isString ? or IsNumber ? in VBA...

[Terry_Orange]

If I have a scalar, then is there a function I can call (or write) to determine if it is a string or a number.

for example a string I would parse, but an integer I would use in a statistical test.

Can anyone offer a solution ?
Thanks
TO.

[Richie(UK)]

Hi Terry,

VBA has a 'IsNumeric' function. See below for the VBE Help examples. Is this what you are after?

This example uses the IsNumeric function to determine if a variable can be evaluated as a number.

Dim MyVar, MyCheck
MyVar = "53" ' Assign value.
MyCheck = IsNumeric(MyVar) ' Returns True.

MyVar = "459.95" ' Assign value.
MyCheck = IsNumeric(MyVar) ' Returns True.

MyVar = "45 Help" ' Assign value.
MyCheck = IsNumeric(MyVar) ' Returns False.

[Terry_Orange]

/@

___>
(__O)
(____@)
(____@)
(__o)_


Exactly what I was after. Thank you.
VERY much appreciated.
T.

[mdalamers]

WATCH OUT!
The solution is not water tight!

MyVar = "459,95" ' Assign value.
MyCheck = IsNumeric(MyVar) ' Returns True.
But if you calculatet with the assumed number e.g.
MYResult = 1* MyVar then you get a strange result: 45995!!!!

[mdalamers]

I came up with the following function.
It will give FALSE for values like:
"", "Four", "2,3", "Now()"
It will give TRUE for values like:
"1", "-0", "2.3" "-2.300", "3/4", "1.2E-8"

If the regional settings are set for countries that have a decimal valueto be "," then "2,3 will be TRUE and "2.3" will be false.

Function fncIsNumber(varInput As Variant) As Boolean
'Checks if the value in the variant is a number. Note this is NOT the case for the IsNumeric function.
'Isnumeric("3,4") returns TRUE and will see 3,4 as 34, while fncIsNumber("3,4") returns False!
'Values like "1.2E-8" are also validates a a valid number.
Dim strDecimalSeparator As String

strDecimalSeparator = Mid(Trim(CStr(1 / 2)), 2, 1) 'Returns ',' or '.' depending on regional settings
If strDecimalSeparator = "." Then
strDecimalSeparator = "," 'Switch to the other decimal separator
Else
strDecimalSeparator = "." 'Switch to the other decimal separator
End If

fncIsNumber = False
If IsNumeric(varInput) Then
If InStr(1, varInput, strDecimalSeparator) = 0 Then fncIsNumber = True
End If
End Function

[Rick Rothstein]

WATCH OUT!
The solution is not water tight!

MyVar = "459,95" ' Assign value.
MyCheck = IsNumeric(MyVar) ' Returns True.
But if you calculatet with the assumed number e.g.
MYResult = 1* MyVar then you get a strange result: 45995!!!!

Here is a message that I originally started posting many years ago in the compiled VB newsgroups and, when applicable, repeated in Excel newsgroups/forums...

I usually try and steer people away from using IsNumeric to "proof" supposedly numeric text. Consider this (also see note below):

ReturnValue = IsNumeric("($1,23,,3.4,,,5,,E67$)")

Most people would not expect THAT to return True. IsNumeric has some "flaws" in what it considers a proper number and what most programmers are looking for.

I had a short tip published by Pinnacle Publishing in their Visual Basic Developer magazine that covered some of these flaws. Originally, the tip was free to view but is now viewable only by subscribers.. Basically, it said that IsNumeric returned True for things like -- currency symbols being located in front or in back of the number as shown in my example (also applies to plus, minus and blanks too); numbers surrounded by parentheses as shown in my example (some people use these to mark negative numbers); numbers containing any number of commas before a decimal point as shown in my example; numbers in scientific notation (a number followed by an upper or lower case "D" or "E", followed by a number equal to or less than 305 -- the maximum power of 10 in VB); and Octal/Hexadecimal numbers (&H for Hexadecimal, &O or just & in front of the number for Octal).

NOTE:
======
In the above example and in the referenced tip, I refer to $ signs and commas and dots -- these were meant to refer to your currency, thousands separator and decimal point symbols as defined in your local settings -- substitute your local regional symbols for these if appropriate.
As for your question about checking numbers, here are two functions that I have posted in the past for similar questions..... one is for digits only and the other is for "regular" numbers (the code is simple enough that it can be pulled from the function "housing" and used directly inside your own code):

Function IsDigitsOnly(Value As String) As Boolean
    IsDigitsOnly = Len(Value) > 0 And _
                   Not Value Like "*[!0-9]*"
End Function
 
Function IsNumber(ByVal Value As String) As Boolean
    '   Leave the next statement out if you don't
    '   want to provide for plus/minus signs
    If Value Like "[+-]*" Then Value = Mid$(Value, 2)
    IsNumber = Not Value Like "*[!0-9.]*" And Not Value Like "*.*.*" And Len(Value) > 0 And Value <> "."
End Function

Here are revisions to the above functions that deal with the local settings for decimal points (and thousand's separators) that are different than used in the US (this code works in the US too, of course).

Function IsNumber(ByVal Value As String) As Boolean
  Dim DP As String
  '   Get local setting for decimal point
  DP = Format$(0, ".")
  '   Leave the next statement out if you don't
  '   want to provide for plus/minus signs
  If Value Like "[+-]*" Then Value = Mid$(Value, 2)
  IsNumber = Not Value Like "*[!0-9" & DP & "]*" And Not Value Like "*" & DP & "*" & DP & "*" And Len(Value) > 0 And Value <> DP
End Function

I'm not as concerned by the rejection of entries that include one or more thousand's separators, but we can handle this if we don't insist on the thousand's separator being located in the correct positions (in other words, we'll allow the user to include them for their own purposes... we'll just tolerate their presence).

Function IsNumber(ByVal Value As String) As Boolean
  Dim DP As String
  Dim TS As String
  '   Get local setting for decimal point
  DP = Format$(0, ".")
  '   Get local setting for thousand's separator
  '   and eliminate them. Remove the next two lines
  '   if you don't want your users being able to
  '   type in the thousands separator at all.
  TS = Mid$(Format$(1000, "#,###"), 2, 1)
  Value = Replace$(Value, TS, "")
  '   Leave the next statement out if you don't
  '   want to provide for plus/minus signs
  If Value Like "[+-]*" Then Value = Mid$(Value, 2)
  IsNumber = Not Value Like "*[!0-9" & DP & "]*" And Not Value Like "*" & DP & "*" & DP & "*" And Len(Value) > 0 And Value <> DP
End Function

[mdalamers]

Hi Rick,

This was very helpful and educational.
Thank you!

ML.