how to check if a string contains only numeric numbers in vba - string

I want to parse out the year info from a string like this one
$8995 Apr 18 2008 Honda Civic Hybrid $8995 (Orem) pic map cars & trucks - by owner
Since I retrieve this string online, sometimes the year element is not at the same place. The way I do it is to split the string by space using split function, then check if each node of the array contains only numeric digits.
However when i use the function IsNumeric, it also returns "$8995" node as true as well.
What is a good way to check if a string contains only numbers, no "$", no ".", not anything else?
Or in my situation, is there a better way to retrieve the year information?
Thanks.

This can be accomplished as a single line of code, using the Like operator
Function StringIsDigits(ByVal s As String) As Boolean
StringIsDigits = Len(s) And (s Like String(Len(s), "#"))
End Function

Will it be the case that all the strings with "years" will have substrings that look like dates? If that is the case, you could just cycle through the string looking for the first group of three that looks like a date, extracting the year from that:
Option Explicit
Function FindYear(S As String) As Long
Dim SS As Variant
Dim sDate As String
Dim I As Long, J As Long
SS = Split(S, " ")
For I = 0 To UBound(SS) - 2
sDate = ""
For J = 0 To 2
sDate = " " & sDate & " " & SS(I + J)
Next J
sDate = Trim(sDate)
If IsDate(sDate) Then
FindYear = Year(sDate)
Exit Function
End If
Next I
End Function

WIthout using Regular Expressions or some very complicated logic, it's going to be difficult to be perfect.
This code will return the pure numeric substrings, but in the case of your example it will return "18" and "2008". You could obviously try to add some more logic to disallow "18" (but allow "13" or "09", etc., but like I said that starts getting complicated. I am happy to help with that, but not knowing exactly what you want, I think it's best to leave that up to you for now.
Const str$ = "$8995 Apr 18 2008 Honda Civic Hybrid $8995 (Orem) pic map cars & trucks - by owner"
Option Explicit
Sub FindNumericValues()
Dim var() As String
Dim numbers As Variant
var = Split(str, " ")
numbers = GetNumerics(var)
MsgBox Join(numbers, ",")
End Sub
Function GetNumerics(words() As String) As Variant
Dim tmp() As Variant
Dim i As Integer
Dim n As Integer
Dim word As Variant
Dim bNumeric As Boolean
For Each word In words
n = 0
bNumeric = True
Do While n < Len(word)
n = n + 1
If Not IsNumeric(Mid(word, n, 1)) Then
bNumeric = False
Exit Do
End If
Loop
If bNumeric Then
ReDim Preserve tmp(i)
tmp(i) = word
i = i + 1
End If
Next
GetNumerics = tmp
End Function

You could parse the year out using RegEx:
Public Function GetYear(someText As String) As Integer
With CreateObject("VBScript.RegExp")
.Global = False
.MultiLine = False
.IgnoreCase = True
.Pattern = " [\d]{4} "
If .Test(testString) Then
GetYear = CInt(.Execute(testString)(0))
Else
GetYear = 9999
End If
End With
End Function
Example code:
Public Const testString As String = "$8995 Apr 18 2008 Honda Civic Hybrid $8995 (Orem) pic map cars & trucks - by owner "
Public Function GetYear(someText As String) As Integer
With CreateObject("VBScript.RegExp")
.Global = False
.MultiLine = False
.IgnoreCase = True
.Pattern = " [\d]{4} "
If .Test(testString) Then
GetYear = CInt(.Execute(testString)(0))
Else
GetYear = 9999
End If
End With
End Function
Sub Foo()
Debug.Print GetYear(testString) '// "2008"
End Sub

