In article ,
says...

would be optimal in terms of regexp performance. [Tangent: participate in
comp.lang.perl.misc and you'll really learn how to use regular expressions.]

Yeah, I probably should do that. At the same time, it would be far
more useful to convince people to use regular expresssions with XL in
the first place. <g

--
Regards,

Tushar Mehta, MS MVP -- Excel
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions

In article ,
says...
"Tushar Mehta" wrote...
Yet another alternative...

If you install the functions at
http://www.tmehta.com/regexp/add_code.htm, and if the string in
question is in column A, you could use the formula

=RegExpSubstitute(A1,"[^\d]*","")
...

While this seems to work, \D it the better/safer/more Perl-compatible way to
express the complement of the character class represented by \d. It also
does more work than it should.

=RegExpSubstitute(A1,"\D+","")

would be optimal in terms of regexp performance. [Tangent: participate in
comp.lang.perl.misc and you'll really learn how to use regular expressions.]

If you have multiple numbers and want only the first set, use
=RegExpSubstitute(A3,"(^[^\d]*)(\d*)(.*$)","$2")

Better to use

=RegExpSubstitute(A3,"^\D*(\d+).*$","$1")

and to pull the last numeral substring,

=RegExpSubstitute(A3,"^.*(\d+)\D*$","$1")

and to pull the n_th numeral substring counting left to right,

=RegExpSubstitute(A3,"^(\D*(\d+)){"&(n-1)&"}\D*(\d+).*$","$3")