Increment character in a string - vb.net

I have a 2 character string composed only of the 26 capital alphabet letters, 'A' through 'Z'.
We have a way of knowing the "highest" used value (e..g "IJ" in {"AB", "AC", "DD", "IH", "IJ"}). We'd like to get the "next" value ("IK" if "IJ" is the "highest").
Function GetNextValue(input As String) As String
Dim first = input(0)
Dim last = input(1)
If last = "Z"c Then
If first = "Z"c Then Return Nothing
last = "A"c
first++
Else
last++
EndIf
Return first & last
End Function
Obviously char++ is not valid syntax in VB.NET. C# apparently allows you to do this. Is there something shorter less ugly than this that'd increment a letter? (Note: Option Strict is on)
CChar(CInt(char)+1).ToString
Edit: As noted in comment/answers, the above line won't even compile. You can't convert from Char -> Integer at all in VB.NET.

The tidiest so far is simply:
Dim a As Char = "a"
a = Chr(Asc(a) + 1)
This still needs handling for the "z" boundary condition though, depending on what behaviour you require.
Interestingly, converting char++ through developerfusion suggests that char += 1 should work. It doesn't. (VB.Net doesn't appear to implicitly convert from char to int16 as C# does).
To make things really nice you can do the increment in an Extension by passing the char byref. This now includes some validation and also a reset back to a:
<Extension>
Public Sub Inc(ByRef c As Char)
'Remember if input is uppercase for later
Dim isUpper = Char.IsUpper(c)
'Work in lower case for ease
c = Char.ToLower(c)
'Check input range
If c < "a" Or c > "z" Then Throw New ArgumentOutOfRangeException
'Do the increment
c = Chr(Asc(c) + 1)
'Check not left alphabet
If c > "z" Then c = "a"
'Check if input was upper case
If isUpper Then c = Char.ToUpper(c)
End Sub
Then you just need to call:
Dim a As Char = "a"
a.Inc() 'a is now = "b"

My answer will support up to 10 characters, but can easily support more.
Private Sub Test
MsgBox(ConvertBase10ToBase26(ConvertBase26ToBase10("AA") + 1))
End Sub
Public Function ConvertBase10ToBase26(ToConvert As Integer) As String
Dim pos As Integer = 0
ConvertBase10ToBase26 = ""
For pos = 10 To 0 Step -1
If ToConvert >= (26 ^ pos) Then
ConvertBase10ToBase26 += Chr((ToConvert \ (26 ^ pos)) + 64)
ToConvert -= (26 ^ pos)
End If
Next
End Function
Public Function ConvertBase26ToBase10(ToConvert As String) As Integer
Dim pos As Integer = 0
ConvertBase26ToBase10 = 0
For pos = 0 To ToConvert.Length - 1
ConvertBase26ToBase10 += (Asc(ToConvert.Substring(pos, 1)) - 64) * (26 ^ pos)
Next
End Function

Unfortunately, there's no easy way -- even CChar(CInt(char)+1).ToString doesn't work. It's even uglier:
CChar(Char.ConvertFromUtf32(Char.ConvertToUtf32(myCharacter, 0) + 1))
but of course you could always put that in a function with a short name or, like Jon E. pointed out, an extension method.

Try this
Private Function IncBy1(input As String) As String
Static ltrs As String = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
Dim first As Integer = ltrs.IndexOf(input(0))
Dim last As Integer = ltrs.IndexOf(input(1))
last += 1
If last = ltrs.Length Then
last = 0
first += 1
End If
If first = ltrs.Length Then Return Nothing
Return ltrs(first) & ltrs(last)
End Function
This DOES assume that the code is only two chars, and are A-Z only.

Dim N as String = ""
Dim chArray As Char = Convert.ToChar(N)
Dim a As String = CChar(Char.ConvertFromUtf32(Char.ConvertToUtf32(chArray, 0) + 1))

