how to find the number of occurrences of a substring within a string vb.net

Question

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

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

Expanding on Sumit Kumar's simple solution (please upvote his answer rather than this one), 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()