Related

Check if Column List Contains Header via Regex - Excel vba

I'm trying to determine if a column has a header or not via VBA. Basically the column will have data following an unknown but identical regex pattern. My plan is to test if A2 has the same type regex string as A1. It would likely even be the same ID + 1. Eg
A1 = X001
A2 = X002
Func IsHeader("A") = True
A1 = ID's
A2 = X001
Func IsHeader("A") = False
I've got an idea to utilize an existing script I made to generate a regex pattern based on an input alphanumerical string, but I'm interested to see what other idea's/ways people might have of solving the issue. I realize there isn't much code, but I know I can do this and I'm working on it now. If you're not interested in answering, thats ok!
Update: Posted Answer, but I'm looking for more than a code review as I realize there is an exchange for that. I'd like to know better ways to achieve goal with a different attack vector.
This is what I got! I'm not sure how SO feels about code reviews, but im interested in what ppl think and how else they could "skin the cat" so please feel free to post an answer.
Sub Test()
If IsHeader = True Then
MsgBox "Has Header"
Else
MsgBox "No Header"
End If
End Sub
Public Function IsHeader() As Boolean
A1Pattern = RegExPattern(Range("A1").Value)
A2Pattern = RegExPattern(Range("A2").Value)
If A1Pattern = A2Pattern Then
IsHeader = True
End If
End Function
Public Function RegExPattern(my_string) As String
RegExPattern = ""
'''Special Character Section'''
Dim special_charArr() As String
Dim special_char As String
special_char = "!,#,#,$,%,^,&,*,+,/,\,;,:"
special_charArr() = Split(special_char, ",")
'''Special Character Section'''
'''Alpha Section'''
Dim regexp As Object
Set regexp = CreateObject("vbscript.regexp")
Dim strPattern As String
strPattern = "([a-z])"
With regexp
.ignoreCase = True
.Pattern = strPattern
End With
'''Alpha Section'''
Dim buff() As String
'my_string = "test1*1#"
ReDim buff(Len(my_string) - 1)
Dim i As Variant
For i = 1 To Len(my_string)
buff(i - 1) = Mid$(my_string, i, 1)
char = buff(i - 1)
If IsNumeric(char) = True Then
'MsgBox char & " = Number"
RegExPattern = RegExPattern & "([0-9])"
End If
For Each Key In special_charArr
special = InStr(char, Key)
If special = 1 Then
If Key <> "*" Then
'MsgBox char & " = Special NOT *"
RegExPattern = RegExPattern & "^[!##$%^&()].*$"
Else
'MsgBox char & " = *"
RegExPattern = RegExPattern & "."
End If
End If
Next
If regexp.Test(char) Then
'MsgBox char & " = Alpha"
RegExPattern = RegExPattern & "([a-z])"
End If
Next
'RegExPattern = Chr(34) & RegExPattern & Chr(34)
'MsgBox RegExPattern
End Function

Returning the numbers in a string as a variable [duplicate]