Related

Math expression parsing: incorrect output when a + followed by a -

I am trying to create a program that parses an expression such as 3x^3 + 7x^2 + 6x - 9. I am using a recursive system to do this. At the moment, I'm just testing to see if I get the correct output for the expression I input by only using constants in the expression. The code works fine with most expressions, but when I have a minus followed by a positive, both terms are subtracted (first term should be subtracted, second term should be added). An example of this is entering 4*3^2-7*3+5*3. The answer to this is 30 but the program does 4*3^2-7*3-5*3 instead and outputs 0. I am unsure how to solve this.
Code:
Private Function ParseExpr(ByRef expression As String)
Dim op, op1 As Integer
op = ParseFactor(expression)
If expression.Length <> 0 Then
If (expression(0) = "+") Then
expression = expression.Substring(1, expression.Length - 1)
op1 = ParseExpr(expression)
op += op1
ElseIf (expression(0) = "-") Then
expression = expression.Substring(1, expression.Length - 1)
op1 = ParseExpr(expression)
op -= op1
End If
End If
Return op
End Function
All is well with raising to a power and multiplication. At that point we are simplified to 36-21+15. Here we must remember that we are adding -21. So we need to go back to -7 . When I build my list of numbers (it handles numbers of more than one digit) I add the minus sign to the number it precedes.
This code does not handle decimal numbers or parenthesis. I think you will be able to add the division operator, if you wish.
Private NumList As New List(Of Double)
Private OperatorList As New List(Of String)
Private Sub OpCode()
Dim Input = "14*3^2-7*3+15*3"
PopulateLists(Input)
Dim OpIndex As Integer
Dim NewNum As Double
Dim operators = {"^", "*", "+"} 'Note: no minus sign, the minus goes with the number
For Each op In operators
Do
OpIndex = OperatorList.IndexOf(op)
If OpIndex = -1 Then
Exit Do
End If
Select Case op
Case "^"
NewNum = NumList(OpIndex) ^ NumList(OpIndex + 1)
Case "*"
NewNum = NumList(OpIndex) * NumList(OpIndex + 1)
Case "+"
NewNum = NumList(OpIndex) + NumList(OpIndex + 1)
End Select
NumList.RemoveAt(OpIndex + 1)
NumList(OpIndex) = NewNum
OperatorList.RemoveAt(OpIndex)
Loop
Next
MessageBox.Show(NumList(0).ToString) 'Displays 150
End Sub
Private Sub PopulateLists(Input As String)
Dim strNum As String = ""
For Each c As Char In Input 'Be careful here, the IDE wants to add () at the end of this line - it doesn't belong
If Char.IsDigit(c) Then
strNum &= c
ElseIf c = "-" Then
OperatorList.Add("+") 'We are adding a negative number
NumList.Add(CDbl(strNum)) 'Add the last number we accumulated so we can start a new one with the minus sign
strNum = "-" 'Start a new number with the minus sign
Else 'The other operators are added to the list
OperatorList.Add(c)
NumList.Add(CDbl(strNum))
strNum = ""
End If
Next
NumList.Add(CInt(strNum)) 'The last number which will not be followed by an operator
End Sub

Add integer to another integer in vb.net?

How can I add an integer to another integer in vb.net?
This is what I need to do:
Given integer: 2187 ->
Converted integer: 2018
I need to add a 0 in between the first and second number, and drop the last digit. This will give me the year.
Here is the code that I have:
Protected Function GetYear(ByVal term As Integer) As Integer
Dim termYear As String = Convert.ToString(term)
termYear.Substring(0, 2)
termYear.Insert(1, "0")
Dim convertedYear As Integer
Int32.TryParse(termYear.ToString, convertedYear)
convertedYear = convertedYear / 10
Return convertedYear
End Function
In general strings are immutable. So you'd have to create a new string out of the addition of substrings. Check this possible solution.
Function GetYear(ByVal term As Integer) As Integer
Dim termYear As String = Convert.ToString(term, Globalization.CultureInfo.InvariantCulture)
Dim result As String = termYear.Substring(0, 1) + "0" + termYear.Substring(1, 2)
Return Int32.Parse(result)
End Function
Strings are immutable, when you do any changes with one of their method, you need to get the returned string.
termYear = termYear.Insert(1, "0")
This question deserves a math based solution. The below code specifies the zero insertion point relative to the number's right side instead of the left as stated in the problem statement. So for a 4 digit number the insertion point is 3 versus 2. It also allows you to change the insertion point.
Private Function GetYear(ByVal term As Integer, Optional zeroDigitPosition As Integer = 3) As Integer
If zeroDigitPosition > 0 Then
Dim divisor As Integer = 1
For i As Integer = 1 To zeroDigitPosition - 1
divisor *= 10
Next
Dim ret As Integer = term \ 10 ' drop one's place digit, remaining digits shift to right
Dim rightShiftedDigits As Integer = ret Mod divisor
Dim remainder As Integer = Math.DivRem(ret, divisor, rightShiftedDigits)
' shift the remainder to the left by divisor * 10
' (remember first right shift invplved \ 10) and add
' rightShiftedDigits to yield result
Return (remainder * divisor * 10) + rightShiftedDigits
Else
Throw New ArgumentOutOfRangeException("zeroDigitPosition must be greater then zero")
End If
End Function

