I want to loop thru an array and extract its delimited values that match every date in a range. For e.g., in the picture below:
I have a date range, say 01-01 to 01-10.
I also have a list of strings (see second pic).
In the array below (see first pic), I have three different values delimited by a semi-colon.
For all matching strings (from second pic) e.g., SISBTXTRPR-(number) and date, I want to extract the last part of the array value.
Picture 1
Picture 2
So, for all array values that match "SISBTXTRPR-4649" (the string from picture 2) and a date (in this case 12-12), I want to extract "2h" from the array. The date range for each string, in this case, "SISBTXTRPR-4649" will be 10 days. I am racking my brain on how to do this :(
This is all I could come up with so far:
While i < UBound(sTimeStamp)
If StrComp(Trim(Format(Now(), "MM-DD")), Trim(Split(sTimeStamp(9), ";")(1))) = 0 And StrComp(Trim(Worksheets("KPIs").Range("AN" & iCounter)), Trim(Split(sTimeStamp(1), ";")(0))) Then
End If
i = i + 1
Wend
Link to file
Sample File
The next code will return occurrences for each string in 'Task' range matching the date from its corresponding 'sTimeStamp Array' string with the one from the 'Date Range Array'. Each occurrence will be add to the next column of 'Task' string column:
Private Sub findOccurrences()
Dim sTask As Worksheet, sStamp As Worksheet, sDate As Worksheet
Dim arrTask As Variant, arrStamp As Variant, arrDate As Variant
Dim i As Long, j As Long, arrS As Variant, El As Variant, dtRef As Date
Set sTask = ThisWorkbook.Sheets("Task")
Set sStamp = ThisWorkbook.Sheets("sTimeStamp Array")
Set sDate = ThisWorkbook.Sheets("Date Range Array")
arrTask = sTask.Range("A2:A" & sTask.Range("A" & sTask.Rows.Count).End(xlUp).Row).Value
arrStamp = sStamp.Range("A2:A" & sStamp.Range("A" & sStamp.Rows.Count).End(xlUp).Row).Value
arrDate = sDate.Range("A2:A" & sDate.Range("A" & sDate.Rows.Count).End(xlUp).Row).Value
'____________________________________________________________________________
sTask.Range("B2:K" & sTask.Range("A" & sTask.Rows.Count).End(xlUp).Row).Clear
Do While i < UBound(arrStamp)
i = i + 1
arrS = Split(arrStamp(i, 1), ";")
For j = 1 To UBound(arrTask)
If arrS(0) = arrTask(j, 1) Then
For Each El In arrDate
dtRef = DateValue(Format(El, "MM-DD"))
If dtRef = DateValue(Format(arrS(1), "MM-DD")) Then
Debug.Print arrS(0) & " (row number " & j + 1 & "), interval """ & _
El & """ exists."
sTask.Cells(j + 1, sTask.Cells(j + 1, _
sTask.Columns.Count).End(xlToLeft).Column).Offset(0, 1).Value = El
End If
Next
End If
Next j
Loop
End Sub
And the short variant working similar to your approach, finding the occurrences for Today date (if I correctly deduced what you intended to achieve), replace the looping part with this:
'______________________________________________________________________________
sStamp.Range("B2:B" & sStamp.Range("A" & sStamp.Rows.Count).End(xlUp).Row).Clear
sTask.Range("A2:A" & sTask.Range("A" & sStamp.Rows.Count).End(xlUp).Row).Interior.ColorIndex = -4142
While i < UBound(arrStamp)
i = i + 1
If StrComp(DateValue(Format(Date, "MM-DD")), DateValue(Split(arrStamp(i, 1), ";")(1))) = 0 And _
Not isMatchErr(CStr(Split(arrStamp(i, 1), ";")(1)), arrDate) Then
Debug.Print "OK for """ & Split(arrStamp(i, 1), ";")(0) & """ of row """ & i & """."
