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#1
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Need help utilizing the "Like" function
Using Excel 2003, here is example code that doesn't work. What am I doing
wrong? My isNumeric test returns "True", but there an "A" in my string. Am I not checking for all the character to be 0 - 9? Sub Test1() Debug.Print isNumeric("A00101") End Sub Function isNumeric(ByVal sInput As String) As Boolean If sInput Like "*[0-9]" Then isNumeric = True Else isNumeric = False End If End Function Thanks, - Pat |
#2
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Need help utilizing the "Like" function
Dreiding -
I am not sure why you aren't using Excel's functions, but here is why your's isn't working. You are only testing for the last character to be 0-9. The asterisk will match to zero or more characters. For example, Like "*6" will match for all of these: abc6 126 6 Z6 That means Like "*[0-9]" will match to any character string ending in a numeric digit. If you have a fixed-length string of length 4 and you want them all numeric, then you could use this: Like "[0-9][0-9][0-9][0-9]" If it is variable-length, then you could loop through each character to test for numerics. -- Daryl S "Dreiding" wrote: Using Excel 2003, here is example code that doesn't work. What am I doing wrong? My isNumeric test returns "True", but there an "A" in my string. Am I not checking for all the character to be 0 - 9? Sub Test1() Debug.Print isNumeric("A00101") End Sub Function isNumeric(ByVal sInput As String) As Boolean If sInput Like "*[0-9]" Then isNumeric = True Else isNumeric = False End If End Function Thanks, - Pat |
#3
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Need help utilizing the "Like" function
Function isNumeric(ByVal sInput As String) As Boolean
isNumeric = VBA.isNumeric(sInput) End Function HTH Bob "Dreiding" wrote in message ... Using Excel 2003, here is example code that doesn't work. What am I doing wrong? My isNumeric test returns "True", but there an "A" in my string. Am I not checking for all the character to be 0 - 9? Sub Test1() Debug.Print isNumeric("A00101") End Sub Function isNumeric(ByVal sInput As String) As Boolean If sInput Like "*[0-9]" Then isNumeric = True Else isNumeric = False End If End Function Thanks, - Pat |
#4
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Need help utilizing the "Like" function
A short-cut for digits is the # symbol; so, you can check for 4 digits like
this... If NumberString Like "####" Then MsgBox "NumberString is 4 digits long" -- Rick (MVP - Excel) "Daryl S" wrote in message ... Dreiding - I am not sure why you aren't using Excel's functions, but here is why your's isn't working. You are only testing for the last character to be 0-9. The asterisk will match to zero or more characters. For example, Like "*6" will match for all of these: abc6 126 6 Z6 That means Like "*[0-9]" will match to any character string ending in a numeric digit. If you have a fixed-length string of length 4 and you want them all numeric, then you could use this: Like "[0-9][0-9][0-9][0-9]" If it is variable-length, then you could loop through each character to test for numerics. -- Daryl S "Dreiding" wrote: Using Excel 2003, here is example code that doesn't work. What am I doing wrong? My isNumeric test returns "True", but there an "A" in my string. Am I not checking for all the character to be 0 - 9? Sub Test1() Debug.Print isNumeric("A00101") End Sub Function isNumeric(ByVal sInput As String) As Boolean If sInput Like "*[0-9]" Then isNumeric = True Else isNumeric = False End If End Function Thanks, - Pat |
#5
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Need help utilizing the "Like" function
If you don't want to use the Like comparison (I think it is the better way
to go), you can call out to the Worksheet ISNUMBER function, but I would not use the VBA IsNumeric function to proof a typed-in value for being a number. Here is a previous posting of mine explaining why... 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$(Value, 2) IsNumber = Not Value Like "*[!0-9.]*" And _ Not Value Like "*.*.*" And _ Len(Value) 0 And Value < "." End Function Here are revisions to the above functions that deal with the local settings for decimal points (and thousand's separators) that are different than used in the US (this code works in the US too, of course). Function IsNumber(ByVal Value As String) As Boolean Dim DP As String ' Get local setting for decimal point DP = Format$(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 -- Rick (MVP - Excel) "Bob Phillips" wrote in message ... Function isNumeric(ByVal sInput As String) As Boolean isNumeric = VBA.isNumeric(sInput) End Function HTH Bob "Dreiding" wrote in message ... Using Excel 2003, here is example code that doesn't work. What am I doing wrong? My isNumeric test returns "True", but there an "A" in my string. Am I not checking for all the character to be 0 - 9? Sub Test1() Debug.Print isNumeric("A00101") End Sub Function isNumeric(ByVal sInput As String) As Boolean If sInput Like "*[0-9]" Then isNumeric = True Else isNumeric = False End If End Function Thanks, - Pat |
#6
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Need help utilizing the "Like" function
... all the character to be 0 - 9?
