Please Help me in creating a replace function.
Problem:
Their is a alphanumeric value of any length (string) and I want to replace its all characters with 'X' except right four characters
Like :
Value : 4111111111111111
Result Should be: XXXXXXXXXXXX1111
I have created a function but got stuck:
public function myfunction(str as string)
str.Replace(str.Substring(0, str.Length - 5), 'X') 'but here I want no of x to be equals to count of length of str - 4
end function
What's a better function to perform such an operation?
Try this on for size.
Public Shared Function ObfuscateCardNumber(ByVal cardNumber As String) As String
If cardNumber.Length <= 4 Then
Return cardNumber
Else
Return cardNumber _
.Substring(cardNumber.Length - 4, 4) _
.PadLeft(cardNumber.Length, "X"c)
End If
End Function
Dim sNumber As String = "4111111111111111"
Dim sResult As String = StrDup(sNumber.Length - 4, "X"c) + Strings.Right(sNumber, 4)
something like
string result for(int i = 0;i > str.length -4;i++)
{
result = result +x
}
result = result + str.substrin(get
last 4)
Related
I have Textboxes Lines:
{ LstScan = 1,100, DrwR2 = 0000000043 }
{ LstScan = 2,200, DrwR2 = 0000000041 }
{ LstScan = 3,300, DrwR2 = 0000000037 }
I should display:
1,100
2,200
3,300
this is a code that I can't bring to a working stage.
Dim data As String = TextBox1.Lines(0)
' Part 1: get the index of a separator with IndexOf.
Dim separator As String = "{ LstScan ="
Dim separatorIndex = data.IndexOf(separator)
' Part 2: see if separator exists.
' ... Get the following part with Substring.
If separatorIndex >= 0 Then
Dim value As String = data.Substring(separatorIndex + separator.Length)
TextBox2.AppendText(vbCrLf & value)
End If
Display as follows:
1,100, DrwR2 = 0000000043 }
This should work:
Function ParseLine(input As String) As String
Const startKey As String = "LstScan = "
Const stopKey As String = ", "
Dim startIndex As String = input.IndexOf(startKey)
Dim length As String = input.IndexOf(stopKey) - startIndex
Return input.SubString(startIndex, length)
End Function
TextBox2.Text = String.Join(vbCrLf, TextBox1.Lines.Select(AddressOf ParseLine))
If I wanted, I could turn that entire thing into a single (messy) line... but this is more readable. If I'm not confident every line in the textbox will match that format, I can also insert a Where() just before the Select().
Your problem is you're using the version of substring that takes from the start index to the end of the string:
"hello world".Substring(3) 'take from 4th character to end of string
lo world
Use the version of substring that takes another number for the length to cut:
"hello world".Substring(3, 5) 'take 5 chars starting from 4th char
lo wo
If your string will vary in length that needs extracting you'll have to run another search (for example, searching for the first occurrence of , after the start character, and subtracting the start index from the newly found index)
Actually, I'd probably use Split for this, because it's clean and easy to read:
Dim data As String = TextBox1.Lines(0)
Dim arr = data.Split()
Dim thing = arr(3)
thing now contains 1,100, and you can use TrimEnd(","c) to remove the final comma
thing = thing.TrimEnd(","c)
You can reduce it to a one-liner:
TextBox1.Lines(0).Split()(3).TrimEnd(","c)
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
Hi I have a function that finds the longest common substring between two strings. It works great except it seems to break when it reaches any single quote mark: '
This causes it to not truly find the longest substring sometimes.
Could anyone help me adjust this function so it includes single quotes in the substring? I know it needs to be escaped someplace I'm just not sure where.
Example:
String 1: Hi there this is jeff's dog.
String 2: Hi there this is jeff's dog.
After running the function the longest common substring would be:
Hi there this is jeff
Edit: seems to also happen with "-" as well.
It will not count anything after the single quote as part of the substring.
