Comparing Text Strings In Access

No, you'll need a function that's taking all possible 5 characters from the first field (assuming it's at least 5 characters or more in length) and uses an INSTR to detect a "hit".

Niv;o)

Nico, your answer won't work in general, as INSTR will return a HIT if those 5 characters are found ANYWHERE in the second string, but the question asked for THE SAME FIVE CHARACTERS in the second string.  You would need to test that the value returned by INSTR, if NOT 0, was equal to the starting position of the 5 Character string from the First field.

Arthur Wood

Here is a code solution that will work for what you need:

''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
'Use the ReturnValues sub to fill an Array with all of the 5 consecutive
'letters possibilities in String1 that are in String2.
'
'Created by Joe Kendall 7/19/2002
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''

Private m_sArray() As String
Private m_nCount As Long

Public Sub ReturnValues(String1 As String, String2 As String)
    '''''''''''''''''''''''''''''''''''''''''''''''''''''''''
    'Usage: Call ReturnValues("ABCDEFGHI", "ZABABCDEFGHIZ")
    '''''''''''''''''''''''''''''''''''''''''''''''''''''''''
    Dim nLen As Long
    Dim nPos As Long
   
    If Len(String1) < 5 Then Exit Sub
    If Len(String2) < 5 Then Exit Sub
   
    m_nCount = 0
    ReDim m_sArray(0)
   
    nLen = Len(String1)
    Do While (nPos + 5) <= nLen
        nPos = nPos + 1
        Call CompareString(String1, String2, nPos)
    Loop
   
    Call PrintValues
End Sub

Private Sub CompareString(String1 As String, String2 As String, Position As Long)
    If InStr(1, String2, MID$(String1, Position, 5)) > 0 Then
        m_nCount = m_nCount + 1
        ReDim Preserve m_sArray(m_nCount)
        m_sArray(m_nCount) = MID$(String1, Position, 5)
    End If
End Sub

Private Sub PrintValues()
    Dim nCount As Long
    Dim nCounter As Long
   
    nCount = UBound(m_sArray())
   
    For nCounter = 0 To nCount
        Debug.Print m_sArray(nCounter)
    Next
End Sub

Let me know if you have any questions.

Thanks!

Joe

You can also make a minor change in mine and it will compare case sensitive letters too. So that ABCDE will not equal abcde.

Thanks!

Joe

I guess you can be right Arthur, it's just how you read the Q.
But INSTR can be used, as it returns the first position and comparing that with the start of the "field1 5 character string" can solve that ;-)

Lets hear what jps9832 needs:
12345a matches a12345
and/or
12345a matches 12345b

Nic;o)

Public function Match5(One as String, Two as String) as boolean

Dim Cnt1 as long
Dim Position1 a

Great solution!!! Exactly what I needed!

A more 'general' solution with NumberOfChar instead of 5:

Public Function MatchX(One As String, Two As String, NumberOfChar As Long) As Boolean

Dim Cnt1 As Long
Dim TestStr As String

If Len(One) < NumberOfChar Or Len(Two) < NumberOfChar Then
   MatchX = False
   Exit Function
End If

For Cnt1 = 1 To Len(One) - NumberOfChar + 1
 TestStr = Mid(One, Cnt1, NumberOfChar)
 If InStr(Two, TestStr) > 0 Then
     MatchX = True
     Exit Function
 End If
Next Cnt1

MatchX = False
End Function

jjv