Check if a string contains another string

You can also use the special word like:

Public Sub Search()
  If "My Big String with, in the middle" Like "*,*" Then
    Debug.Print ("Found ','")
  End If
End Sub

There is also the InStrRev function which does the same type of thing, but starts searching from the end of the text to the beginning.

Per @rene's answer...

Dim pos As Integer
pos = InStrRev("find the comma, in the string", ",")

...would still return 15 to pos, but if the string has more than one of the search string, like the word "the", then:

Dim pos As Integer
pos = InStrRev("find the comma, in the string", "the")

...would return 20 to pos, instead of 6.

Building on Rene's answer, you could also write a function that returned either TRUE if the substring was present, or FALSE if it wasn't:

Public Function Contains(strBaseString As String, strSearchTerm As String) As Boolean
'Purpose: Returns TRUE if one string exists within another
On Error GoTo ErrorMessage
    Contains = InStr(strBaseString, strSearchTerm)
Exit Function
ErrorMessage:
MsgBox "The database has generated an error. Please contact the database administrator, quoting the following error message: '" & Err.Description & "'", vbCritical, "Database Error"
End
End Function

You wouldn't really want to do this given the existing Instr/InstrRev functions but there are times when it is handy to use EVALUATE to return the result of Excel worksheet functions within VBA

Option Explicit

Public Sub test()

    Debug.Print ContainsSubString("bc", "abc,d")

End Sub
Public Function ContainsSubString(ByVal substring As String, ByVal testString As String) As Boolean
    'substring = string to test for; testString = string to search
    ContainsSubString = Evaluate("=ISNUMBER(FIND(" & Chr$(34) & substring & Chr$(34) & ", " & Chr$(34) & testString & Chr$(34) & "))")

End Function

Just to complete the possibilities listed, I would like to demonstrate how you can use Split() for an all-round function with the following variants depending on the optional argument n passed:

• a) show whether a substring was found at all (-1 or omitted as default value)
• b) show how many substrings were found (0) ,
• c) show at which position the nth substring was found (1 .. n).

Function StrIncludes( _
      ByVal s As String, _
      Optional ByVal IncludeString As String = ",", _
      Optional n As Long = -1 _
    ) As Long
'Purp.: find specified substring based on numeric value n
'Note : 2nd argument IncludeString is optional (default value is comma if omitted)
'       3rd argument n:  -1~~>only boolean; 0~~>count(s); 1..n ~~>position
    Dim tmp: tmp = Split(s, IncludeString)
    StrIncludes = UBound(tmp) > 0           ' a) boolean return value indicating a found substring
    
    Select Case n                           ' individual numeric values:
        Case 0                              ' b) return Count(s), not boolean value
            StrIncludes = UBound(tmp)
        Case 1
            StrIncludes = IIf(StrIncludes, Len(tmp(n - 1)) + n, 0)
        Case Is > 1                        ' c) return Position of nth finding
            If n > UBound(tmp) Then StrIncludes = 0: Exit Function
            StrIncludes = IIf(StrIncludes, Len(tmp(0)) + n, 0)
            Dim i As Long
            For i = 2 To n: StrIncludes = StrIncludes + Len(tmp(i - 1)): Next
    End Select
End Function

Example call

Sub ExampleCall()
'   define base string
    Dim s As String
    s = "Take this example string, does it contain a comma, doesn't it?"
'a) check if base string contains indicated search string, e.g. a comma (default value)
    Debug.Print "Is Found: " & CBool(StrIncludes(s)) ' ~~> Is Found: True
'b) get number of substrings
    Debug.Print "Count(s): " & StrIncludes(s, , 0)   ' ~~> Count(s): 2
'c) get position of nth substring
    Debug.Print "~~~ Findings of nth substring ~~~ "
    Dim n As Long
    For n = 1 To 3
        Debug.Print n & ordinalSuffix(n) & " substring at Pos.:  " & StrIncludes(s, , n)
    Next
End Sub

Function ordinalSuffix(ByVal number As Long) As String
    Dim suffixes: suffixes = Split(" st nd rd th")
    ordinalSuffix = suffixes(Abs(number))
End Function

Debugging results in immediate window

Is Found: Wahr
Count(s): 2
~~~ Findings of nth substring ~~~ 
1st substring at Pos.:  25
2nd substring at Pos.:  50
3rd substring at Pos.:  0  ' no finding at all