lets say
In a range A1
8cm book + 13cm ruler + 0.34cm paper
then how i sum up 8 + 13 + 0.34 inside the cell then = 21.34cm
then put the sumup in A2
——————-updates question———————
Sorry, I must make the question clearer!
8cm book + 13cm ruler + ruler version#2 +0.34cm paper
And I was wondering is it still working for the code from all master?
this problem bothers me 1 week.
hope can get the anwser!
thanks!!
Try this simple regex based UDF.
Option Explicit
Function sumNums(str As String) As Double
Dim n As Long
Static rgx As Object, cmat As Object
If rgx Is Nothing Then
Set rgx = CreateObject("VBScript.RegExp")
End If
With rgx
.Global = True
.MultiLine = True
.Pattern = "(\-?\d*\.?\d+)"
If .test(str) Then
Set cmat = .Execute(str)
For n = 0 To cmat.Count - 1
sumNums = sumNums + CDbl(cmat.Item(n))
Next n
End If
End With
End Function
Here is a User Defined Function, just using String indexing:
Function SumNumbers(s As String) As Currency
Dim i As Long
Dim j As Long
For i = 1 To Len(s)
If InStr("-0123456789.", Mid(s, i, 1)) Then
j = i
Do
i = i + 1
Loop While InStr("0123456789.", Mid(s, i, 1))
SumNumbers = SumNumbers + CCur(Mid(s, j, i - j))
End If
Next
End Function
Put it in the Visual Basic Editor (Shift+F11) in a new Module (right-click project, choose "Insert..." and "Module").
Use the formula in your sheet, for example in cell B1:
=SumNumbers(A1)
NB: I prefer to use Currency and CCur to avoid floating point errors. It expects numbers to have at the most 4 decimals, and supports up to about 14 digits before the decimal separator.
another solution is:
Function cal_cm(cl As Range) As Double
Dim i%, testVal$, result$
result = "="
For i = 1 To Len(cl)
testVal = cl.Characters(i, 1).Text
If "0123456789+." Like "*" & testVal & "*" Then
result = result & testVal
End If
Next i
cal_cm = Evaluate(result)
End Function
test
Here is a way to do it without VBA, but with some limitations:
It expects every number to be followed by "cm"
It expects every number to have a space in front of it (except for the first number which may be at the very start of the string)
You need to know what the maximum count of numbers is in the input
Assuming the input string is in cell A1, you could proceed as follows:
Leave B1 blank
In C1 put the formula =IFERROR(MID($A1,1,FIND("cm",$A1,LEN(B1)+2)-1), "0")
Copy C1 to D1, E1, and F1
In G1 put the formula =VALUE(IFERROR(RIGHT(C1,LEN(C1)-FIND("#",SUBSTITUTE(C1," ","#",LEN(C1)-LEN(SUBSTITUTE(C1," ",""))),1)),C1))
Copy G1 to H1, I1, and J1
In K1 put the formula =SUM(G1:J1)
You would get this result. The color coding identifies each of the groups that are copies:
K1 will have the desired output.
Note that F1 (and J1) contain 0. This would get the value of a fourth value in the input if it existed.
You'll understand how to make room for more cells if your input could have 5 or more numerical values embedded in it.
Related
Quick question, if I want to delete everything after the second occurrence of a number:
i.e -
I have:
1105 Bracket Ave. Suite 531 Touche
5201 Used St. 1351 Bored Today
I want:
1105 Bracket Ave. Suite 531
5201 Used St. 1351
is there a simple formula or VBA I would use for this?
Here is a UDF using VBA's regular expression engine to remove all after the second integer.
Option Explicit
Function FirstTwoNumbers(S As String) As String
Dim RE As Object
Set RE = CreateObject("vbscript.regexp")
With RE
.Pattern = "(\d+\D+\d+).*"
FirstTwoNumbers = .Replace(S, "$1")
End With
End Function
If there is only a single integer, it will return the entire string.