Function which Removes Only Non-ASCII characters in a column in access table

I have a access table and i am writing a vba code to remove non-ascii characters from the table, i have tried using below two functions
Public Function removeall(stringData As String) As String
Dim letter As Integer
Dim final As String
Dim i As Integer
For i = 1 To Len(stringData) 'loop thru each char in stringData
letter = Asc(Mid(stringData, i, 1)) 'find the char and assign asc value
Select Case letter 'Determine what type of char it is
Case Is < 91 And letter > 64 'is an upper case char
final = final & Chr(letter)
Case Is < 123 And letter > 96 'is an lower case char
final = final & Chr(letter)
Case Is = 32 'is a space
final = final & Chr(letter)
End Select
Next i
removeall = final
End Function
And also tried using below function
Public Function Clean(InString As String) As String
'-- Returns only printable characters from InString
Dim x As Integer
For x = 1 To Len(InString)
If Asc(Mid(InString, x, 1)) > 31 And Asc(Mid(InString, x, 1)) < 127 Then
Clean = Clean & Mid(InString, x, 1)
End If
Next x
End Function
But the problem is : In removeall function it removes everything including # and space characters.. And In Clean function also removes special characters as well.
I need a correct function which retains key board characters and removes all other characters
Examples of strings in tables are :
1) "ATTACHMENT FEEDING TUBE FITS 5-18 ºFR# "
2) "CATHETER FOLEY 3WAY SILI ELAST 20FR 30ML LATEXº"
Any help would be greatly appreciated
Output should be like
1) "ATTACHMENT FEEDING TUBE FITS 5-18 FR"
2) "CATHETER FOLEY 3WAY SILI ELAST 20FR 30ML LATEX"
One approach would be to use a whitelist of accepted characters. e.g.
' You can set up your domain specific list:
Const Whitelist = "1234567890" & _
"qwertyuiopasdfghjklzxcvbnm" & _
"QWERTYUIOPASDFGHJKLZXCVBNM" & _
" `~!##$%^&*()_-=+[]{};:""'|\<>?/ –"
Public Sub test()
Debug.Print Clean("ATTACHMENT FEEDING TUBE FITS 5-18 ºFR#")
Debug.Print Clean("CATHETER FOLEY 3WAY SILI ELAST 20FR 30ML LATEXº")
End Sub
Public Function isAllowed(char As String) As Boolean
isAllowed = InStr(1, Whitelist, char, vbBinaryCompare) > 0
End Function
Public Function Clean(dirty As String) As String
'-- Returns only printable characters from dirty
Dim x As Integer
Dim c As String
For x = 1 To Len(dirty)
c = Mid(dirty, x, 1)
If isAllowed(c) Then
Clean = Clean & c
End If
Next x
End Function
Alternate approach that preserves ALL ASCII characters, without working with a whitelist, in a single function:
Public Function RemoveNonASCII(str As String) As String
Dim i As Integer
For i = 1 To Len(str)
If AscW(Mid(str, i, 1)) < 127 Then 'It's an ASCII character
RemoveNonASCII = RemoveNonASCII & Mid(str, i, 1) 'Append it
End If
Next i
End Function

