I have seen a lot of Topics to the "unable to get the match property of the Worksheetfunction class" problem. But I can't get my code fixed.
Why isn't this code work?
rowNum = Application.WorksheetFunction.Match(aNumber, Sheet5.Range("B16:B615"), 0)
But a few rows higher this code works:
rowNum2 = Application.WorksheetFunction.Match(originCode, Sheet7.Range("B10:B17"), 0)
The only difference between my two lines is that in rowNum2 I used a String for look up and in rowNum a integer. Is it possible that the look up Value needs to be a String?
@Update on my Problem
Select Case service
Case "Low Cost"
MsgBox Sheet5.Cells(16, "B") 'Gets value 0.5
Set Rng = Sheet5.Range("B16:B615")
If Not IsError(Application.Match("0.5", Rng, 0)) Then 'But jumps to Else
rowNum = Application.Match(Weight, Rng, 0) 'Weight = 0.5
MsgBox rowNum
Else
MsgBox "error"
End If
Case "Standard"
Case "Express"
Case Else
End Select
@UPDATE 2
!!! Take care that "0.5" is a String and not 0.5 So 0.5 is not "0.5" (that was my error in the code)
Match returns the position of the matched value within lookup_array, not the value itself. For example, MATCH("b",{"a","b","c"},0) returns 2, the relative position of "b" within the array {"a","b","c"} . Match does not distinguish between uppercase and lowercase letters when matching text values.
The MATCH function looks for the position of the lookup value in the selected array table. The MATCH function is mainly used with the VLOOKUP function to supply the column index number using the column heading automatically. The MATCH function is available as a worksheet function in VBA.
VBA Match Function looks for the position or row number of the lookup value in the table array i.e. in the main excel table. For example, VLOOKUP, HLOOKUP, MATCH, INDEX, etc. These are the lookup functions that are more important than others.
The MATCH function searches for a specified item in a range of cells, and then returns the relative position of that item in the range. For example, if the range A1:A3 contains the values 5, 25, and 38, then the formula =MATCH(25,A1:A3,0) returns the number 2, because 25 is the second item in the range.
You are getting this error because the value cannot be found in the range. String or integer doesn't matter. Best thing to do in my experience is to do a check first to see if the value exists.
I used CountIf below, but there is lots of different ways to check existence of a value in a range.
Public Sub test()
Dim rng As Range
Dim aNumber As Long
aNumber = 666
Set rng = Sheet5.Range("B16:B615")
If Application.WorksheetFunction.CountIf(rng, aNumber) > 0 Then
rowNum = Application.WorksheetFunction.Match(aNumber, rng, 0)
Else
MsgBox aNumber & " does not exist in range " & rng.Address
End If
End Sub
ALTERNATIVE WAY
Public Sub test()
Dim rng As Range
Dim aNumber As Variant
Dim rowNum As Long
aNumber = "2gg"
Set rng = Sheet5.Range("B1:B20")
If Not IsError(Application.Match(aNumber, rng, 0)) Then
rowNum = Application.Match(aNumber, rng, 0)
MsgBox rowNum
Else
MsgBox "error"
End If
End Sub
OR
Public Sub test()
Dim rng As Range
Dim aNumber As Variant
Dim rowNum As Variant
aNumber = "2gg"
Set rng = Sheet5.Range("B1:B20")
rowNum = Application.Match(aNumber, rng, 0)
If Not IsError(rowNum) Then
MsgBox rowNum
Else
MsgBox "error"
End If
End Sub
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