I need to find numbers from a string. How does one find numbers from a string in VBA Excel?
Assuming you mean you want the non-numbers stripped out, you should be able to use something like:
Function onlyDigits(s As String) As String
' Variables needed (remember to use "option explicit"). '
Dim retval As String ' This is the return string. '
Dim i As Integer ' Counter for character position. '
' Initialise return string to empty '
retval = ""
' For every character in input string, copy digits to '
' return string. '
For i = 1 To Len(s)
If Mid(s, i, 1) >= "0" And Mid(s, i, 1) <= "9" Then
retval = retval + Mid(s, i, 1)
End If
Next
' Then return the return string. '
onlyDigits = retval
End Function
Calling this with:
Dim myStr as String
myStr = onlyDigits ("3d1fgd4g1dg5d9gdg")
MsgBox (myStr)
will give you a dialog box containing:
314159
and those first two lines show how you can store it into an arbitrary string variable, to do with as you wish.
Regular expressions are built to parse. While the syntax can take a while to pick up on this approach is very efficient, and is very flexible for handling more complex string extractions/replacements
Sub Tester()
MsgBox CleanString("3d1fgd4g1dg5d9gdg")
End Sub
Function CleanString(strIn As String) As String
Dim objRegex
Set objRegex = CreateObject("vbscript.regexp")
With objRegex
.Global = True
.Pattern = "[^\d]+"
CleanString = .Replace(strIn, vbNullString)
End With
End Function
Expanding on brettdj's answer, in order to parse disjoint embedded digits into separate numbers:
Sub TestNumList()
Dim NumList As Variant 'Array
NumList = GetNums("34d1fgd43g1 dg5d999gdg2076")
Dim i As Integer
For i = LBound(NumList) To UBound(NumList)
MsgBox i + 1 & ": " & NumList(i)
Next i
End Sub
Function GetNums(ByVal strIn As String) As Variant 'Array of numeric strings
Dim RegExpObj As Object
Dim NumStr As String
Set RegExpObj = CreateObject("vbscript.regexp")
With RegExpObj
.Global = True
.Pattern = "[^\d]+"
NumStr = .Replace(strIn, " ")
End With
GetNums = Split(Trim(NumStr), " ")
End Function
Use the built-in VBA function Val, if the numbers are at the front end of the string:
Dim str as String
Dim lng as Long
str = "1 149 xyz"
lng = Val(str)
lng = 1149
Val Function, on MSDN
I was looking for the answer of the same question but for a while I found my own solution and I wanted to share it for other people who will need those codes in the future. Here is another solution without function.
Dim control As Boolean
Dim controlval As String
Dim resultval As String
Dim i as Integer
controlval = "A1B2C3D4"
For i = 1 To Len(controlval)
control = IsNumeric(Mid(controlval, i, 1))
If control = True Then resultval = resultval & Mid(controlval, i, 1)
Next i
resultval = 1234
This a variant of brettdj's & pstraton post.
This will return a true Value and not give you the #NUM! error. And \D is shorthand for anything but digits. The rest is much like the others only with this minor fix.
Function StripChar(Txt As String) As Variant
With CreateObject("VBScript.RegExp")
.Global = True
.Pattern = "\D"
StripChar = Val(.Replace(Txt, " "))
End With
End Function
This is based on another answer, but is just reformated:
Assuming you mean you want the non-numbers stripped out, you should be able to use something like:
'
' Skips all characters in the input string except digits
'
Function GetDigits(ByVal s As String) As String
Dim char As String
Dim i As Integer
GetDigits = ""
For i = 1 To Len(s)
char = Mid(s, i, 1)
If char >= "0" And char <= "9" Then
GetDigits = GetDigits + char
End If
Next i
End Function
Calling this with:
Dim myStr as String
myStr = GetDigits("3d1fgd4g1dg5d9gdg")
Call MsgBox(myStr)
will give you a dialog box containing:
314159
and those first two lines show how you can store it into an arbitrary string variable, to do with as you wish.
Alternative via Byte Array
If you assign a string to a Byte array you typically get the number equivalents of each character in pairs of the array elements. Use a loop for numeric check via the Like operator and return the joined array as string:
Function Nums(s$)
Dim by() As Byte, i&, ii&
by = s: ReDim tmp(UBound(by)) ' assign string to byte array; prepare temp array
For i = 0 To UBound(by) - 1 Step 2 ' check num value in byte array (0, 2, 4 ... n-1)
If Chr(by(i)) Like "#" Then tmp(ii) = Chr(by(i)): ii = ii + 1
Next i
Nums = Trim(Join(tmp, vbNullString)) ' return string with numbers only
End Function
Example call
Sub testByteApproach()
Dim s$: s = "a12bx99y /\:3,14159" ' [1] define original string
Debug.Print s & " => " & Nums(s) ' [2] display original string and result
End Sub
would display the original string and the result string in the immediate window:
a12bx99y /\:3,14159 => 1299314159
Based on #brettdj's answer using a VBScript regex ojbect with two modifications:
The function handles variants and returns a variant. That is, to take care of a null case; and
Uses explicit object creation, with a reference to the "Microsoft VBScript Regular Expressions 5.5" library
Function GetDigitsInVariant(inputVariant As Variant) As Variant
' Returns:
' Only the digits found in a varaint.
' Examples:
' GetDigitsInVariant(Null) => Null
' GetDigitsInVariant("") => ""
' GetDigitsInVariant(2021-/05-May/-18, Tue) => 20210518
' GetDigitsInVariant(2021-05-18) => 20210518
' Notes:
' If the inputVariant is null, null will be returned.
' If the inputVariant is "", "" will be returned.
' Usage:
' VBA IDE Menu > Tools > References ...
' > "Microsoft VBScript Regular Expressions 5.5" > [OK]
' With an explicit object reference to RegExp we can get intellisense
' and review the object heirarchy with the object browser
' (VBA IDE Menu > View > Object Browser).
Dim regex As VBScript_RegExp_55.RegExp
Set regex = New VBScript_RegExp_55.RegExp
Dim result As Variant
result = Null
If IsNull(inputVariant) Then
result = Null
Else
With regex
.Global = True
.Pattern = "[^\d]+"
result = .Replace(inputVariant, vbNullString)
End With
End If
GetDigitsInVariant = result
End Function
Testing:
Private Sub TestGetDigitsInVariant()
Dim dateVariants As Variant
dateVariants = Array(Null, "", "2021-/05-May/-18, Tue", _
"2021-05-18", "18/05/2021", "3434 ..,sdf,sfd 444")
Dim dateVariant As Variant
For Each dateVariant In dateVariants
Debug.Print dateVariant & ": ", , GetDigitsInVariant(dateVariant)
Next dateVariant
Debug.Print
End Sub
Public Function ExtractChars(strRef$) As String
'Extract characters from a string according to a range of charactors e.g'+.-1234567890'
Dim strA$, e%, strExt$, strCnd$: strExt = "": strCnd = "+.-1234567890"
For e = 1 To Len(strRef): strA = Mid(strRef, e, 1)
If InStr(1, strCnd, strA) > 0 Then strExt = strExt & strA
Next e: ExtractChars = strExt
End Function
In the immediate debug dialog:
? ExtractChars("a-5d31.78K")
-531.78

