I'm a newbie on Excel.
So I have a list of some names ending with Hexa decimals. And some names, that doesn't have any.
My mission is to see only those names with Hexa decimals. (Mabye somehow filter them out)
Column:
BFAXSPOINTDEVBAUHOFLAN2AD
BFAXSQLBAUHOFLAN207
BFAXSQLDEVBAUHOFLAN27A
BFREPDEVBAUHOFLAN258
BFREPORTINGBAUHOFLAN20B
COBALTSEA02900
COBALTSEAVHOST900
DIRECTO8000
DIRECTO9000
DIRECTODCDIRECTOLA009
DYNAMAEBSSISE006
SURVEYEBSSISE006
KVMSRV00",
KVMSRV01",
KVMSRV02",
ASR
CACTI
DBSYNC",
DTV
and so on...
The Function HEX2DEC will help you achieve what you want - it attempts to convert a number as a hexidecimal, into a decimal. If it is not a valid Hex input, it will produce an error.
The key is understanding how many digits you expect your decimal to be - is it the last 5 characters; the last 10; etc. Also note that there is a risk that random text / numbers will be seen as hexidecimal when really that's not what it represents [but that's a problem with the question as you have laid it out; going solely based on the text provided, all we can see is whether a particular cell creates a valid Hexidecimal].
The full formula would look like this[assuming your data starts in A1, and that your Hexidecimal numbers are expected to be 6 characters long, this goes in B1 and is copied down]:
=ISERROR(HEX2DEC(RIGHT(A1,6)))
This takes the 6 rightmost characters of a cell, and attempts to convert it from Hex to Decimal. If it fails, it will produce TRUE [because of ISERROR]; if it succeeds, it will produce FALSE.
Then simply filter on your column to see the subset of results you care about.
Consider the following UDF:
Public Function EndsInHex(r As Range) As Boolean
Dim s As String, CH As String
s = r(1).Text
CH = Right(s, 1)
If CH Like "[A-F]" Or CH Like "[0-9]" Then
EndsInHex = True
Else
EndsInHex = False
End If
End Function
For the string to end in a hex, the last character must be a hex.
Related
I'm working on a formula that checks a cell for a "£" sign, then checks the value after it. If the value is anything other than a rounded value, e.g. "£30" or "£30.00", it would be an error.
My problem is, the text has a full stop after the decimal value. This results in the 'DECIMAL' value being returned. How do I not count £30. as an error.
Below is the formula as input through VBA
=IF(ISERROR(SEARCH(""."",LEFT((RIGHT(RC[-2],LEN(RC[-2])-SEARCH(""£"",RC[-2]))),SEARCH("" "",(RIGHT(RC[-2],LEN(RC[-2])-SEARCH(""£"",RC[-2])))))))=FALSE,IF(ISERROR(SEARCH("".00"",LEFT((RIGHT(RC[-2],LEN(RC[-2])-SEARCH(""£"",RC[-2]))),SEARCH("" "",(RIGHT(RC[-2],LEN(RC[-2])-SEARCH(""£"",RC[-2])))))))=FALSE,"""",""DECIMAL""),"""")
Here is a User Defined Function that should do what you want. It looks for a substring that starts with the € currency sign and then some number of digits. If there is a dot following, it looks at the dot and the next two digits (or nothing if no digits). If the value of that sequence is zero, then the function returns TRUE, else FALSE
Obviously, you can modify this to return whatever you want, or use it within an IF formula on the worksheet.
Note that it will only work on strings, and not on values formatted as currency.
Option Explicit
Function ValidCurrency(S As String) As Boolean
Dim RE As Object, MC As Object
Const sPattern As String = "€\s*\d+(\.\d{2})?"