If the numbers might be decimal numbers, will need to modify .Pattern
And here is another UDF using only native VBA methods:
Function FirstTwo(S As String) As String
Dim V
Dim tS As String
Dim I As Long, numNumbers As Long
V = Split(S)
Do Until numNumbers = 2
tS = tS & Space(1) & V(I)
I = I + 1
If IsNumeric(V(I - 1)) Then numNumbers = numNumbers + 1
Loop
FirstTwo = Mid(tS, 2)
End Function
and finally, a formula with no particular assumptions:
=LEFT(A1,FIND(CHAR(1),SUBSTITUTE(A1," ",CHAR(1),LOOKUP(2,1/ISNUMBER(-TRIM(MID(SUBSTITUTE(A1," ",REPT(" ",99)),seq_99,99))),seq))))
seq and seq99 are Named Formulas Formula ► Define Name
seq Refers to: =ROW(INDEX($1:$255,1,1):INDEX($1:$255,255,1))
seq_99 Refers to: =IF(ROW(INDEX($1:$255,1,1):INDEX($1:$255,255,1))=1,1,(ROW(INDEX($1:$255,1,1):INDEX($1:$255,255,1))-1)*99)
This solution is with these assumptions:-
First occurrence of a number will not have a length > 10
There will atleast a distance of 10 or 10 alphabets including spaces between first and second number
There will always be a 'space' existing after second number
There will always be a second number present in the string
Try this:-
=TRIM(MID(A1,1,FIND(" ",A1,MIN(FIND({0,1,2,3,4,5,6,7,8,9},A1&"0123456789",MIN(FIND({0,1,2,3,4,5,6,7,8,9},A1&"0123456789"))+10)))))
Here is a VBA approach, amend range to suit. It puts the answer in the adjacent column
Sub x()
Dim oMatches As Object, r As Range
With CreateObject("VBScript.RegExp")
.Global = True
.Pattern = "\d+"
For Each r In Range("A1:A5")
If .Test(r) Then
Set oMatches = .Execute(r)
If oMatches.Count > 1 Then
r.Offset(, 1).Value = Left(r, oMatches(1).firstindex + oMatches(1).Length)
Else
r.Offset(, 1).Value = r.Value
End If
Else
r.Offset(, 1).Value = r.Value
End If
Next r
End With
End Sub
You can use the following formula,if A1 is your string,in B1 write:
=LEFT(A1,MAX(IFERROR(ISNUMBER(VALUE(MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1)))*ROW(INDIRECT("1:"&LEN(A1))),0)))
press Ctrl+Shift+Enter at the same time Array Formula
This will read the length of the string and return the Maximum place of numbers (last number in the string) and return the Left() string till this number
How do I extract only the first two letters and the 6 digit number from one cell to another? ie. Column 1 will have aa111111, bb222222, ccccc, dd12, eeee1
I only want to copy aa111111 and bb222222 in this case.
Thanks,
Alex
Try this short macro:
Sub KopyKat()
Dim N As Long, i As Long, K As Long
Dim s As String
N = Cells(Rows.Count, 1).End(xlUp).Row
K = 1
For i = 1 To N
s = Cells(i, 1).Value
If Len(s) = 8 _
And Mid(s, 1, 1) Like "[a-zA-Z]" _
And Mid(s, 2, 1) Like "[a-zA-Z]" _
And IsNumeric(Mid(s, 3)) Then
Cells(K, 2).Value = s
K = K + 1
End If
Next i
End Sub
If your strings are in A:A then in B1 (and copy down):
=IF(LEN(A1)>7,IF(AND(CODE(LOWER(MID(A1,{1,2},1)))<>CODE(UPPER(MID(A1,{1,2},1))),ISNUMBER(MID(A1,{3;4;5;6;7;8},1)*1)),LEFT(A1,8),""),"")
This is an array formula and must be confirmed with ctrl + shift + enter.
Running evaluate formula shows what happens and how it works :)
Using VBA then this will do:
Sub CopyMe()
Dim x As Variant, i As Long
For Each x In Range([A1], Cells(Rows.Count, 1).End(xlUp)).Value2
If x Like "[a-zA-Z][a-zA-Z]######" Then
i = i + 1
Cells(i, 2) = x
End If
Next
End Sub
Here is another way to do it. The formula loops through your string looking for a 6 digit number, at which point it takes the previous two characters (as long as they exist) and returns the 8 character string. Otherwise it returns an empty string.
=IFERROR(MID(A1,SUMPRODUCT(ROW($A$1:$A$100),--ISNUMBER(VALUE(MID(A1,ROW($A$1:$A$100),6))))-2,8),"")
Loops 100 times, so will work with strings up to a maximum length of 106 characters. To use the formula place your string in column A, and place this formula in cell B1 (for example) and drag down
I sometimes work with symmetric matrices in MS-Excel (both v2007 and v2003).
Is there an option to help me to copy expressions from the lower triangle to the upper one?
It should be something like copy and paste/transponse but those functions normally work only with rectangular areas.
in the added picture you can see an exemple of an expression that I have to replicate by linking the symmetric value in the superior triangle of the matrix.