Here's is the function:
Public Shared Function LongestCommonSubstring(str1 As String, str2 As String, ByRef subStr As String)
Try
subStr = String.Empty
If String.IsNullOrEmpty(str1) OrElse String.IsNullOrEmpty(str2) Then
Return 0
End If
Dim num As Integer(,) = New Integer(str1.Length - 1, str2.Length - 1) {}
Dim maxlen As Integer = 0
Dim lastSubsBegin As Integer = 0
Dim subStrBuilder As New StringBuilder()
For i As Integer = 0 To str1.Length - 1
For j As Integer = 0 To str2.Length - 1
If str1(i) <> str2(j) Then
num(i, j) = 0
Else
If (i = 0) OrElse (j = 0) Then
num(i, j) = 1
Else
num(i, j) = 1 + num(i - 1, j - 1)
End If
If num(i, j) > maxlen Then
maxlen = num(i, j)
Dim thisSubsBegin As Integer = i - num(i, j) + 1
If lastSubsBegin = thisSubsBegin Then
subStrBuilder.Append(str1(i))
Else
lastSubsBegin = thisSubsBegin
subStrBuilder.Length = 0
subStrBuilder.Append(str1.Substring(lastSubsBegin, (i + 1) - lastSubsBegin))
End If
End If
End If
Next
Next
subStr = subStrBuilder.ToString()
Return subStr
Catch e As Exception
Return ""
End Try
End Function
I tried it with dotnetfiddle and there it is working with your Code you posted. Please activate your warnings in your project. You have function with no return value and you return an integer or a string. This is not correct. How are you calling your function?
Here is my example I tested for you:
https://dotnetfiddle.net/mVBDQp
Your code works perfectly like Regex! As far as I can see, there is really nothing wrong with your code.
Here I even tested it under more severe case:
Public Sub Main()
Dim a As String = ""
Dim str1 As String = "Hi there this is jeff''s dog.-do you recognize this?? This__)=+ is m((a-#-&&*-ry$##! <>Hi:;? the[]{}re this|\ is jeff''s dog." 'Try to trick the logic!
Dim str2 As String = "Hi there this is jeff''s dog. ^^^^This__)=+ is m((a-#-&&*-ry$##! <>Hi:;? the[]{}re this|\ is jeff''s dog."
LongestCommonSubstring(str1, str2, a)
Console.WriteLine(a)
Console.ReadKey()
End Sub
Note that I put '-$#^_)=+&|\{}[]?!;:.<> all there. Plus I tried to trick your code by giving early result.
But the result is excellent!
You could probably put more actual samples on the inputs which give you problems. Else, you could possibly describe the environment that you use/deploy your code into. Maybe the problem lies elsewhere and not in the code.
The quickest way to solve this would be to use an escape code and replace all the ' with whatever escape code you use
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))
I have a string (for example: "Hello there. My name is John. I work very hard. Hello there!") and I am trying to find the number of occurrences of the string "hello there". So far, this is the code I have:
Dim input as String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase as String = "hello there"
Dim Occurrences As Integer = 0
If input.toLower.Contains(phrase) = True Then
Occurrences = input.Split(phrase).Length
'REM: Do stuff
End If
Unfortunately, what this line of code seems to do is split the string every time it sees the first letter of phrase, in this case, h. So instead of the result Occurrences = 2 that I would hope for, I actually get a much larger number. I know that counting the number of splits in a string is a horrible way to go about doing this, even if I did get the correct answer, so could someone please help me out and provide some assistance?
Yet another idea:
Dim input As String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase As String = "Hello there"
Dim Occurrences As Integer = (input.Length - input.Replace(phrase, String.Empty).Length) / phrase.Length
You just need to make sure that phrase.Length > 0.
the best way to do it is this:
Public Function countString(ByVal inputString As String, ByVal stringToBeSearchedInsideTheInputString as String) As Integer
Return System.Text.RegularExpressions.Regex.Split(inputString, stringToBeSearchedInsideTheInputString).Length -1
End Function
str="Thisissumlivinginsumgjhvgsum in the sum bcoz sum ot ih sum"
b= LCase(str)
array1=Split(b,"sum")
l=Ubound(array1)
msgbox l
the output gives u the no. of occurences of a string within another one.
You can create a Do Until loop that stops once an integer variable equals the length of the string you're checking. If the phrase exists, increment your occurences and add the length of the phrase plus the position in which it is found to the cursor variable. If the phrase can not be found, you are done searching (no more results), so set it to the length of the target string. To not count the same occurance more than once, check only from the cursor to the length of the target string in the Loop (strCheckThisString).
Dim input As String = "hello there. this is a test. hello there hello there!"
Dim phrase As String = "hello there"
Dim Occurrences As Integer = 0
Dim intCursor As Integer = 0
Do Until intCursor >= input.Length
Dim strCheckThisString As String = Mid(LCase(input), intCursor + 1, (Len(input) - intCursor))
Dim intPlaceOfPhrase As Integer = InStr(strCheckThisString, phrase)
If intPlaceOfPhrase > 0 Then
Occurrences += 1
intCursor += (intPlaceOfPhrase + Len(phrase) - 1)
Else
intCursor = input.Length
End If
Loop
You just have to change the input of the split function into a string array and then delare the StringSplitOptions.
