Thanks.

"Bob Phillips" <bob.phillips@notheretiscali.co.uk> wrote in message
news:%239ZVCoGlFHA.3568@tk2msftngp13.phx.gbl...
> myFormula = "=COUNTIF(E2:E" & LastRow - 1 & ",""" & currentCDBID & """)"
>
>
> --
>
> HTH
>
> RP
> (remove nothere from the email address if mailing direct)
>
>
> "Xavier Minet" <xminet@hotmail.com> wrote in message
> news:%23roepXGlFHA.1204@TK2MSFTNGP12.phx.gbl...
>> Hi,
>>
>> I am trying to identify duplicate values in an Excel 2003 column. If I
>> am
>> using the following code:
>>
>> myFormula = "=COUNTIF(E2:E" & LastRow - 1 & ",""=" & currentCDBID & """)"
>> countDUP = Evaluate(myFormula)
>> If countDUP > 1 Then ... rest of my code...
>>
>> countDUP always returns "0" as result of the evaluate call (and in my
>> spreadsheet there are obviously duplicate values)
>>
>> I thought I spotted the cause of the issue: countif requires a ";" as
>> separator between the range and the criteria. So I transformed my code
> into:
>>
>> myFormula = "=COUNTIF(E2:E" & LastRow - 1 & ";""=" & currentCDBID & """)"
>> countDUP = Evaluate(myFormula)
>> If countDUP > 1 Then ... rest of my code...
>>
>> And now countDUP always contains "Error 2015" which refers to a type
>> mismatch.
>>
>> Does anyone know what is the cause of my problem ?
>>
>> Any help would be appreciated.
>>
>>
>> Xavier
>>
>>
>
>