ValidCurrency = False
Set RE = CreateObject("vbscript.regexp")
With RE
.Pattern = sPattern
If .test(S) = True Then
Set MC = .Execute(S)
ValidCurrency = (MC(0).submatches(0) = 0)
End If
End With
End Function
Assuming the string you want to check is in F1, I think you could do something like this:
=IF(ISNUMBER(FIND("£",F1)),IF(TRIM(MID(F1&"00",FIND("£",F1)+1,5))="30.00",TRUE,FALSE),FALSE)
This:
assumes the £ number you want to inspect is the first occurrence of that "£" character in the string.
Assumes that an occurrence of "£30." would occur at the end of the sentence, as you assert in your comments above.
But perhaps you left out some details for examples in which it won't work.
For your follow up question, the following would work for all two-digit whole number values (10 thru 99). Note that I'm comparing the value in the string to its ROUND()ed value to see if the two match, and thus make sure there's nothing after the decimal. This should be enough for you to adapt it further on your own if you need to cover a broader range of values, but at this point, I think I've answered your original question and then some.
=IF(ISNUMBER(FIND("£",F1)),IF(VALUE(TRIM(MID(F1&"00",FIND("£",F1)+1,5)))=ROUND(VALUE(TRIM(MID(F1&"00",FIND("£",F1)+1,5))),0),TRUE,FALSE),FALSE)
I am trying to convert 14 bit hex numbers to decimal.
I have this VBA code.
Option Explicit
Public Function HexadecimalToDecimal(HexValue As String) As Double
Dim ModifiedHexValue As String
ModifiedHexValue = Replace(HexValue, "0x", "&H")
HexadecimalToDecimal = CDec(ModifiedHexValue)
End Function
With numbers like this to convert to decimal
0x047B1142591E80
0x044A81325A1E80
0x047B7542591E80
I keep getting random results across large amounts of data. Sometimes spot on other times the numbers are off by 6 or 2.
Try changing the return type of the function from Double to Variant. Double has only about 15 decimal digits of precision, so can't, for example, capture the value 1261213964639872 (which has 16 digits) exactly. The closest it can get is 1261213964639870. By changing the return type to Variant, the full precision returned by CDec will be preserved. You can't use a Decimal return type, because VBA for some reason does not support this.
The problem isn't with VBA. Excel cells can only hold 15 digits in number format. So the "number" 1234567891234567 will always display 1234567891234560. This can be avoided by converting items to text AND/OR changing the cell format to text.
But this doesn't always work.
The only surefire way to make sure it will retain all digits is to append something onto the string that isn't a number.
This code will append an apostrophe before the number, but return the entire string.
Public Function HexadecimalToDecimal(HexValue As String) As String
Dim ModifiedHexValue As String
ModifiedHexValue = Replace(HexValue, "0x", "&H")
HexadecimalToDecimal = "'" & CDec(ModifiedHexValue)
End Function
Unfortunately, not a perfect solution.
If I am having G4ED7883666 and I want the output to be 7883666
and I have to apply this on a range of cells and they are not the same length and the only common thing is that I have to delete anything before the number that lies before the alphabet?
This formula finds the last number in a string, that is, all digits to the right of the last alpha character in the string.
=RIGHT(A1,MATCH(99,IFERROR(1*MID(A1,LEN(A1)+1-ROW($1:$25),1),99),0)-1)
Note that this is an array formula and must be entered with the Control-Shift-Enter keyboard combination.
How the formula works
Let's assume that the target string is fairly simple: "G4E78"
Working outward from the middle of the formula, the first thing to do is create an array with the elements 1 through 25. (Although this might seem to limit the formula to strings with no more than 25 characters, it actually places a limit of 25 digits on the size of the number that may be extracted by the formula.
ROW($1:$25) = {1;2;3;4;5;6;7; etc.}
Subtracting from this array the value of (1 + the length of the target string) produces a new array, the elements of which count down from the length of string. The first five elements will correspond to the position of the characters of the string - in reverse order!
LEN(A1)+1-ROW($1:$25) = {5;4;3;2;1;0;-1;-2;-3;-4; etc.}
The MID function then creates a new array that reverses the order of the characters of the string.
