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)
Related
I wanted your help, I'm currently working in extracting some data, now the thing is that I have to count an specific amount of Call IDs a call ID format is the following 9129572520020000711. The pattern is 19 characters that starts with 9 and ends in 1.
and I want to count how many times this pattern appears in one cell
I.E. this is the value in one cell and I want to count how many times the pattern appears.
1912957252002000071129129545183410000711391295381628700007114912959791875000071159129597085000000711691295892838400007117912958908933000071189129452513730000711
To solve this with formulae you need to know:
The starting character
The ending character
The length of your Call ID
Finding all possible Call IDs
Let B1 be your number string and B2 be the call ID (or pattern) you are looking for. In B5 enter the formula =MID($B$2,1,1) to find the starting character you are looking for. In B6 enter =RIGHT($B$2,1) for the end character. In B7 enter =LEN($B$2) for the length of the call ID.
In Column A we'll enter the position of every starting character. The first formula will be a simple Find() formula in B10 as =FIND($B$5,$B$1,1). To find the other starting characters start the Find() at the location after the last starting character: =FIND($B$5,$B$1,$A10+1) in B11. Copy this down the column a few dozen times (or more).
In Column B we'll see if the next X characters (where X is the length of the Call ID) meets the criteria for a Call ID:
=IF(MID($B$1,$A10+($B$7-1),1)=$B$6,TRUE,FALSE)
The MID($B$1,$A10+($B$7-1),1)=$B$6 checks if the character at the end of the character at the end of this possible Call ID is the end character we're looking for. $A10+($B$7) calculates the position of the possible Call ID and $B$6 is the end character.
In Column C we can return the actual Call ID if there is a match. This isn't necessary to find the count, but will be useful later. Simply check if the value in Column B is True and, if yes, return the calculated string: =IF(B10,MID($B$1,$A10,$B$7),"").
To actually count the number of valid Call IDs, do a CountIf() of the Call ID column to check for the number of True values: =IF(B10,MID($B$1,$A10,$B$7),"").
If you don't want all the #Values! just wrap everything in IFERROR(,"") formulas.
Finding all consecutive Call IDs
However , some of these Call IDs overlap. Operating on the assumption that Call IDs cannot overlap, we simply have to start our search after the end character of a found ID, not the start. Insert an "Ending Position" column in Column B with the formulae: =$A10+($C$7-1), starting in B11. Alter A11 to =FIND($C$5,$C$1,$B10+1) and copy down. Don't change A10 as this finds the first starting position and is not depending on anything but the original text.
Which ones are valid?
I don't know, that depends on other criteria for your Call IDs. If you receive them consecutively, then the second method is best and the other possible ones found are by coincidence. If not, then you'll have to apply some other validation criteria to the first method, hence why we identified each ID.
You can solve this simply with a UDF using a regular expression.
Option Explicit
Function callIDcount(S As String) As Long
Dim RE As Object, MC As Object
Const sPat As String = "9\d{17}1"
Set RE = CreateObject("vbscript.regexp")
With RE
.Global = True
.Pattern = sPat
Set MC = .Execute(S)
callIDcount = MC.Count
End With
End Function
Using your example, this returns a count of 8
The regular expression engine captures all of the matches that match the pattern, into the match collection. To see how many are there, we merely return the count of that collection.
Trivial modifications would allow one to return the actual ID's also, should that be necessary.
The regex:
9\d{17}1
9\d{17}1
Match the character “9” literally 9
Match a single character that is a “digit” (ASCII 0–9 only) \d{17}
Exactly 17 times {17}
Match the character “1” literally 1
Created with RegexBuddy
EDIT Reading through TheFizh's post, he considered that you might want the count to include overlapping CallID's. In other words, given:
9129572520020000711291
We see that includes:
9129572520020000711
9572520020000711291
where the second overlaps with the first, but both meet your requirements.
Should that be what you want, merely change the regex so it does not "consume" the match:
Const sPat As String = "9(?=\d{17}1)"
and you will return the result of 15 instead of 8, which would be non-overlapping pattern.
Do you mean something like what's following?
Sub CallID_noPatterns()
Dim CallID As String, CallIDLen As Integer
CallID = "9#################1"
CallIDLen = Len(CallID) 'the CallID's length
'Say that you want to get the value of "A1" cell and deal with its value
Dim CellVal As String, CellLen As Integer
CellVal = CStr(Range("A1").Text) 'get its value as a string
CellLen = Len(CellVal) 'get its length
'You Have 2 options:-
'1-The value is smaller than your CallID length. (Not Applicable)
'2-The value is longer than or equal to your CallID length
'So just run your code for the 2nd option
Dim i As Integer, num_checks, num_patterns
i = 0
num_patterns = 0
'imagine both of them as 2 arrays, every array consists of sequenced elements
'and your job is to take a sub-array from your value, of a length
' equals to CallID's length
'then compare your sub-array with CallID
num_checks = CellLen - CallIDLen + 1
If CellLen >= CallIDLen Then
For i = 0 To num_checks - 1 Step 19
For j = i To num_checks - 1
If Mid(CellVal, (j + 1), CallIDLen) Like CallID Then
num_patterns = num_patterns + 1
Exit For
End If
Next j
Next i
End If
'Display your result
MsgBox "Number of Patterns: " & Str(num_patterns)
End Sub
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.
