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InStr Question

Hello,

VB6(sp5) ADO

How would I find a string within a string and flag it with a message box as in the following?

Dim strDude As String
strFind = "DSMan"
strDude = "DSMan.mdb"

If strFind is in the string: strDude then

msgbox "Found in string"

End IF

Thanks,

ADawn
0
ADawn
Asked:
ADawn
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1 Solution
 
hesCommented:
Try
If InStr(strDude,strFind) then
  msgbox "Found in string"
End If
0
 
rspahitzCommented:
Is this a trick Q?  If you knew to use the Instr function, why not just check the syntax in the help files?
0
 
amebaCommented:
Yes, in help files (F1), or in Object Browser (F2), or just type "Instr(" to see a list of arguments.
This also works:
    If UBound(Split(strDude, strfind)) > 0 Then
        MsgBox "Found in string"
    End If
0
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amebaCommented:
or:
    If strDude Like "*" & strFind & "*" Then
        MsgBox "Found in string"
    End If
0
 
amebaCommented:
or (VB6 only):
    If UBound(Filter(Array(strDude), strfind)) > -1 Then
        MsgBox "Found in string"
    End If
0
 
amebaCommented:
want more?  :-)
    If InStrRev(strDude, strfind) > 0 Then
        MsgBox "Found in string"
    End If
0
 
ADawnAuthor Commented:
I had the right answer all along. The problem was with the letter CASE when I did the search. DSDUDE wasn't the same as dsdude. So, I converteed both values to lower case, then ran the code. It works fine.

Is there a function that will convert to proper case, then make the search for a particular item within the string?

ADawn
0
 
ADawnAuthor Commented:
I had the right answer all along. The problem was with the letter CASE when I did the search. DSDUDE wasn't the same as dsdude. So, I converteed both values to lower case, then ran the code. It works fine.

Is there a function that will convert to proper case, then make the search for a particular item within the string?

ADawn
0
 
rspahitzCommented:
If InStr(strDude,strFind,,vbTextCompare) then
...
:this will compare without regard to text case and would give the same results as this:

If InStr(lcase(strDude),lcase(strFind)) then
0

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