sStamp.Range("B" & i + 1).Value = "OK"
If Not isMatchErr(CStr(Split(arrStamp(i, 1), ";")(0)), arrTask) Then
rowOK = WorksheetFunction.Match(Split(arrStamp(i, 1), ";")(0), arrTask, 0) + 1
sTask.Range("A" & rowOK).Interior.ColorIndex = 3
End If
End If
Wend
And add the next function:
Function isMatchErr(strTime As String, arrDate As Variant) As Boolean
Dim k As Long
On Error Resume Next
k = WorksheetFunction.Match(strTime, arrDate, 0)
If Err.Number <> 0 Then
Err.Clear: On Error GoTo 0: isMatchErr = True
End If
On Error GoTo 0
End Function
Besides the message in Immediate Window, an "OK" will be put on column B:B for all occurrences (in 'sTimeStamp Array' sheet) and background of the matching cell (in 'Task' sheet will be colored in red... In order to do that, I added a new record and modified an existing cell, for "Today" ("01-12"). Please do the same in order to obtain at least two results in column B:B.
Please confirm that this is what you wanted. If not, please better clarify the need...
I'm making an heuristic analyse and i have the fallowing problem : I want to find in column D numbers that match with column J and replace them by a "0". You can see what I'm trying to do on this image :
Problem : Column D have multiples values per cell and column J have one value per cell.
some part of the code:
Dim i,j As Integer
Dim temp As String
Dim x As Integer
Dim d As String
i = Application.CountA(Range("E:E")) + 10
'number of cell with values
j = Application.CountA(Range("J:J")) + 10
For j = 11 To j
temp = Range("J" & j).Value
For i = 11 To i
d = Range("D" & i).Value
*For x = LBound(vec) To UBound(vec)
If vec(x) = temp Then
vec(x) = 0
Range("D" & i).Value = vec(x)
End If
Next
Next
Next
*-> Here it is the problem, i cant figured out how to pass over the coma "," in column D,and store the data. I want to compare the temp with value on "d", but "d" can i have multiple numbers on the same cell, like " 3, 2, 1", and if there is any match like temp = 3, then d= "0,2,1".
English is not my native language so i hope you can understand what i want.
Thanks!
I think your almost there you just need to split up each cell and then search then recreate and replace the string in the cell. Please note I've not tested this.
Dim i,j As Integer
Dim temp As String
Dim x As Integer
Dim d As String
i = Application.CountA(Range("E:E")) + 10
'number of cell with values
j = Application.CountA(Range("J:J")) + 10
For j = 11 To j
temp = Range("J" & j).Value
For i = 11 To i
d = Range("D" & i).Value
Vec = split(d, ",") 'split the cell
d = "" 'clear the string
For x = LBound(vec) To UBound(vec)
If vec(x) = temp Then
vec(x) = 0
End If
d = d & vec(x) & "," 'recreate the string
Next
Range("D" & i).Value = left(d, len(d) - 1) 'save the string without the last ,
Next
Next
you can do this with the below formula - no need for VBA
Make a new column somewhere near column D.
In your new column, use this FIND and Substitute formula:
=IF(NOT(ISERROR(FIND(J:J,D:D))),SUBSTITUTE(D:D,J:J,"0"),D:D)
This formula looks for the value in column J within the cells in column D.
If a match is found, 0 is substituted for the found number. Otherwise, column D is returned.
If you want, you can hide column D to avoid confusion.
in VBA you have a string function called Split(vString, delimiter) that will split a string into tokens using the specified delimiter. See
MSDN Library: VB Split Function
examples:
Mr Spreadsheet John Walkenbach: Split Function examples
VB-Helper
Does anyone have an Excel VBA function which can return the column letter(s) from a number?
For example, entering 100 should return CV.
This function returns the column letter for a given column number.