So I have a column called chemical formula for like 40,000 entries, and what I want to be able to do is count up how many elements are contained in the chemical formula. So for example:-
EXACT_MASS FORMULA
626.491026 C40H66O5
275.173274 C13H25NO5
For this, I need some kind of formula that will return with the result of
C H O
40 66 5
13 25 5
all as separate columns for the different elements and in rows for the different entries. Is there a formula that can do this?
You could make your own formula.
Open the VBA editor with ALT and F11 and insert a new module.
Add a reference to Microsoft VBScript Regular Expressions 5.5 by clicking Tools, then references.
Now add the following code:
Public Function FormulaSplit(theFormula As String, theLetter As String) As String
Dim RE As Object
Set RE = CreateObject("VBScript.RegExp")
With RE
.Global = True
.MultiLine = False
.IgnoreCase = False
.Pattern = "[A-Z]{1}[a-z]?"
End With
Dim Matches As Object
Set Matches = RE.Execute(theFormula)
Dim TheCollection As Collection
Set TheCollection = New Collection
Dim i As Integer
Dim Match As Object
For i = (Matches.Count - 1) To 0 Step -1
Set Match = Matches.Item(i)
TheCollection.Add Mid(theFormula, Match.FirstIndex + (Len(Match.Value) + 1)), UCase(Trim(Match.Value))
theFormula = Left(theFormula, Match.FirstIndex)
Next
FormulaSplit = "Not found"
On Error Resume Next
FormulaSplit = TheCollection.Item(UCase(Trim(theLetter)))
On Error GoTo 0
If FormulaSplit = "" Then
FormulaSplit = "1"
End If
Set RE = Nothing
Set Matches = Nothing
Set Match = Nothing
Set TheCollection = Nothing
End Function
Usage:
FormulaSplit("C40H66O5", "H") would return 66.
FormulaSplit("C40H66O5", "O") would return 5.
FormulaSplit("C40H66O5", "blah") would return "Not found".
You can use this formula directly in your workbook.
I've had a stab at doing this in a formula nad come up with the following:
=IFERROR((MID($C18,FIND(D17,$C18)+1,2))*1,IFERROR((MID($C18,FIND(D17,$C18)+1,1))*1,IFERROR(IF(FIND(D17,$C18)>0,1),0)))
It's not very neat and would have to be expanded further if any of your elements are going to appear more than 99 times - I also used a random placement on my worksheet so the titles H,C and O are in row 17. I would personally go with Jamie's answer but just wanted to try this to see if I could do it in a formula possible and figured it was worth sharing just as another perspective.
Even though this has an excellent (and accepted) VBA solution, I couldn't resist the challenge to do this without using VBA.
I posted a solution earlier, which wouldn't work in all cases. This new code should always work:
=MAX(
IFERROR(IF(FIND(C$1&ROW($1:$99),$B2),ROW($1:$99),0),0),
IFERROR(IF(FIND(C$1&CHAR(ROW($65:$90)),$B2&"Z"),1,0),0)
)
Enter as an array formula: Ctrl + Shift + Enter
Output:
The formula outputs 0 when not found, and I simply used conditional formatting to turn zeroes gray.
How it works
This part of the formula looks for the element, followed by a number between 1 and 99. If found, the number of atoms is returned. Otherwise, 0 is returned. The results are stored in an array:
IFERROR(IF(FIND(C$1&ROW($1:$99),$B2),ROW($1:$99),0),0)
In the case of C13H25NO5, a search for "C" returns this array:
{1,0,0,0,0,0,0,0,0,0,0,0,13,0,0,0,...,0}
1 is the first array element, because C1 is a match. 13 is the thirteenth array element, and that's what we're interested in.
The next part of the formula looks for the element, followed by an uppercase letter, which indicates a new element. (The letters A through Z are characters 65 through 90.) If found, the number 1 is returned. Otherwise, 0 is returned. The results are stored in an array:
IFERROR(IF(FIND(C$1&CHAR(ROW($65:$90)),$B2&"Z"),1,0),0)
"Z" is appended to the chemical formula, so that a match will be found when its last element has no number. (For example, "H2O".) There is no element "Z" in the Periodic Table, so this won't cause a problem.