Excel VBA insert character between number and letter

I would like some VBA code that would allow me to detect if a string contains any instances of a number followed by a letter and then insert a new character between them. For example:
User enters the following string:
4x^2+3x
Function returns:
4*x^2+3*x
Thanks in advance.
Edit: Thanks for the advice guys, I think I have it working but I'd like to see if you can improve what I've got:
Sub insertStr()
On Error Resume Next
Dim originalString As String
Dim newLeft As String
Dim newRight As String
originalString = Cells(1, 1).Value
Repeat:
For i = 1 To Len(originalString)
If IsNumeric(Mid(originalString, i, 1)) = True Then
Select Case Asc(Mid(originalString, i + 1, 1))
Case 65 To 90, 97 To 122
newLeft = Left(originalString, i)
newRight = Right(originalString, Len(originalString) - i)
originalString = newLeft & "*" & newRight
GoTo Repeat
Case Else
GoTo Nexti
End Select
End If
Nexti:
Next i
End Sub
And just to show how it might be done using Regular expressions, and also allowing you to specify any particular character to insert:
Option Explicit
Function InsertChar(S As String, Insert As String) As String
Dim RE As Object
Set RE = CreateObject("vbscript.regexp")
With RE
.Global = True
.Pattern = "(\d)(?=[A-Za-z])"
InsertChar = .Replace(S, "$1" & Insert)
End With
End Function
The pattern is interpreted as
\d Find any number and capture it
(?=[A-Za-z]) that is followed by a letter
And the replacement is
$1 return the capturing group
& concatenated with
Insert (the string to be inserted)
Following Ron's suggestion:
Public Function InsertStar(sIn As String) As String
Dim L As Long, temp As String, CH As String
L = Len(sIn)
temp = Left(sIn, 1)
For i = 2 To L
CH = Mid(sIn, i, 1)
If IsLetter(CH) And IsNumeric(Right(temp, 1)) Then
temp = temp & "*"
End If
temp = temp & CH
Next i
InsertStar = temp
End Function
Public Function IsLetter(sIn As String) As Boolean
If sIn Like "[a-zA-Z]" Then
IsLetter = True
Else
IsLetter = False
End If
End Function

