You added two more character to the string when you chaged the year from two
digits to 4 digits

from
Range("Bithday").Value = Left(Idnumber, 8)
to
Range("Bithday").Value = Left(Idnumber, 10)

You will either get
52-02-08-5034-087 : 17 characters
or
52-02-2008-5034-087 : 19 characters


Os the code would look like this

if len(IdNumber) = 17 then
Range("Bithday").Value = Left(Idnumber, 8)
else
Range("Bithday").Value = Left(Idnumber, 10)
end if
"Hennie Neuhoff" wrote:

Hi. All,
Easy one for the experts.
My country use a unique [13 number-digit] Identification system.
The first 6-digits represent the date of birth [yymmdd]
In my code the ID number is entered on a userform as text and
displayed as follows 52-02-08-5034-087 [in this example the d.of b
is 8-Feb-1952]
Range("Bithday").Value = Left(Idnumber, 8)
displays the required birthday in Range("Bithday")
With the note "This cell contains a date string represented with
only 2 digits for the year."
If I select the option "Convert XX to 19XX" it gives me exactly
what I want.
I tried macro record but it doesnt give me any result.
How do I include the "Convert XX to 19XX" in my code ?

--
HJN