>Oh, posh! < bg >
>You all seem to be insisting on a whole lot more work than is necessary.
>If you run a character string thru VAL() and end up with a non-zero number, you have a number without any distracting inner characters.
>

x='2.5ab3'
>y=val(x) ?y && 2.50

>
>So if you have a string with a number that could be imbedded, you just have to strip off leading chars to get to it.
>
>I'd say Craig's suggestion would be needed if you want to find more than one imbedded number.

Yup. You're method works the best for the leading numbers.