    SecondMinimum = secondVal
 
    ' Expr1: SecondMinimum([nPP1SHF],[nPP2SHF],[nPP3SHF],[nPP4SHF],[nPP5SHF],[nPP6SHF],[nPP7SHF],[nPP8SHF],[nPP9SHF],[nPP0SHF])
 
End Function
 

[twinnyfo]

Did you adjust your code to match what I provided? It worked perfectly for me with all values.

[liamthequietman]

I just ran the code in post #23. When I place the expression into the query I receive, error wrong number of arguments:

Expr1: Minimum([nPP1SHF],[nPP2SHF],[nPP3SHF],[nPP4SHF],[nPP5SHF],[nPP6SHF],[nPP7SHF],[nPP8SHF],[nPP9SHF],[nPP0SHF])

[twinnyfo]

Did you change line 1 as discussed in post #27?