On Aug 23, 4:00*pm, GS wrote:
GS presented the following explanation :





Try...

Sub ParseAmount()
* Dim vTemp As Variant
* vTemp = Split(RemoveCharacters(Range("A1").Value, ","), ".")
* With Range("B1")
* * .Value = vTemp(0)
* * With .Offset(, 1)
* * * .NumberFormat = "00": .Value = vTemp(1)
* * End With '.Offset(, 1)
* End With 'Range("B1")
End Sub

Function RemoveCharacters(Amount As Double, Char As String) As Variant
* RemoveCharacters = Replace(CStr(Amount), Char, "")
End Function

Assumes ColA is where the amounts are listed, ColB and ColC is where
the parsed values go.

Assumes you want a 2-digit result for the cents in ColC, AND you will
format ColB to include thousands separator. Otherwise, the function is
not needed...

Sub ParseAmount2()
* Dim vTemp As Variant
* vTemp = Split(Range("A1").Value, ".")
* With Range("B1")
* * .Value = vTemp(0)
* * With .Offset(, 1)
* * * .NumberFormat = "00": .Value = vTemp(1)
* * End With '.Offset(, 1)
* End With 'Range("B1")
End Sub

OR...
If you want the result to be formatted as text (ie: left aligned):

Sub ParseAmount3()
* With Range("A1")
* * .Offset(, 1).Resize(1, 2) = Split(.Value, ".")
* End With 'Range("A1")
End Sub

--
Garry

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Ok, I liked the approach above best for spreading the number between
two columns.

Now, how do I get the 2nd part of my puzzle to work where 27651.09
becomes split one digit to a column?

Somehow when I get this looping and putting numbers in different
places, the routine needs to start in the right column. For example,
if the next row has 107150.25, then the 1 in this example must start
in the appropriate column so that the placeholders line up properly .
It would be best for it to count from the right most column; I am
allowed 9 columns in the table. These digits all represent dollars and
cents.

1 0 7 1 5 0 2 5
2 7 6 5 1 0 9
7 0 2 5 1 1