Try out this line of code:
Occurrences = input.Split({phrase}, StringSplitOptions.None).Length
I haven't checked this, but I'm thinking you'll also have to account for the fact that occurrences would be too high due to the fact that you're splitting using your string and not actually counting how many times it is in the string, so I think Occurrences = Occurrences - 1
Hope this helps
You could create a recursive function using IndexOf. Passing the string to be searched and the string to locate, each recursion increments a Counter and sets the StartIndex to +1 the last found index, until the search string is no longer found. Function will require optional parameters Starting Position and Counter passed by reference:
Function InStrCount(ByVal SourceString As String, _
ByVal SearchString As String, _
Optional ByRef StartPos As Integer = 0, _
Optional ByRef Count As Integer = 0) As Integer
If SourceString.IndexOf(SearchString, StartPos) > -1 Then
Count += 1
InStrCount(SourceString, _
SearchString, _
SourceString.IndexOf(SearchString, StartPos) + 1, _
Count)
End If
Return Count
End Function
Call function by passing string to search and string to locate and, optionally, start position:
Dim input As String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase As String = "hello there"
Dim Occurrences As Integer
Occurrances = InStrCount(input.ToLower, phrase.ToLower)
Note the use of .ToLower, which is used to ignore case in your comparison. Do not include this directive if you do wish comparison to be case specific.
One more solution based on InStr(i, str, substr) function (searching substr in str starting from i position, more info about InStr()):
Function findOccurancesCount(baseString, subString)
occurancesCount = 0
i = 1
Do
foundPosition = InStr(i, baseString, subString) 'searching from i position
If foundPosition > 0 Then 'substring is found at foundPosition index
occurancesCount = occurancesCount + 1 'count this occurance
i = foundPosition + 1 'searching from i+1 on the next cycle
End If
Loop While foundPosition <> 0
findOccurancesCount = occurancesCount
End Function
As soon as there is no substring found (InStr returns 0, instead of found substring position in base string), searching is over and occurances count is returned.
Looking at your original attempt, I have found that this should do the trick as "Split" creates an array.
Occurrences = input.split(phrase).ubound
This is CaSe sensitive, so in your case the phrase should equal "Hello there", as there is no "hello there" in the input
Expanding on Sumit Kumar's simple solution, here it is as a one-line working function:
Public Function fnStrCnt(ByVal str As String, ByVal substr As String) As Integer
fnStrCnt = UBound(Split(LCase(str), substr))
End Function
Demo:
Sub testit()
Dim thePhrase
thePhrase = "Once upon a midnight dreary while a man was in a house in the usa."
If fnStrCnt(thePhrase, " a ") > 1 Then
MsgBox "Found " & fnStrCnt(thePhrase, " a ") & " occurrences."
End If
End Sub 'testit()
I don't know if this is more obvious?
Starting from the beginning of longString check the next characters up to the number characters in phrase, if phrase is not found start looking from the second character etc. If it is found start agin from the current position plus the number of characters in phrase and increment the value of occurences
Module Module1
Sub Main()
Dim longString As String = "Hello there. My name is John. I work very hard. Hello there! Hello therehello there"
Dim phrase As String = "hello There"
Dim occurences As Integer = 0
Dim n As Integer = 0
Do Until n >= longString.Length - (phrase.Length - 1)
If longString.ToLower.Substring(n, phrase.Length).Contains(phrase.ToLower) Then
occurences += 1
n = n + (phrase.Length - 1)
End If
n += 1
Loop
Console.WriteLine(occurences)
End Sub
End Module
I used this in Vbscript, You can convert the same to VB.net as well
Dim str, strToFind
str = "sdfsdf:sdsdgs::"
strToFind = ":"
MsgBox GetNoOfOccurranceOf( strToFind, str)
Function GetNoOfOccurranceOf(ByVal subStringToFind As String, ByVal strReference As String)
Dim iTotalLength, newString, iTotalOccCount
iTotalLength = Len(strReference)
newString = Replace(strReference, subStringToFind, "")
iTotalOccCount = iTotalLength - Len(newString)
GetNoOfOccurranceOf = iTotalOccCount
End Function
I know this thread is really old, but I got another solution too:
Function countOccurencesOf(needle As String, s As String)
Dim count As Integer = 0
For i As Integer = 0 to s.Length - 1
If s.Substring(i).Startswith(needle) Then
count = count + 1
End If
Next
Return count
End Function