Function Col_Letter(lngCol As Long) As String
Dim vArr
vArr = Split(Cells(1, lngCol).Address(True, False), "$")
Col_Letter = vArr(0)
End Function
testing code for column 100
Sub Test()
MsgBox Col_Letter(100)
End Sub
If you'd rather not use a range object:
Function ColumnLetter(ColumnNumber As Long) As String
Dim n As Long
Dim c As Byte
Dim s As String
n = ColumnNumber
Do
c = ((n - 1) Mod 26)
s = Chr(c + 65) & s
n = (n - c) \ 26
Loop While n > 0
ColumnLetter = s
End Function
Something that works for me is:
Cells(Row,Column).Address
This will return the $AE$1 format reference for you.
For example: MsgBox Columns( 9347 ).Address returns .
To return ONLY the column letter(s): Split((Columns(Column Index).Address(,0)),":")(0)
For example: MsgBox Split((Columns( 2734 ).Address(,0)),":")(0) returns .
And a solution using recursion:
Function ColumnNumberToLetter(iCol As Long) As String
Dim lAlpha As Long
Dim lRemainder As Long
If iCol <= 26 Then
ColumnNumberToLetter = Chr(iCol + 64)
Else
lRemainder = iCol Mod 26
lAlpha = Int(iCol / 26)
If lRemainder = 0 Then
lRemainder = 26
lAlpha = lAlpha - 1
End If
ColumnNumberToLetter = ColumnNumberToLetter(lAlpha) & Chr(lRemainder + 64)
End If
End Function
Just one more way to do this. Brettdj's answer made me think of this, but if you use this method you don't have to use a variant array, you can go directly to a string.
ColLtr = Cells(1, ColNum).Address(True, False)
ColLtr = Replace(ColLtr, "$1", "")
or can make it a little more compact with this
ColLtr = Replace(Cells(1, ColNum).Address(True, False), "$1", "")
Notice this does depend on you referencing row 1 in the cells object.
This is a version of robartsd's answer (with the flavor of Jan Wijninckx's one line solution), using recursion instead of a loop.
Public Function ColumnLetter(Column As Integer) As String
If Column < 1 Then Exit Function
ColumnLetter = ColumnLetter(Int((Column - 1) / 26)) & Chr(((Column - 1) Mod 26) + Asc("A"))
End Function
I've tested this with the following inputs:
1 => "A"
26 => "Z"
27 => "AA"
51 => "AY"
702 => "ZZ"
703 => "AAA"
-1 => ""
-234=> ""
This is available through using a formula:
=SUBSTITUTE(ADDRESS(1,COLUMN(),4),"1","")
and so also can be written as a VBA function as requested:
Function ColName(colNum As Integer) As String
ColName = Split(Worksheets(1).Cells(1, colNum).Address, "$")(1)
End Function
robertsd's code is elegant, yet to make it future-proof, change the declaration of n to type long
In case you want a formula to avoid macro's, here is something that works up to column 702 inclusive
=IF(A1>26,CHAR(INT((A1-1)/26)+64),"")&CHAR(MOD(A1-1,26)+65)
where A1 is the cell containing the column number to be converted to letters.
LATEST UPDATE: Please ignore the function below, #SurasinTancharoen managed to alert me that it is broken at n = 53.
For those who are interested, here are other broken values just below n = 200:
Please use #brettdj function for all your needs. It even works for Microsoft Excel latest maximum number of columns limit: 16384 should gives XFD
END OF UPDATE
The function below is provided by Microsoft:
Function ConvertToLetter(iCol As Integer) As String
Dim iAlpha As Integer
Dim iRemainder As Integer
iAlpha = Int(iCol / 27)
iRemainder = iCol - (iAlpha * 26)
If iAlpha > 0 Then
ConvertToLetter = Chr(iAlpha + 64)
End If
If iRemainder > 0 Then
ConvertToLetter = ConvertToLetter & Chr(iRemainder + 64)
End If
End Function
Source: How to convert Excel column numbers into alphabetical characters
APPLIES TO
Microsoft Office Excel 2007
Microsoft Excel 2002 Standard Edition
Microsoft Excel 2000 Standard Edition
Microsoft Excel 97 Standard Edition
This is a function based on #DamienFennelly's answer above. If you give me a thumbs up, give him a thumbs up too! :P
Function outColLetterFromNumber(iCol as Integer) as String
sAddr = Cells(1, iCol).Address
aSplit = Split(sAddr, "$")
outColLetterFromNumber = aSplit(1)
End Function
There is a very simple way using Excel power: Use Range.Cells.Address property, this way:
strCol = Cells(1, lngRow).Address(xlRowRelative, xlColRelative)
This will return the address of the desired column on row 1. Take it of the 1:
strCol = Left(strCol, len(strCol) - 1)
Note that it so fast and powerful that you can return column addresses that even exists!
