Not a lot to go on there Alfredo_CPA. Perhaps it would help you to know that
VB has an IsNumeric function that can tell you if the argument passed into
is a number or not. Now, with that said, the IsNumeric function is quite
"generous" in what it considers a number, so you will have to decide whether
it will work for you or not. Here is a previous post of mine describing the
problem with IsNumeric and some possible solutions...
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:
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$(Value, 2)
IsNumber = Not Value Like "*[!0-9.]*" And _
Not Value Like "*.*.*" And _
Len(Value) 0 And Value < "." And _
Value < vbNullString
End Function
Here are revisions to the above functions that deal with the local settings
for decimal points (and thousand's separators) that are different than used
in the US (this code works in the US too, of course).
Function IsNumber(ByVal Value As String) As Boolean
Dim DP As String
' Get local setting for decimal point
DP = Format$(0, ".")
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _
Not Value Like "*" & DP & "*" & DP & "*" And _
Len(Value) 0 And Value < DP And _
Value < vbNullString
End Function
I'm not as concerned by the rejection of entries that include one or more
thousand's separators, but we can handle this if we don't insist on the
thousand's separator being located in the correct positions (in other words,
we'll allow the user to include them for their own purposes... we'll just
tolerate their presence).
Function IsNumber(ByVal Value As String) As Boolean
Dim DP As String
Dim TS As String
' Get local setting for decimal point
DP = Format$(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 And _
Value < vbNullString
End Function
--
Rick (MVP - Excel)
"Alfredo_CPA" .(donotspam) wrote in message
...
I'm using excel 2003
I have an input box that allows the suer to enter a variable.
I need a code that validates the "type of entry" the user types. I.e. if
the
input is a number I need the code to do an specific action, but if the
user
types a "legend/word", I want the code to do a different action.
My problem is not with the if structure, my problem is the code to
validate
the type of entry...
--
thanks