For example, the first element of the new array is the result of MID(A1, 5, 1), the second of MID(A1, 4, 1) and so on. The #VALUE! errors reflect the fact that MID cannot evaluate 0 or negative values as the position of a string, e.g., MID(A1,0,1) = #VALUE!.
MID(A1,LEN(A1)+1-ROW($1:$25),1) = {"8";"7";"E";"4";"G";#VALUE!;#VALUE!; etc.}
Multiplying the elements of the array by 1 turns the character elements of that array to #VALUE! errors as well.
=1*MID(A1,LEN(A1)+1-ROW($1:$25),1) = {"8";"7";#VALUE!;"4";#VALUE!;#VALUE!;#VALUE!; etc.}
And the IFERROR function turns the #VALUES into 99, which is just an arbitrary number greater than the value of a single digit.
IFERROR(1*MID(A1,LEN(A1)+1-ROW($1:$25),1),99) = {8;7;99;4;99;99;99; etc.}
Matching on the 99 gives the position of the first non-digit character counting from the right end of the string. In this case, "E" is the first non-digit in the reversed string "87E4G", at position 3. This is equivalent to saying that the number we are looking for at the end of the string, plus the "E", is 3 characters long.
MATCH(99,IFERROR(1*MID(A1,LEN(A1)+1-ROW($1:$25),1),99),0) = 3
So, for the final step, we take 3 - 1 (for the "E) characters from the right of string.
RIGHT(A1,MATCH(99,IFERROR(1*MID(A1,LEN(A1)+1-ROW($1:$25),1),99),0)-1) = "78"
One more submission for you to consider. This VBA function will get the right most digits before the first non-numeric character
Public Function GetRightNumbers(str As String)
Dim i As Integer
For i = Len(str) To 0 Step -1
If Not IsNumeric(Mid(str, i, 1)) Then
Exit For
End If
Next i
GetRightNumbers = Mid(str, i + 1)
End Function
You can write some VBA to format the data (just starting at the end and working back until you hit a non-number.)
Or you could (if you're happy to get an addin like Excelicious) then you can use regular expressions to format the text via a formula. An expression like [0-9]+$ would return all the numbers at the end of a string IIRC.
NOTE: This uses the regex pattern in James Snell's answer, so please upvote his answer if you find this useful.
Your best bet is to use a regular expression. You need to set a reference to VBScript Regular Expressions for this to work. Tools --> References...
Now you can use regex in your VBA.
This will find the numbers at the end of each cell. I am placing the result next to the original so that you can verify it is working the way you want. You can modify it to replace the cell as soon as you feel comfortable with it. The code works regardless of the length of the string you are evaluating, and will skip the cell if it doesn't find a match.
Sub GetTrailingNumbers()
Dim ws As Worksheet
Dim rng As Range
Dim cell As Range
Dim result As Object, results As Object
Dim regEx As New VBScript_RegExp_55.RegExp
Set ws = ThisWorkbook.Sheets("Sheet1")
' range is hard-coded here, but you can define
' it programatically based on the shape of your data
Set rng = ws.Range("A1:A3")
' pattern from James Snell's answer
regEx.Pattern = "[0-9]+$"
For Each cell In rng
If regEx.Test(cell.Value) Then
Set results = regEx.Execute(cell.Value)
For Each result In results
cell.Offset(, 1).Value = result.Value
Next result
End If
Next cell
End Sub
Takes the first 4 digits from the right of num:
num1=Right(num,4)
Takes the first 5 digits from the left of num:
num1=Left(num,5)
First takes the first ten digits from the left then takes the first four digits from the right:
num1=Right(Left(num, 10),4)
In your case:
num=G4ED7883666
num1=Right(num,7)
I am putting a string into excel. The string is often only numeric digits but can have alpha characters or hypens etc.