To get the number in the appropriate cell, we can use OFFSET and the cell address the forms the base of the table. Note that the formula will produce a *Circular Reference` error if entered in on the diagonal. The formula will work for both sides of the diagonal - you just have to decide which one will hold the data, and which will hold the formula.
Offset takes Row and Column to decide the target. By subtracting the base cell row and column from the current position, we can invert the row and columns, and get the data.
Using your example, with the origin of the table in B2, we end up with the following formula:
=OFFSET($B$2,COLUMN()-COLUMN($B$2),ROW()-ROW($B$2))
you can copy this formula into the cells, and get the reflection. Now you have the number, you can do any calculation you require on the reflection. Using your example, this would make the formula:
=10-OFFSET($B$2,COLUMN()-COLUMN($B$2),ROW()-ROW($B$2))
Result:
Using INDEX to make it non volatile would change the formula slightly. First, we would need a reference to the entire table, not just the top cell. Second, we would need to add 1 to the row/column calculation, as it refers to the first cell as row/column 1, not an offset of 0 as the previous formula.
=INDEX($B$2:$K$11,COLUMN()-COLUMN($B$2)+1,ROW()-ROW($B$2)+1)
and your example of 10-Cell would become:
=10-INDEX($B$2:$K$11,COLUMN()-COLUMN($B$2)+1,ROW()-ROW($B$2)+1)
As one of the above answers demonstrates, this can be done by using Excel formulas. I however find this to be a very tedious procedure. Especially if this is something you need to do on a regular basis. In that case VBA could save you a lot of time.
The following code will work on a square selection and fill the rest of the matrix no matter if it is the lower- or upper part of the matrix that is pre-filled.
Option Explicit
Sub FillSymetricMatrix()
Dim i As Integer, j As Integer
Dim SelRng As Range
Dim FillArea As String
Dim FRow As Integer
Dim FCol As Integer
Set SelRng = Selection
FRow = SelRng.Rows(1).Row
FCol = SelRng.Columns(1).Column
'Returns information about which area to fill
If ActiveSheet.Cells(FRow + SelRng.Rows.Count - 1, FCol).Value <> vbNullString Then 'Lower filled
If ActiveSheet.Cells(FRow, FCol + SelRng.Columns.Count - 1).Value = vbNullString Then 'Upper empty
FillArea = "Upper"
Else
FillArea = "Error"
End If
Else
If ActiveSheet.Cells(FRow, FCol + SelRng.Columns.Count - 1).Value <> vbNullString Then 'Upper filled
FillArea = "Lower"
Else
FillArea = "Error"
End If
End If
'Determines if the selection is square
If SelRng.Rows.Count <> SelRng.Columns.Count Then FillArea = "Error"
'Fills empty area of the square (symetric) matrix
Select Case FillArea
Case Is = "Upper"
For i = 0 To SelRng.Rows.Count - 1 Step 1
For j = 0 To SelRng.Columns.Count - 1 Step 1
If i <= j Then ActiveSheet.Cells(i + FRow, j + FCol).Value = ActiveSheet.Cells(j + FRow, i + FCol).Value
Next j
Next i
Case Is = "Lower"
For i = 0 To SelRng.Rows.Count - 1 Step 1
For j = 0 To SelRng.Columns.Count - 1 Step 1
If i <= j Then ActiveSheet.Cells(j + FRow, i + FCol).Value = ActiveSheet.Cells(i + FRow, j + FCol).Value
Next j
Next i
Case Else
MsgBox "The procedure cannot be performed on the current selection!"
End Select
End Sub
I guess what you need is a function which returns the "diagonal" value of a square matrix, e.g. for any X(j,k) return X(k,j)
Try this:
Function DIAGONAL(Arg As Range, Reference As Range) As Variant
Dim MyRow As Long, MyCol As Long
If Reference.Rows.Count <> Reference.Columns.Count Then
DIAGONAL = CVErr(xlErrRef)
Else
MyRow = Arg.Row - Reference.Row + 1
MyCol = Arg.Column - Reference.Column + 1
If MyRow < 1 Or MyCol < 1 Or MyRow > Reference.Rows.Count Or MyCol > Reference.Columns.Count Then
DIAGONAL = CVErr(xlErrNA)
Else
DIAGONAL = Reference(MyCol, MyRow)
End If
End If
End Function
once you entered this function in VBA, you can use it inside or outside your square matrix ... you just need to ensure that your argument (parameter: Arg) is within the matrix (parameter: Reference) ... or you get an #N/A error. Or you get a #REF error if the matrix isn't square.
