CitySec
asked on
How do I spell out a number in VB.NET/ASP.NET 1.1?
I have an integer that will always be a whole number between 1 and 20, but I'd like to display it for my users in it's spelled-out form. "1" would be "one", "2" would be "two", and so on up to 20. How would I go about accomplishing this in ASP.NET 1.1? Seems so simple but I could not find it on the Google.
i think you just need to convert number to string , if so this simple example to do this
Dim arra(,) As String = {{"0", "zero"}, {"1", "one"}, {"2", "two"}, {"3", "three"}}
Dim sentence As String = "112213"
Dim sentenceToView As String = ""
For i As Integer = 0 To sentence.Length - 1
sentenceToView = sentenceToView & arra(Convert.ToUInt32(sentence(i).ToString()), 1) & ","
Next
ASKER CERTIFIED SOLUTION
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The routine below, GetSpelledNumber, converts a real number into a spelled out number (English only). See the comments in the routine for instructions on how to use.
Examples:
GetSpelledNumber(1234) = One Thousand Two Hundred Thirty Four
GetSpelledNumber(1234.56) = One Thousand Two Hundred Thirty Four Point Five Six
GetSpelledNumber(1234.56, TRUE) = One Thousand Two Hundred Thirty Four Dollars and Fifty Six Cents
[Begin Code Segment]
Public Function GetSpelledNumber( _
ByVal Number As Double, _
Optional ByVal ReturnCurrency As Boolean, _
Optional ByVal ReturnFraction As Boolean _
) As String
' Return the number as a text string with each digit spelled out.
'
' Syntax
'
' GetSpelledNumber(Number, (ReturnCurrency])
'
' Number - Any double number.
'
' ReturnCurrency - Pass True to return the numnber as dollars and cents.
' Optional. If omitted then False is assumed.
'
' ReturnFraction - Pass True to return any decimal number as a numeric
' fraction versus words. When using this option, if there is no decimal
' portion then the word "exactly" is appended to the result. Optional. If
' omitted then False is assumed. Ignored if ReturnCurrency is True.
Dim NumberParts As Variant
Dim NumberText As String
Dim Index As Long
Dim Part As String
Dim Result As String
If Number = 0 Then
If ReturnCurrency Then
GetSpelledNumber = "No Dollars and No Cents"
Else
GetSpelledNumber = "Zero"
End If
Exit Function
End If
NumberParts = Split(Number, ".")
NumberText = NumberParts(0)
NumberText = Right(" " & NumberText, Int((Len(NumberText) + 2) / 3) * 3)
For Index = 1 To Len(NumberText) / 3
Part = GetSpelledNumberPart(CLng(
Select Case Index
Case 1: Result = Part
Case 2: Result = Part & " Thousand " & Result
Case 3: Result = Part & " Million " & Result
Case 4: Result = Part & " Billion " & Result
Case 5: Result = Part & " Trillion " & Result
End Select
Next Index
If ReturnCurrency Then Result = Result & " Dollar" & IIf(Not Result = "One", "s", vbNullString)
If ReturnCurrency Then
Result = Result & " and "
If UBound(NumberParts) > 0 Then
Part = GetSpelledNumberPart(Round
Else
Part = GetSpelledNumberPart(0)
End If
If Len(Part) = 0 Then
Result = Result & "No Cents"
Else
Result = Result & Part & " Cent" & IIf(Not Part = "One", "s", vbNullString)
End If
Else
If UBound(NumberParts) > 0 Then
If ReturnFraction Then
Result = Result & " and " & CLng(NumberParts(1)) & "/1" & String(Len(NumberParts(1))
Else
Result = Result & " Point"
For Index = 1 To Len(NumberParts(1))
Result = Result & " " & GetSpelledNumberPart(Mid(N
Next Index
End If
Else
If ReturnFraction Then Result = Result & " Exactly"
End If
End If
GetSpelledNumber = Result
End Function
Private Function GetSpelledNumberPart( _
ByVal NumberPart As String _
) As String
' Return the spelled out one to three digit number. Used by GetSpelledNumber.
'
' Syntax
'
' GetSpelledNumberPart(Numbe
'
' NumberPart - A one to three character string containing only numeric digits.
Dim Index As Long
Dim Result As String
Dim Part As String
NumberPart = Mid(NumberPart, 1, 3)
For Index = 1 To Len(NumberPart)
Part = vbNullString
Select Case Index
Case 1, 3
Select Case Mid(NumberPart, Len(NumberPart) - Index + 1, 1)
Case 1: Part = "One"
Case 2: Part = "Two"
Case 3: Part = "Three"
Case 4: Part = "Four"
Case 5: Part = "Five"
Case 6: Part = "Six"
Case 7: Part = "Seven"
Case 8: Part = "Eight"
Case 9: Part = "Nine"
End Select
If Index = 3 Then Part = Part & " Hundred"
Case 2
Select Case Mid(NumberPart, Len(NumberPart) - Index + 1, 1)
Case 1
Result = vbNullString
Select Case Mid(NumberPart, Len(NumberPart), 1)
Case 0: Part = "Ten"
Case 1: Part = "Eleven"
Case 2: Part = "Twelve"
Case 3: Part = "Thirteen"
Case 4: Part = "Fourteen"
Case 5: Part = "Fifteen"
Case 6: Part = "Sixteen"
Case 7: Part = "Seventeen"
Case 8: Part = "Eighteen"
Case 9: Part = "Nineteen"
End Select
Case 2: Part = "Twenty"
Case 3: Part = "Thirty"
Case 4: Part = "Forty"
Case 5: Part = "Fifty"
Case 6: Part = "Sixty"
Case 7: Part = "Seventy"
Case 8: Part = "Eighty"
Case 9: Part = "Ninety"
End Select
End Select
If Len(Part) > 0 Then Result = Part & IIf(Len(Result) = 0, vbNullString, " ") & Result
Next Index
GetSpelledNumberPart = Result
End Function
[End Code Segment]
Kevin