When I don't set the number format or set it like this
(Where xlSheet(0) is Excel.Worksheet)
xlSheet(0).Columns("N:N").EntireColumn.Columns.NumberFormat = "#"
It outputs in scientific notation.
When I use this code:
xlSheet(0).Columns("N:N").EntireColumn.Columns.NumberFormat = "0"
It rounds up the number to the nearest 100,000 so that the last five digits are 0's when they shouldn't be.
Should be: 1539648751235678942
But is: 1539648751235600000
The cells that have a hyphen or a letter aren't affected and work fine.
Any help would be greatly appreciated.
EDIT:
I add the data like this:
I loop through and put in xlSheet(0).Cells(i, 14) = rs!value_number
Where rs is my ADODB.Recordset
EDIT2: Herbert Sitz got it by adding an apostrophe before the text! Thanks everyone.
I think problem is that the number you're trying to enter can't be accommodated exactly by Excel. Excel has limitations on what numbers it display/represent because of the way numbers are stored internally. In Excel's case numbers are limited to 15 digit precision (see http://office.microsoft.com/en-us/excel-help/excel-specifications-and-limits-HP010073849.aspx ), which is not enough to represent your number.
You can enter the number as a string ("152..42") and all digits will be displayed, but you won't be able to perform exact mathematical operations with it.
For numbers, Excel can only handle 15 significant digits.
If you want to store a number that is more than 15 digits long without losing data, you have to store the data as text.
Doing what you've been doing will resolve the issue:
You can do either of the following to add your numbers as text:
xlSheet(0).Cells(i, 14).Numberformat = "#"
xlSheet(0).Cells(i, 14) = rs!value_number
Or
xlSheet(0).Cells(i, 14) = "'" & rs!value_number
I have a sheet with a list of names in Column B and an ID column in A. I was wondering if there is some kind of formula that can take the value in column B of that row and generate some kind of ID based on the text? Each name is also unique and is never repeated in any way.
It would be best if I didn't have to use VBA really. But if I have to, so be it.
Solution Without VBA.
Logic based on First 8 characters + number of character in a cell.
= CODE(cell) which returns Code number for first letter
= CODE(MID(cell,2,1)) returns Code number for second letter
= IFERROR(CODE(MID(cell,9,1)) If 9th character does not exist then return 0
= LEN(cell) number of character in a cell
Concatenating firs 8 codes + adding length of character on the end
If 8 character is not enough, then replicate additional codes for next characters in a string.
Final function:
=CODE(B2)&IFERROR(CODE(MID(B2,2,1)),0)&IFERROR(CODE(MID(B2,3,1)),0)&IFERROR(CODE(MID(B2,4,1)),0)&IFERROR(CODE(MID(B2,5,1)),0)&IFERROR(CODE(MID(B2,6,1)),0)&IFERROR(CODE(MID(B2,7,1)),0)&IFERROR(CODE(MID(B2,8,1)),0)&LEN(B2)
Sorry, I didn't found a solution with formula only even if this thread might help (trying to calculate the points in a scrabble game) but I didn't find a way to be sure the generated hash would be unique.
Yet, here is my solution, based on a UDF (Used-Defined Function):
Put the code in a module:
Public Function genId(ByVal sName As String) As Long
'Function to create a unique hash by summing the ascii value of each character of a given string
Dim sLetter As String
Dim i As Integer
For i = 1 To Len(sName)
genId = Asc(Mid(sName, i, 1)) * i + genId
Next i
End Function
And call it in your worksheet like a formula:
=genId(A1)
[EDIT] Added the * i to take into account the order. It works on my unit tests
May be OTT for your needs, but you can use a call to CoCreateGuid to get a real GUID
Private Declare Function CoCreateGuid Lib "ole32" (ID As Any) As Long
Function GUID() As String
Dim ID(0 To 15) As Byte
Dim i As Long
If CoCreateGuid(ID(0)) = 0 Then
For i = 0 To 15
GUID = GUID & Format(Hex$(ID(i)), "00")
Next
Else
GUID = "Error while creating GUID!"