So in your example you would enter into B4: =10-DIAGONAL(B4,$B$2:$K$11) and copy this throughout the lower triangle.
You can even transpose a complete matrix ... in your screen shot, move to cell B13, enter =DIAGONAL(B2,$B$2:$K$11) and copy 9x down & right
No buttons, no need to explicitely start a Sub ... any size of n x n matrix, handles strings and numbers, ...
Here is an example with VBA. Start with an un-filled table and a button.
Then make the button run the code:
Option Explicit
Private Sub symmButton_Click()
MakeSymmetric Range("B2")
End Sub
Public Sub MakeSymmetric(ByRef r As Range)
Dim M As Long
M = CountCols(r)
Dim vals() As Variant
vals = r.Resize(M, M).Value2
Dim i As Long, j As Long
For i = 2 To M
For j = 1 To i - 1
vals(i, j) = vals(j, i)
Next j
Next i
r.Resize(M, M).Value2 = vals
End Sub
Public Function CountCols(ByRef r As Range) As Long
If IsEmpty(r) Then
CountCols = 0
ElseIf IsEmpty(r.Offset(0, 1)) Then
CountCols = 1
Else
CountCols = r.Worksheet.Range(r, r.End(xlToRight)).Columns.Count
End If
End Function
and finally observe the results
Similar to Sean's solution, I would also use formulas. In order to get the transposed value, use this formula:
=INDEX($B$2:$G$7,COLUMN()-COLUMN($B$2)+1,ROW()-ROW($B$2)+1)
If you want to do a more complex operation (e.g. =10-[transposedValue]), I'd recommend you use a named range: Insert a new name, e.g. TransposedValuein the Name Manager. Instead of a cell link, provide the above formula. Now you can literally write the following formula in your matrix:
=10-TransposedValue
I have this way. As you said copy paste transpose work on rectangular range. And your problem is that you have a triangular range.
You will love this....
1). Select the square range containing your upper triangular matrix and Copy.
2). Select a cell in an empty place and do the following two steps
a.) Paste Special - Values
b.) Paste Special - Values - Transpose - Skip Blanks
And you have got your symmetric matrix :-)
Anil.
Mixing together Ja72's fill code with SeanC c's Excel function code, I think I can make a generic matrix template that is properly prefilled with the dynamic Excel formula. So dynamic, and can be reused without any copy and paste.
Public Sub MakeSymmetric(ByRef r As Range)
Dim M As Long
M = 300
' Was CountCols(r), but I just limited to 300 columns for now
Dim vals() As Variant
vals = r.Resize(M, M).Value2
Dim i As Long, j As Long
For i = 2 To M
For j = 1 To i - 1
vals(j, i) = "=OFFSET($B$2,COLUMN()-COLUMN($B$2),ROW()-ROW($B$2))"
Next j
'Make diagonal down the middle show ---
vals(j, i) = "---"
Next i
vals(1, 1) = "---"
r.Resize(M, M).Value2 = vals
End Sub
Sub FillSymmetric()
MakeSymmetric Range("B2")
End Sub
I don't really know any VB though, so I haven't quite figured out how to fill the header yet. I don't know Stackoverflow yet either, but I will try to add a picture.
Original List to Matrixize
Dynamically transposing values typed in SouthWest half to NorthEast half
Short answer: INDIRECT(ADDRESS(COLUMN(D2), ROW(D2)))
Explnation: you may remember we use coordinates with numbers to represent a location in Cartesian Coordinates System. So, it's easy to get a diagonal symmetric value e.g. just change (2, 3) to (3, 2).
But in Excel, we need a wordaround if we want to do so. Because, address is marked by a combination of a letter and a digit, say B2. You can't just change B2 to 2B.
Luckily, we can still use numbers to represent a cell by leveraging the power of COW() and COLUMN().
In the image below, C2 and B3 are symmetrical. This shows how to put the value of C2 to B3.
Making the formula from C.W. more generic (similar to Peter Albert), this will help when your matrix is not starting at A1 but e.g. in C10:
=INDIRECT(ADDRESS(COLUMN(C11)-COLUMN($C$10)+1,ROW(C11)-ROW($C$10)+1))
So, subtract the origin row/column and add 1.
I'm trying to generate words in Column B from a list of given words in Column A.
Right now my code in Excel VBA does this:
Function GetText()
Dim GivenWords
GivenWords = Sheets(1).Range(Sheets(1).[a1], Sheets(1).[a20])
GetText = A(Application.RandBetween(1, UBound(A)), 1)
End Function
This generates a word from the list I have provided in A1:A20, but I don't want any duplicates.
