Thanks Mike,
I see Char(7) must be code for blank and by changing the last # I can find
any number of blank spaces, e.g., 4 for the 4th blank, 5 for the 5th blank
etc.
=FIND(CHAR(7),SUBSTITUTE($A27," ",CHAR(7),4))

Your disclaimer sounds like a version of Occams Razor principle. i.e., is
the principle that "entities must not be multiplied beyond necessity" and the
conclusion thereof, that the simplest explanation or strategy tends to be the
best one..
....When competing hypotheses are otherwise equal, adopt the hypothesis
that introduces the fewest assumptions while still sufficiently answering the
question.

"Mike H" wrote:

Hi,

Try this

=FIND(CHAR(7),SUBSTITUTE(A1," ",CHAR(7),3))
--
Mike

When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.


"Steve Stad" wrote:

Can you tell me how to find the position number of the third space in a
string of varying length and text.
e.g.,
123 456 789 111 222
need to find position number of space between 789 and 111
or in:
abcd efghi jklmnopqr stuvxy
need position number of space between "r" and 's'.

I am familiar with =find(" ",a1) to find a space but need help with logic
for third or second space.