Ken,
Apology from me ... I had a similar problem but ASSUMED (and I
know I shouldn't) it was simply due to me copying/pasting/"text-to-columns"
from the NG so I simply manually deleted extra characters from the end of the
strings.

Further investigation indicates there are CHAR(160) (with CHAR(32) - blanks)
characters at the end of most strings so these need to be removed. [Was the
date extracted from a non-Excel source?)

The following VBA macro does this and appears to clear the problem. To
insert into the spreadsheet:

Alt+F11
Right click in VBA Project window
Insert==Module
Copy/paste code below

To use, highlight the range/column to be changed and RUN the macro.



Sub ReplaceChr160()
Selection.Replace What:=Chr(160), _
Replacement:=Chr(32), LookAt:=xlPart, _
SearchOrder:=xlByRows, MatchCase:=False
End Sub

HTH

"Ken" wrote:

Toppers -

I didn't mean to imply your formula didn't work, just that it didn't work
correctly for me, in my spreadsheet. Below, I have list the results from
your 1st and 2nd formulas, running it on the same list both times. As you
can see, in my spreadsheet, the results aren't what I expected. It seems
where not capturing all the numbers. Any thoughts?

Column "A" (raw data) 1st Formula Results 2nd
Formula Results
8675-US (59) 59
1 (should be 59)
8675-EU x 5 #VALUE!
5
6330-02(18) 18
1 (should be 18)
6330-02CE (x10 ) #VALUE! 1
(should be 10)
6305-02 (1) 1
1
58516 (x117) #VALUE!
1 (should be 117)
6312-05(X30) #VALUE!
1 (should be 30)
604-800006-002 #VALUE! 1
5054-SUA-US #VALUE! 1
6312-05(X20 ) #VALUE!
1 (should be 20)
6420-05(30) 30
1 (should be 30)
6420-05 ( 9 ) 9
1 (should be 9)

I really appreciate your help!

Thanks,
//Ken

"Toppers" wrote in message
...
Sorry but it DOES capture (X4), (X10), (X117). These are a direct copy
of
my results using the supplied formula.

700001103 (x4) 4
8675-US (59) 59
8675-EU x 5 5
6330-02(18) 18
6330-02CE (x10 ) 10
6305-02 (1) 1
58516 (x117) 117

It doesn't capture the situation where no quantity was present but this
will:

=IF(ISERROR(VALUE(TRIM(MID(SUBSTITUTE(SUBSTITUTE(S UBSTITUTE(A8,"(","
"),"x"," "),")",""),FIND(" ",SUBSTITUTE(SUBSTITUTE(A8,"("," "),"x","
")),255)))),1,VALUE(TRIM(MID(SUBSTITUTE(SUBSTITUTE (SUBSTITUTE(A8,"(","
"),"x"," "),")",""),FIND(" ",SUBSTITUTE(SUBSTITUTE(A8,"("," "),"x","
")),255))))

700001103 (x4) 4
8675-US (59) 59
8675-EU x 5 5
6330-02(18) 18
6330-02CE (x10 ) 10
6305-02 (1) 1
58516 (x117) 117
58517 1

HTH

"Ken" wrote:

Richard,

I think your on to something here. Toppers response worked great except,
as
you pointed out, it only captures the numbers in (##) or ( ## ). It does
not
capture numbers in (x##) or ( x## ) or (X##), etc... Ultimately, I need
to
capture all scenarios. If the part number does not have a quantity at
the
end (usually indicated by the "x/X" or "(" ), then the result should be
the
number "1", indicating a single part number.

EXAMPLE:
Cell "A"
700001103 (x4)
(part number) 700001103 (quantity of) (x4) < (i need just the
number indicating the quantity, regardless of what it is/is not wrapped
in)
4

Is this possible to do?

Thanks again, to you and Toppers
//Ken

"Richard Buttrey" wrote in
message ...
What do you mean by foolproof?

It is foolproof for the limited set of 6 records you give as an
example.

Have you any other formats that are not covered by
the rules covered.

i.e
include characters between "(" and ")" unless the character is "x" or
" " (space)
or if null result include all characters following the last space

Regards

On Fri, 1 Sep 2006 11:27:02 -0700, Toppers
wrote:

Ken,
I tested this in your sample and it appears to work BUT I
cannot
guarantee it is "foolproof"!

=VALUE(TRIM(MID(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE( A1,"("," "),"x","
"),")",""),FIND(" ",SUBSTITUTE(SUBSTITUTE(A1,"("," "),"x","
")),255)))

HTH

"Ken" wrote:

I need to search for then extract a specific portion of cell data.
Below, I
have provided an example of what I'm trying to do.

Column "A" (raw data) Column "B" (end result)
700001103 (x4) 4
8675-US (59) 59
8675-EU x 5 5
6330-02(18) 18
6330-02CE (x10 ) 10
6305-02 (1) 1
58516 (x117) 117

I hope someone can help me.

Thanks.
//Ken




__
Richard Buttrey
Grappenhall, Cheshire, UK
__________________________