In the case of C13H25NO5, a search for "N" returns this array:
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0}
1 is the 15th element in the array. That's because it found the letters "NO", and O is the 15th letter of the alphabet.
Taking the maximum value from each array gives us the number of atoms as desired.
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'm having problem making the match() work in excel VBA. The code is:
x = Application.Match("Sep 2008", Range("F1:F1"), 0)
The value in cell F1 is 9/1/2008.
Even if I changed Sep 2008 to 9/1/2008, it still doesn't return any value.
Any idea how to fix it?
The reason why Even if I changed Sep 2008 to 9/1/2008, it still doesn't return any value.
Is because when there is a Date in excel, Excel automatically converts that date to a numeric value, What you really want to search for is:
39692
This number is the number of days between 9/1/2008 and excel default of 1/1/1900
every date in excel is stored with a value like this. So the easiest way to handle this would be to convert what you see as a date to what excel sees as a date using CDate().
This by itself will give you an unuseful error that vba can't get the property.
That is because the Lookup_value can be a value (number, text, or logical value) or a cell reference to a number, text, or logical value. Not a date so simply convert the now date value to a number to search for the matching number in the list using CLng()
Give this a shot it will also be much faster then using the Find alternative:
x = WorksheetFunction.Match(CLng(CDate("Sep 2008")), Range("F1:F1"), 0)
This should give you the result expected
To handle when no match is found try this Sub:
Sub MatchDate()
Dim myvalue As Double
Dim LastRow As Long
LastRow = Cells(Rows.Count, "F").End(xlUp)
On Error GoTo NotFound
myvalue = WorksheetFunction.Match(CLng(CDate("Sep 2008")), Range("F1:F" & LastRow), 0)
MsgBox (myvalue)
End
NotFound:
MsgBox ("No Match Was Found")
End
End:
End Sub
Your best bet is to use .Find(). This will return a range if found or nothing if not.
Set x = Range("F1:F1").Find(CDate("Sept 2008"), , , xlWhole)
If you wanted the column number:
x = Range("F1:F1").Find(CDate("Sept 2008"), , , xlWhole).Column
With capture of not found
Sub test()
Dim y As Date, x As Variant, c As Long
y = CDate("Sep 2008")
Set x = Range("1:1").Find(y, , , xlWhole)
If Not x Is Nothing Then
c = x.Column '<~~found
Else
Exit Sub 'not found
End If
End Sub
Bottom line:
use WorksheetFunction.Match(CDbl(date), range, 0)
Alternatively, use a Date cell's Value2 property (which will also be a Double) instead of Value for the search key.
CLng suggested in other answers would discard the time part of the date.
The same problem exists for the Currency data type but you can't use CDbl for it (see below for options).
Range.Value2 Property (Excel) article suggests that Date and Currency types are "special" in that they have an "internal representation" that's in stark contrast with displayed value. Indeed:
Date is internally represented as IEEE 64-bit (8-byte) floating-point numbers where the integer part is the date and fractional part is the time
Currency is also 8-byte but is treated as a fixed-point number with 4 fractional digits (an integer scaled by 10'000)
Apparently, Match compares these internal values for performance reasons. So, we must ensure that they, rather than the readable representations, match exactly.
Since Date is already floating-point internally, CDbl(date) doesn't actually change the data.
For the Currency type, CDbl does change data, so it's out of question. So either
use the exact representation of the key (to 4 fractional digits) this way or another if you require exact match, or
make the cells in the range actually be formulas with Round) if the value to compare with comes from elsewhere and/or you only require equality to 2 fractional digits
This way it works using this method:
Nbr,L, C as Integer
Datedeb as date
nbr = WorksheetFunction.Match(CLng(CDate(Datedeb)), Range(Cells(L, C), Cells(L + 100, C)), 0)
I think I can safely assume that the value in F1 is a date. In you code "Sep 2008" is a string. You will never be able to get a successful match as long as your datatypes are inconsistent.
If you are looking for a date, then make sure that the first parameter is a date.
Dim dSearchSDate As Date
dSearchSDate = "01/Sept/2008"
x = Application.Match(dSearchSDate, Range("F1:F1"), 0)
Here is another possible approach.
Sub temp()
Dim x
Dim dSearchSDate As Date
dSearchSDate = "01/Sept/2008"
If ThisWorkbook.Worksheets(1).Range("F1:F1").Value = dSearchSDate Then
Debug.Print "Found it!"
Else
Debug.Print "Doh!!"
End If
End Sub
I know this post is old, but I had the same issue, and did find the answer.
To make it work, you first need to make VBA see the same data formatting as it appears in your excel spreadsheet :
YourVar = Format("YourDate","mmm-yyyy")
YourResult = Application.match(Clng(Cdate(YourVar)), YourRange, 0)
Regards
Gilles