End If
End Function
Test using
Sub testGUID()
MsgBox GUID
End Sub
How to best implement depends on your needs. One way would be to write a macro to get a GUID populate a column where names exist. (note, using it as a udf as is is no good, since it will return a new GUID when recalculated)
EDIT
See this answer for creating a SHA1 hash of a string
Do you just want an incrementing numeric id column to sit next to your values? If so, and if your values will always be unique, you can very easily do this with formulae.
If your values were in column B, starting in B2 underneath your headers for example, in A2 you would type the formula "=IF(B2="","",1+MAX(A$1:A1))". You can copy and paste that down as far as your data extends, and it will increment a numeric identifier for each row in column B which isn't blank.
If you need to do anything more complicated, like identify and re-identify repeating values, or make identifiers 'freeze' once they're populated, let me know. Currently, when you clear or add values to your list the identifers will toggle themselves up and down, so you need to be careful if your data changes.
Unique identifier based on the number of specific characters in text. I used an identifier based on vowels and numbers.
=LEN($J$14)-LEN(SUBSTITUTE($J$14;"a";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"e";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"i";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"j";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"o";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"u";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"y";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"1";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"2";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"3";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"4";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"5";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"6";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"7";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"8";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"9";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"0";""))
You say you are confident that there are no duplicate values in your words. To push it further, are you confident that the first 8 characters in any word would be unique?
If so, you can use the below formula. It works by individually taking each character's ASCII code - 40 [assuming normal characters, this puts numbers at between 8 & 57, and letters at between 57 & 122], and multiplying that characters code by 10 ^ [that character's digit placement in the word]. Basically it takes that character code [-40], and concatenates each code onto the next.
EDIT Note that this code no longer requires that at least 8 characters exist in your word to prevent an error, as the actual word to be coded has 8 "0"'s appended to it.
=TEXT(SUM((CODE(MID(LOWER(RIGHT(REPT("0",8)&A3,8)),{1,2,3,4,5,6,7,8},1))-40)*10^{0,2,4,6,8,10,12,14}),"#")
Note that as this uses the ASCII values of the characters, the ID # could be used to identify the name directly - this does not really create anonymity, it just turns 8 unique characters into a unique number. It is obfuscated with the -40, but not really 'safe' in that sense. The -40 is just to get normal letters and numbers in the 2 digit range, so that multiplying by 10^0,2,4 etc. will create a 2 digit unique add-on to the created code.
EDIT FOR ALTERNATIVE METHOD
I had previously attempted to do this so that it would look at each letter of the alphabet, count the number of times it appears in the word, and then multiply that by 10*[that letter's position in the alphabet]. The problem with doing this (see comment below for formula) is that it required a number of 10^26-1, which is beyond Excel's floating point precision. However, I have a modified version of that method:
By limiting the number of allowed characters in the alphabet, we can get the max total size possible to 10^15-1, which Excel can properly calculate. The formula looks like this:
=RIGHT(REPT("0",15)&TEXT(SUM(LEN(A3)*10^{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14}-LEN(SUBSTITUTE(A3,MID(Alphabet,{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15},1),""))*10^{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14}),"#"),15)
[The RIGHT("00000000000000"... portion of the formula is meant to keep all codes the same number of characters]
Note that here, Alphabet is a named string which holds the characters: "abcdehilmnorstu". For example, using the above formula, the word "asdf" counts the instances of a, s, and d, but not 'f' which isn't in my contracted alphabet. The code of "asdf" would be:
001000000001001
This only works with the following assumptions:
The letters not listed (nor numbers / special characters) are not required to make each name unique. For example, asdf & asd would have the same code in the above method.
And,
The order of the letters is not required to make each name unique. For example, asd & dsa would have the same code in the above method.