Substitute lngRow for the desired column number using Selection.Column property!
Here is a simple one liner that can be used.
ColumnLetter = Mid(Cells(Row, LastColA).Address, 2, 1)
It will only work for a 1 letter column designation, but it is nice for simple cases. If you need it to work for exclusively 2 letter designations, then you could use the following:
ColumnLetter = Mid(Cells(Row, LastColA).Address, 2, 2)
This will work regardless of what column inside your one code line for cell thats located in row X, in column Y:
Mid(Cells(X,Y).Address, 2, instr(2,Cells(X,Y).Address,"$")-2)
If you have a cell with unique defined name "Cellname":
Mid(Cells(1,val(range("Cellname").Column)).Address, 2, instr(2,Cells(1,val(range("Cellname").Column)).Address,"$")-2)
So I'm late to the party here, but I want to contribute another answer that no one else has addressed yet that doesn't involve arrays. You can do it with simple string manipulation.
Function ColLetter(Col_Index As Long) As String
Dim ColumnLetter As String
'Prevent errors; if you get back a number when expecting a letter,
' you know you did something wrong.
If Col_Index <= 0 Or Col_Index >= 16384 Then
ColLetter = 0
Exit Function
End If
ColumnLetter = ThisWorkbook.Sheets(1).Cells(1, Col_Index).Address 'Address in $A$1 format
ColumnLetter = Mid(ColumnLetter, 2, InStr(2, ColumnLetter, "$") - 2) 'Extracts just the letter
ColLetter = ColumnLetter
End Sub
After you have the input in the format $A$1, use the Mid function, start at position 2 to account for the first $, then you find where the second $ appears in the string using InStr, and then subtract 2 off to account for that starting position.
This gives you the benefit of being adaptable for the whole range of possible columns. Therefore, ColLetter(1) gives back "A", and ColLetter(16384) gives back "XFD", which is the last possible column for my Excel version.
Easy way to get the column name
Sub column()
cell=cells(1,1)
column = Replace(cell.Address(False, False), cell.Row, "")
msgbox column
End Sub
I hope it helps =)
The solution from brettdj works fantastically, but if you are coming across this as a potential solution for the same reason I was, I thought that I would offer my alternative solution.
The problem I was having was scrolling to a specific column based on the output of a MATCH() function. Instead of converting the column number to its column letter parallel, I chose to temporarily toggle the reference style from A1 to R1C1. This way I could just scroll to the column number without having to muck with a VBA function. To easily toggle between the two reference styles, you can use this VBA code:
Sub toggle_reference_style()
If Application.ReferenceStyle = xlR1C1 Then
Application.ReferenceStyle = xlA1
Else
Application.ReferenceStyle = xlR1C1
End If
End Sub
Furthering on brettdj answer, here is to make the input of column number optional. If the column number input is omitted, the function returns the column letter of the cell that calls to the function. I know this can also be achieved using merely ColumnLetter(COLUMN()), but i thought it'd be nice if it can cleverly understand so.
Public Function ColumnLetter(Optional ColumnNumber As Long = 0) As String
If ColumnNumber = 0 Then
ColumnLetter = Split(Application.Caller.Address(True, False, xlA1), "$")(0)
Else
ColumnLetter = Split(Cells(1, ColumnNumber).Address(True, False, xlA1), "$")(0)
End If
End Function
The trade off of this function is that it would be very very slightly slower than brettdj's answer because of the IF test. But this could be felt if the function is repeatedly used for very large amount of times.