Shift letters to the end of a string Visual Basic

I'm trying to shift letters to the end of the word. Like the sample output I have in the image.
Using getchar and remove function, I was able to shift 1 letter.
mychar = GetChar(word, 1) 'Get the first character
word = word.Remove(0, 1) 'Remove the first character
input.Text = mychar
word = word & mychar
output.Text = word
This is my code for shifting 1 letter.
I.E. for the word 'Star Wars', it currently shifts 1 letter, and says 'tar WarsS'
How can I make this move 3 characters to the end? Like in the sample image.
intNumChars = input.text
output.text = mid(word,4,len(word)) & left(word,3)
I wanted it to be easy for you to read but you can set the intNumChars variable to the value in your text box and replace the 4 with intNumChars + 1 and the 3 with intNumChars.
The mid() function can return a section of text in the middle of a string mid(string,start,finish). The len() function returns the length of a string so that the code will work on texts that are different lengths. The left function returns characters from the left() of a string.
I hope this is of some help.
You could write a that as a sort of permute, which maps each char-index to a new place in the range [0, textLength[
In order to do that you'll have to write a custom modulus as the Mod operator is more a remainder than a modulus (from a mathematical point of view, regarding how negative are handled)
With that you just need to loop over your string indexes and map each one to it's "offsetted" value modulo the length of the text
' Can be made local to Shift method if needed
Function Modulus(dividend As Integer, divisor As Integer) As Integer
dividend = dividend Mod divisor
Return If(dividend < 0, dividend + divisor, dividend)
End Function
Function Shift(text As String, offset As Integer) As String
' validation omitted
Dim length = text.Length
Dim arr(length - 1) As Char
For i = 0 To length - 1
arr(Modulus(i + offset, length) Mod length) = text(i)
Next
Return new String(arr)
End Function
That way you can easily handle negative values or values greater than the length of the text.
Note, the same thing is possible with a StringBuilder instead of an array ; I'm not sure which one is "better"
Function Shift(text As String, offset As Integer) As String
Dim builder As New StringBuilder(text)
Dim length = text.Length
For i = 0 To length - 1
builder(Modulus(i + offset, length) Mod length) = text(i)
Next
Return builder.ToString
End Function
Using Gordon's code did the trick. The left function visual studio tried to create a stub of the function, so I used the fully qualified function name when calling it. But this worked perfectly.
intNumChars = shiftnumber.Text
output.Text = Mid(word, intNumChars + 1, Len(word)) & Microsoft.VisualBasic.Left(word, intNumChars)
n = 3
output.Text = Right(word, Len(word) - n) & Left(word, n)

Get last number from string and increase it by one

Have some problems with my function. In database i could have diffrent numbers. For instance below: ( i know it looks strange )
12 312323.3
013.43.9
3.23.14353.55 WHATEVER 345.193
728937.3
87.3 ojojo 23.434blabla 24.424.7
What i need to do is increase number after LAST DOT so just make + 1.
The problem is its not working when it comes after dot more than one digit then.
here is my current code:
Dim inputValue as String = "34.234234.6.12"
'--Get Last char from string and add 1 to it
Dim lastChar As String = CInt(CStr(inputValue.Last)) + 1
'--Remove last char and add lastChar
Dim nextCombinNummer As String = lastValue.Nummer.Substring(0, lastValue.Nummer.Length - 1) & lastChar
Return nextCombinNummer
I think the problem is lastValue.Last + 1 as it will take only one digit, and also when i remove by substring last digit but only 2 will be removed.
Can you help me out with this? How to always take number after last dot from string and then increase that number by 1 and return new entire number?
EDIT:
I think i am able to get and increase the number but still dont know how to remove and put it at the end:
Think that's ok:
Dim inputValue as String = "34.234234.6.12"
Dim number As String = inputValue .Substring(inputValue .LastIndexOf("."c) + 1)
Dim numberIncreased as integer = CInt(number) + 1
'How to do this correctly? :
Dim nextCombinNummer As String = lastValue.Nummer.Substring(0, lastValue.Nummer.Length - 1) & numberIncreased
An easy solution is to cast as Integer the last part of the string, add one, then recompose your string :
'Original Value
Dim val As String = "123.456.789"
'We take only the last part and add one
Dim nb = Integer.Parse(val.Substring(val.LastIndexOf(".") + 1)) + 1
'We recompose the string
Dim FinalVal As String = val.Substring(0, val.LastIndexOf(".") + 1) & nb.ToString()
I'd use following which uses String.Split, Int32.TryParse and String.Join:
Dim numbers As New List(Of String) From {"12.312323.3", "013.43.9", "3.231435355345.193", "728937.3", "87.323.43424.424.7"}
for i As Int32 = 0 To numbers.Count -1
Dim num = numbers(i)
Dim token = num.Split("."c)
dim lastNum = token.Last() ' or token(token.Length-1)
Dim n As Int32
If int32.TryParse(lastNum, n)
n += 1
token(token.Length-1) = n.ToString()
End If
numbers(i) = string.Join(".", token)
Next