Retrieve alpha characters from alphanumeric string

How can I split up AB2468123 with excel-vba
I tried something along these lines:
myStr = "AB2468123"
split(myStr, "1" OR "2" OR "3"......."9")
I want to get only alphabet (letters) only.
Thanks.
How about this to retrieve only letters from an input string:
Function GetLettersOnly(str As String) As String
Dim i As Long, letters As String, letter As String
letters = vbNullString
For i = 1 To Len(str)
letter = VBA.Mid$(str, i, 1)
If Asc(LCase(letter)) >= 97 And Asc(LCase(letter)) <= 122 Then
letters = letters + letter
End If
Next
GetLettersOnly = letters
End Function
Sub Test()
Debug.Print GetLettersOnly("abc123") // prints "abc"
Debug.Print GetLettersOnly("ABC123") // prints "ABC"
Debug.Print GetLettersOnly("123") // prints nothing
Debug.Print GetLettersOnly("abc123def") // prints "abcdef"
End Sub
Edit: for completeness (and Chris Neilsen) here is the Regex way:
Function GetLettersOnly(str As String) As String
Dim result As String, objRegEx As Object, match As Object
Set objRegEx = CreateObject("vbscript.regexp")
objRegEx.Pattern = "[a-zA-Z]+"
objRegEx.Global = True
objRegEx.IgnoreCase = True
If objRegEx.test(str) Then
Set match = objRegEx.Execute(str)
GetLettersOnly = match(0)
End If
End Function
Sub test()
Debug.Print GetLettersOnly("abc123") //prints "abc"
End Sub
Simpler single shot RegExp
Sub TestIt()
MsgBox CleanStr("AB2468123")
End Sub
Function CleanStr(strIn As String) As String
Dim objRegex As Object
Set objRegex = CreateObject("vbscript.regexp")
With objRegex
.Pattern = "[^a-zA-Z]+"
.Global = True
CleanStr = .Replace(strIn, vbNullString)
End With
End Function
This is what i have found out that works the best. It may be somewhat basic, but it does the job :)
Function Split_String(Optional test As String = "ABC111111") As Variant
For i = 1 To Len(test)
letter = Mid(test, i, 1)
If IsNumeric(letter) = True Then
justletters = Left(test, i - 1)
justnumbers = Right(test, Len(test) - (i - 1))
Exit For
End If
Next
'MsgBox (justnumbers)
'MsgBox (justletters)
'just comment away the answer you want to have :)
'Split_String = justnumbers
'Split_String = justletters
End Function
Possibly the fastest way is to parse a Byte String:
Function alpha(txt As String) As String
Dim b, bytes() As Byte: bytes = txt
For Each b In bytes
If Chr(b) Like "[A-Za-z]" Then alpha = alpha & Chr(b)
Next b
End Function
More information here.

How to find numbers from a string?