Hi. One of many ideas if you want to use 'Like' Function AllDigits(s) AllDigits = s Like WorksheetFunction.Rept("[0-9]", Len(s)) End Function Sub TestIt() Debug.Print AllDigits("123.45") Debug.Print AllDigits("678") End Sub Returns: False True = = = = = = = HTH :) Dana DeLouis On 2/10/2010 12:21 PM, Dreiding wrote: Using Excel 2003, here is example code that doesn't work. What am I doing wrong? My isNumeric test returns "True", but there an "A" in my string. Am I not checking for all the character to be 0 - 9? Sub Test1() Debug.Print isNumeric("A00101") End Sub Function isNumeric(ByVal sInput As String) As Boolean If sInput Like "*[0-9]" Then isNumeric = True Else isNumeric = False End If End Function Thanks, - Pat |
#7
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Need help utilizing the "Like" function
Function AllDigits(s)
AllDigits = s Like WorksheetFunction.Rept("[0-9]", Len(s)) End Function Noting that you can use the # sign in place of [0-9] when searching for digits, you do not have to call out to the Worksheet to use its REPT function... VB has the String function that you can use instead. Function AllDigits(s) AllDigits = s Like String(Len(s), "#") End Function If, however, you are more used to using [0-9] instead of the # sign and want to continue doing so, you can still do this using native VB functions only... Function AllDigits(s) AllDigits = s Like Replace(String(Len(s), "x"), "x", "[0-9]") End Function where you can use any character in place of the "x" characters I used. Because of this, we can simplify this code by using a space instead of the "x" and then noting that VB has a Space function which returns the number of space specified in its argument... Function AllDigits(s) AllDigits = s Like Replace(Space(Len(s)), " ", "[0-9]") End Function -- Rick (MVP - Excel) "Dana DeLouis" wrote in message ... ... all the character to be 0 - 9? Hi. One of many ideas if you want to use 'Like' Function AllDigits(s) AllDigits = s Like WorksheetFunction.Rept("[0-9]", Len(s)) End Function Sub TestIt() Debug.Print AllDigits("123.45") Debug.Print AllDigits("678") End Sub Returns: False True = = = = = = = HTH :) Dana DeLouis On 2/10/2010 12:21 PM, Dreiding wrote: Using Excel 2003, here is example code that doesn't work. What am I doing wrong? My isNumeric test returns "True", but there an "A" in my string. Am I not checking for all the character to be 0 - 9? Sub Test1() Debug.Print isNumeric("A00101") End Sub Function isNumeric(ByVal sInput As String) As Boolean If sInput Like "*[0-9]" Then isNumeric = True Else isNumeric = False End If End Function Thanks, - Pat |
#8
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Need help utilizing the "Like" function
Function AllDigits(s)
AllDigits = s Like String(Len(s), "#") End Function Thanks Rick. I totally forgot about that. :~ Thanks for the catch. Dana DeLouis On 2/10/2010 11:01 PM, Rick Rothstein wrote: Function AllDigits(s) AllDigits = s Like WorksheetFunction.Rept("[0-9]", Len(s)) End Function Noting that you can use the # sign in place of [0-9] when searching for digits, you do not have to call out to the Worksheet to use its REPT function... VB has the String function that you can use instead. Function AllDigits(s) AllDigits = s Like String(Len(s), "#") End Function If, however, you are more used to using [0-9] instead of the # sign and want to continue doing so, you can still do this using native VB functions only... Function AllDigits(s) AllDigits = s Like Replace(String(Len(s), "x"), "x", "[0-9]") End Function where you can use any character in place of the "x" characters I used. Because of this, we can simplify this code by using a space instead of the "x" and then noting that VB has a Space function which returns the number of space specified in its argument... Function AllDigits(s) AllDigits = s Like Replace(Space(Len(s)), " ", "[0-9]") End Function -- = = = = = = = HTH :) Dana DeLouis |
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