Here is a late answer, just for simplistic approach using Int() and If in case of 1-3 character columns:
Function outColLetterFromNumber(i As Integer) As String
If i < 27 Then 'one-letter
col = Chr(64 + i)
ElseIf i < 677 Then 'two-letter
col = Chr(64 + Int(i / 26)) & Chr(64 + i - (Int(i / 26) * 26))
Else 'three-letter
col = Chr(64 + Int(i / 676)) & Chr(64 + Int(i - Int(i / 676) * 676) / 26)) & Chr(64 + i - (Int(i - Int(i / 676) * 676) / 26) * 26))
End If
outColLetterFromNumber = col
End Function
Function fColLetter(iCol As Integer) As String
On Error GoTo errLabel
fColLetter = Split(Columns(lngCol).Address(, False), ":")(1)
Exit Function
errLabel:
fColLetter = "%ERR%"
End Function
Here, a simple function in Pascal (Delphi).
function GetColLetterFromNum(Sheet : Variant; Col : Integer) : String;
begin
Result := Sheet.Columns[Col].Address; // from Col=100 --> '$CV:$CV'
Result := Copy(Result, 2, Pos(':', Result) - 2);
end;
This formula will give the column based on a range (i.e., A1), where range is a single cell. If a multi-cell range is given it will return the top-left cell. Note, both cell references must be the same:
MID(CELL("address",A1),2,SEARCH("$",CELL("address",A1),2)-2)
How it works:
CELL("property","range") returns a specific value of the range depending on the property used. In this case the cell address.
The address property returns a value $[col]$[row], i.e. A1 -> $A$1.
The MID function parses out the column value between the $ symbols.
Sub GiveAddress()
Dim Chara As String
Chara = ""
Dim Num As Integer
Dim ColNum As Long
ColNum = InputBox("Input the column number")
Do
If ColNum < 27 Then
Chara = Chr(ColNum + 64) & Chara
Exit Do
Else
Num = ColNum / 26
If (Num * 26) > ColNum Then Num = Num - 1
If (Num * 26) = ColNum Then Num = ((ColNum - 1) / 26) - 1
Chara = Chr((ColNum - (26 * Num)) + 64) & Chara
ColNum = Num
End If
Loop
MsgBox "Address is '" & Chara & "'."
End Sub
Column letter from column number can be extracted using formula by following steps
1. Calculate the column address using ADDRESS formula
2. Extract the column letter using MID and FIND function
Example:
1. ADDRESS(1000,1000,1)
results $ALL$1000
2. =MID(F15,2,FIND("$",F15,2)-2)
results ALL asuming F15 contains result of step 1
In one go we can write
MID(ADDRESS(1000,1000,1),2,FIND("$",ADDRESS(1000,1000,1),2)-2)
this is only for REFEDIT ... generaly use uphere code
shortly version... easy to be read and understood /
it use poz of $
Private Sub RefEdit1_Change()
Me.Label1.Caption = NOtoLETTER(RefEdit1.Value) ' you may assign to a variable var=....'
End Sub
Function NOtoLETTER(REFedit)
Dim First As Long, Second As Long
First = InStr(REFedit, "$") 'first poz of $
Second = InStr(First + 1, REFedit, "$") 'second poz of $
NOtoLETTER = Mid(REFedit, First + 1, Second - First - 1) 'extract COLUMN LETTER
End Function
Cap A is 65 so:
MsgBox Chr(ActiveCell.Column + 64)
Found in: http://www.vbaexpress.com/forum/showthread.php?6103-Solved-get-column-letter
what about just converting to the ascii number and using Chr() to convert back to a letter?
col_letter = Chr(Selection.Column + 96)
Here's another way:
{
Sub find_test2()
alpha_col = "A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,W,Z"
MsgBox Split(alpha_col, ",")(ActiveCell.Column - 1)
End Sub
}