Jason

You could use this Function on a copy of the cell(s) with numbers/text.

Function DeleteNonNumerics(ByVal sStr As String) As Long
   Dim i As Long
   If sStr Like "*[0-9]*" Then
       For i = 1 To Len(sStr)
           If Mid(sStr, i, 1) Like "[0-9]" Then
               DeleteNonNumerics = DeleteNonNumerics & Mid(sStr, i, 1)
           End If
       Next i
   Else
       DeleteNonNumerics = sStr
   End If
End Function

Usage is:  =DeleteNonNumerics(cellref)

Gord Dibben  MS Excel MVP

You could use this UDF which makes use of Regular Expressions.

<alt-F11> opens the VB Editor.

Ensure your project is highlighted in the Project Explorer window, then
Insert/Module and paste the code below into the window that opens.

You must also set a reference to Microsoft VBScript Regular Expressions 5.5
(Under Tools/References).

=====================================================
Option Explicit

Function Get15(str As String) As String
Dim objRegExp As RegExp
Dim objMatch As Match
Dim colMatches   As MatchCollection
Const Pattern As String = "(^|\D)(\d{15})(\D|$)"

' Create a regular expression object.
  Set objRegExp = New RegExp

  'Set the pattern by using the Pattern property.
  objRegExp.Pattern = Pattern

  'Set global applicability.
  objRegExp.Global = True

  'Test whether the String can be compared.
  If (objRegExp.Test(str) = True) Then
   Set colMatches = objRegExp.Execute(str)
   Set objMatch = colMatches.Item(0)
   Get15 = objMatch.SubMatches(1)
  End If
End Function
====================================================

"Pattern" sets the specifications.  The Pattern that needs to be matched
consists of:

1.  Either a non-Digit character or the start of the string
followed by
2.  any 15 consecutive digits
followed by
3.  a non-Digit character or the end of the string

To use this, enter the UDF  =Get15(str) where str is either a string or a cell
reference containing the string.
--ron

Do any numbers have leading 0's:

012345678901234

This formula will extract the number string but will drop any leading 0's:

=LOOKUP(99^99,--MID(A1,MIN(FIND({0,1,2,3,4,5,6,7,8,9},A1&"0123456789")),ROW(INDIRECT("1:"&LEN(A1)))))

Excel insists on formatting the result as SCIENTIFIC so you'll want to
format as NUMBER 0 decimal places. Or, if you do have leading 0's you can
use a custom format of: 000000000000000

Biff

Hi Jason,

A suggestion:
Function XtractNum(s As String) As Double
Dim i As Long, d As Double, c As String, f As Double

i = 0: d = 0#: f = 1#

StateStart:
i = i + 1
If i > Len(s) Then GoTo StateEnd
c = Mid(s, i, 1)
Select Case c
Case "0" To "9"
   d = CDbl(c)
   GoTo StatePreComma
Case "."
   d = 0#
   GoTo StatePostComma
Case Else
   GoTo StateStart
End Select

StatePreComma:
i = i + 1
If i > Len(s) Then GoTo StateEnd
c = Mid(s, i, 1)
Select Case c
Case "0" To "9"
   d = 10# * d + CDbl(c)
   GoTo StatePreComma
Case "."
   GoTo StatePostComma
Case Else
   GoTo StateEnd
End Select

StatePostComma:
i = i + 1
If i > Len(s) Then GoTo StateEnd
c = Mid(s, i, 1)
Select Case c
Case "0" To "9"
   f = f / 10#
   d = d + CDbl(c) * f
   GoTo StatePostComma
Case Else
   GoTo StateEnd
End Select

StateEnd:
XtractNum = d

'Small specialty: Accept only 15 digit numbers:
If Len(CStr(d)) <> 15 Then
   XtractNum = CVErr(xlErrNum)
End If

End Function

Please notice that leading zeros will be swallowed and that the
decimal point will be counted as a digit, too. Change the "small
specialty" at the end of my UDF if necessary.

Regards,
Bernd