I need to find numbers from a string. How does one find numbers from a string in VBA Excel?
Assuming you mean you want the non-numbers stripped out, you should be able to use something like:
Function onlyDigits(s As String) As String
' Variables needed (remember to use "option explicit"). '
Dim retval As String ' This is the return string. '
Dim i As Integer ' Counter for character position. '
' Initialise return string to empty '
retval = ""
' For every character in input string, copy digits to '
' return string. '
For i = 1 To Len(s)
If Mid(s, i, 1) >= "0" And Mid(s, i, 1) <= "9" Then
retval = retval + Mid(s, i, 1)
End If
Next
' Then return the return string. '
onlyDigits = retval
End Function
Calling this with:
Dim myStr as String
myStr = onlyDigits ("3d1fgd4g1dg5d9gdg")
MsgBox (myStr)
will give you a dialog box containing:
314159
and those first two lines show how you can store it into an arbitrary string variable, to do with as you wish.
Regular expressions are built to parse. While the syntax can take a while to pick up on this approach is very efficient, and is very flexible for handling more complex string extractions/replacements
Sub Tester()
MsgBox CleanString("3d1fgd4g1dg5d9gdg")
End Sub
Function CleanString(strIn As String) As String
Dim objRegex
Set objRegex = CreateObject("vbscript.regexp")
With objRegex
.Global = True
.Pattern = "[^\d]+"
CleanString = .Replace(strIn, vbNullString)
End With
End Function
Expanding on brettdj's answer, in order to parse disjoint embedded digits into separate numbers:
Sub TestNumList()
Dim NumList As Variant 'Array
NumList = GetNums("34d1fgd43g1 dg5d999gdg2076")
Dim i As Integer
For i = LBound(NumList) To UBound(NumList)
MsgBox i + 1 & ": " & NumList(i)
Next i
End Sub
Function GetNums(ByVal strIn As String) As Variant 'Array of numeric strings
Dim RegExpObj As Object
Dim NumStr As String
Set RegExpObj = CreateObject("vbscript.regexp")
With RegExpObj
.Global = True
.Pattern = "[^\d]+"
NumStr = .Replace(strIn, " ")
End With
GetNums = Split(Trim(NumStr), " ")
End Function
Use the built-in VBA function Val, if the numbers are at the front end of the string:
Dim str as String
Dim lng as Long
str = "1 149 xyz"
lng = Val(str)
lng = 1149
Val Function, on MSDN
I was looking for the answer of the same question but for a while I found my own solution and I wanted to share it for other people who will need those codes in the future. Here is another solution without function.
Dim control As Boolean
Dim controlval As String
Dim resultval As String
Dim i as Integer
controlval = "A1B2C3D4"
For i = 1 To Len(controlval)
control = IsNumeric(Mid(controlval, i, 1))
If control = True Then resultval = resultval & Mid(controlval, i, 1)
Next i
resultval = 1234
This a variant of brettdj's & pstraton post.
This will return a true Value and not give you the #NUM! error. And \D is shorthand for anything but digits. The rest is much like the others only with this minor fix.
Function StripChar(Txt As String) As Variant
With CreateObject("VBScript.RegExp")
.Global = True
.Pattern = "\D"
StripChar = Val(.Replace(Txt, " "))
End With
End Function
This is based on another answer, but is just reformated:
Assuming you mean you want the non-numbers stripped out, you should be able to use something like:
'
' Skips all characters in the input string except digits
'
Function GetDigits(ByVal s As String) As String
Dim char As String
Dim i As Integer
GetDigits = ""
For i = 1 To Len(s)
char = Mid(s, i, 1)
If char >= "0" And char <= "9" Then
GetDigits = GetDigits + char
End If
Next i
End Function
Calling this with:
Dim myStr as String
myStr = GetDigits("3d1fgd4g1dg5d9gdg")
Call MsgBox(myStr)
will give you a dialog box containing:
314159
and those first two lines show how you can store it into an arbitrary string variable, to do with as you wish.
Alternative via Byte Array
If you assign a string to a Byte array you typically get the number equivalents of each character in pairs of the array elements. Use a loop for numeric check via the Like operator and return the joined array as string:
Function Nums(s$)
Dim by() As Byte, i&, ii&
by = s: ReDim tmp(UBound(by)) ' assign string to byte array; prepare temp array
For i = 0 To UBound(by) - 1 Step 2 ' check num value in byte array (0, 2, 4 ... n-1)
If Chr(by(i)) Like "#" Then tmp(ii) = Chr(by(i)): ii = ii + 1
Next i
Nums = Trim(Join(tmp, vbNullString)) ' return string with numbers only
End Function
Example call
Sub testByteApproach()
Dim s$: s = "a12bx99y /\:3,14159" ' [1] define original string
Debug.Print s & " => " & Nums(s) ' [2] display original string and result
End Sub
would display the original string and the result string in the immediate window:
a12bx99y /\:3,14159 => 1299314159
Based on #brettdj's answer using a VBScript regex ojbect with two modifications:
The function handles variants and returns a variant. That is, to take care of a null case; and
Uses explicit object creation, with a reference to the "Microsoft VBScript Regular Expressions 5.5" library
Function GetDigitsInVariant(inputVariant As Variant) As Variant
' Returns:
' Only the digits found in a varaint.
' Examples:
' GetDigitsInVariant(Null) => Null
' GetDigitsInVariant("") => ""
' GetDigitsInVariant(2021-/05-May/-18, Tue) => 20210518
' GetDigitsInVariant(2021-05-18) => 20210518
' Notes:
' If the inputVariant is null, null will be returned.
' If the inputVariant is "", "" will be returned.
' Usage:
' VBA IDE Menu > Tools > References ...
' > "Microsoft VBScript Regular Expressions 5.5" > [OK]
' With an explicit object reference to RegExp we can get intellisense
' and review the object heirarchy with the object browser
' (VBA IDE Menu > View > Object Browser).
Dim regex As VBScript_RegExp_55.RegExp
Set regex = New VBScript_RegExp_55.RegExp
Dim result As Variant
result = Null
If IsNull(inputVariant) Then
result = Null
Else
With regex
.Global = True
.Pattern = "[^\d]+"
result = .Replace(inputVariant, vbNullString)
End With
End If
GetDigitsInVariant = result
End Function
Testing:
Private Sub TestGetDigitsInVariant()
Dim dateVariants As Variant
dateVariants = Array(Null, "", "2021-/05-May/-18, Tue", _
"2021-05-18", "18/05/2021", "3434 ..,sdf,sfd 444")
Dim dateVariant As Variant
For Each dateVariant In dateVariants
Debug.Print dateVariant & ": ", , GetDigitsInVariant(dateVariant)
Next dateVariant
Debug.Print
End Sub
Public Function ExtractChars(strRef$) As String
'Extract characters from a string according to a range of charactors e.g'+.-1234567890'
Dim strA$, e%, strExt$, strCnd$: strExt = "": strCnd = "+.-1234567890"
For e = 1 To Len(strRef): strA = Mid(strRef, e, 1)
If InStr(1, strCnd, strA) > 0 Then strExt = strExt & strA
Next e: ExtractChars = strExt
End Function
In the immediate debug dialog:
? ExtractChars("a-5d31.78K")
-531.78

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