I am trying to use the Right([PartID],5)command to pull the last 5
digits from a field, but I have some "PartID" that I need the last 6
or 7 characters. All the characters I need have "ZZZ" in front of
them. For example:

ZZZ00013
ZZZ000444
ZZZ0064185

Is there a way for access to get all the characters to the right of
"ZZZ", no matter how many characters?

Thanks for your help ,

Eric

well, a quick and dirty way would be to work backwards...

let's assume your value is in a variable called tt

dim xs as integer, tpcar as string, resu as string

resu = ""

for xs = len(tt) to 1 step -1

tpcar = right(tt,xs,1)

if tpcar <> "Z" then

resu = tpcar & resu

else

exit for

end if

next xs



At the end of the loop, resu will contain the digits at the right of the
rightmost "Z"

should work,

Bernard

"ERIC" <DATAE2000 (AT) YAHOO (DOT) COM> wrote

If you know that the field will always start with the three characters
you want to ignore, then you could use:

MsgBox Mid$("ZZZ000444", 4)

Note that by omitting the 3rd parameter of the MID$ function, you are
selecting all characters starting with the 4th.

On 28 Jul 2003 06:09:37 -0700, DATAE2000 (AT) YAHOO (DOT) COM (ERIC) wrote:

On 28 Jul 2003 06:09:37 -0700, DATAE2000 (AT) YAHOO (DOT) COM (ERIC) wrote:

Try this:
Mid$([PartID], 4)

-Tom.

Tom,
Thanks for your reply. I was going to use that command, but I will not
know if the char length will be 3-4 characters in front of the center.
Thanks!


Tom van Stiphout <tom7744 (AT) no (DOT) spam.cox.net> wrote

The following code will search for the first occurance of a numeric in
the string, and will then return the numeric portion, thus allowing
for any number of characters before the numeric:

Dim iWork As Integer
Dim sWork as String
sWork = "ZZZ12345"

For iWork = 1 to Len(sWork)
if IsNumeric(Mid$(sWork, iWork, 1)) Then Exit For
Next

MsgBox "The numeric portion is: " & Mid$(sWork, iWork)




On 29 Jul 2003 04:24:38 -0700, DATAE2000 (AT) YAHOO (DOT) COM (ERIC) wrote: