>>>Assume you have information in MyString and sure that USERID is there:
>>>TempSting = SUBSTR(MyString, ATC('USERID', MyString)+LEN('USERID'))
>>>Result = LEFT(TempSting, AT(' ', TempSting) - 1)
>>>
>>
>>Vlad: When I tried your code I got an empty string returned for the second test condition ("USERIDAA" at end of MyString). Changing 2nd line to
>>
>>Result = LEFT(TempSting, AT(' ', TempSting + " ") - 1)
>>
>>seemed to work
>>
>>............Rich
>
> Correct. I assumed that space always exist in string.

The Vbscript.RegExp object shoud handle this:

oRegExp = CREATEOBJ('vbscript.regexp')
oRegExp.global = .t.
oRegExp.Pattern = '\sUSERID\w{2}\s'
oMatches = oRegExp.Execute(' USERID 1 USERIDAA USER32 USERID99 USERID42')
FOR EACH Match in oMatches
   ? Match.Value
ENDFOR

>
>>
>>>>I have a string that has ' USERID?? ' somewhere in the string. i want returned the characters directly after the USERID and before the next space or at end of string. 'the is USERIDABC code' I want ABC. 'The USERIDAA' I want AA.
>>>>
>>>>As always, I have a terrible time with string functions. Have been looking in my Help Character Function Favorites now for 30 minutes and only have a hodge podge of silly looking code.
>>>>
>>>>Thanks
>>>>
>>>>Brenda