GetText() will be run 15 times in Column B from B1:B15.
How can I check for any duplicates in Column B, or more efficiently, remove the words temporarily from the list once it has been used?
For example,
Select Range A1:A20
Select one value randomly (e.g A5)
A5 is in Column B1
Select Range A1:A4 and A6:A20
Select one value randomly (e.g A7)
A7 is in Column B2
Repeat, etc.
This was trickier than I thought. The formula should be used as a vertical array eg. select the cells where you want the output, press f2 type =gettext(A1:A20) and press ctrl+shift+enter
This means that you can select where your input words are in the worksheet, and the output can be upto as long as that list of inputs, at which point you'll start getting #N/A errors.
Function GetText(GivenWords as range)
Dim item As Variant
Dim list As New Collection
Dim Aoutput() As Variant
Dim tempIndex As Integer
Dim x As Integer
ReDim Aoutput(GivenWords.Count - 1) As Variant
For Each item In GivenWords
list.Add (item.Value)
Next
For x = 0 To GivenWords.Count - 1
tempIndex = Int(Rnd() * list.Count + 1)
Aoutput(x) = list(tempIndex)
list.Remove tempIndex
Next
GetText = Application.WorksheetFunction.Transpose(Aoutput())
End Function
Here's how I would do it, using 2 extra columns, and no VBA code...
A B C D
List of words Rand Rank 15 Words
Apple =RAND() =RANK(B2,$B$2:$B$21) =INDEX($A$2:$A$21,MATCH(ROW()-1,$C$2:$C$21,0))
copy B2 and C2 down as far as the list, and drag D down for however many words you want.
Copy the word list somewhere, as every time you change something on the sheet (or recalculate), you will get a new list of words
Using VBA:
Sub GetWords()
Dim Words
Dim Used(20) As Boolean
Dim NumChosen As Integer
Dim RandWord As Integer
Words = [A1:A20]
NumChosen = 0
While NumChosen < 15
RandWord = Int(Rnd * 20) + 1
If Not Used(RandWord) Then
NumChosen = NumChosen + 1
Used(RandWord) = True
Cells(NumChosen, 2) = Words(RandWord, 1)
End If
Wend
End Sub
Here is the code. I am deleting the cell after using it. Please make a backup of your data before using this as it will delete the cell contents (it will not save automatically...but just in case). You need to run the 'main' sub to get the output.
Sub main()
Dim i As Integer
'as you have put 15 in your question, i am using 15 here. Change it as per your need.
For i = 15 To 1 Step -1
'putting the value of the function in column b (upwards)
Sheets(1).Cells(i, 2).Value = GetText(i)
Next
End Sub
Function GetText(noofrows As Integer)
'if noofrows is 1, the rand function wont work
If noofrows > 1 Then
Dim GivenWords
Dim rowused As Integer
GivenWords = Sheets(1).Range(Sheets(1).Range("A1"), Sheets(1).Range("A" & noofrows))
'getting the randbetween value to a variable bcause after taking the value, we can delete the cell.
rowused = (Application.RandBetween(1, UBound(GivenWords)))
GetText = Sheets(1).Range("A" & rowused)
Application.DisplayAlerts = False
'deleting the cell as we have used it and the function should not use it again
Sheets(1).Cells(rowused, 1).Delete (xlUp)
Application.DisplayAlerts = True
Else
'if noofrows is 1, there is only one value left. so we just use it.
GetText = Sheets(1).Range("A1").Value
Sheets(1).Cells(1, 1).Delete (xlUp)
End If
End Function
Hope this helps.
In a program that I'm trying to write now I take two columns of numbers and perform calculations on them. I don't know where these two columns are located until the user tells me (they input the column value in a cell in the workbook that my code is located in).
For example, if the user inputted "A" and "B" as the columns where all the information is in I can perform calculations based on those values. Likewise if they wanted to analyze another worksheet (or workbook) and the columns are in "F" and "G" they could input those. The problem is that I'm asking the user to input those two columns as well as four others (the last four are the result columns). I did this in hopes that I would be able to make this flexible, but now inflexibility is acceptable.
My question is, if I'm given a value of where some information will be (let's say "F") how can I figure out what the column will be after or before that inputted value. So if I'm only given "F" I'll be able to create a variable to hold the "G" column.
Below are examples of how the variables worked before I needed to do this new problem:
Dim first_Column As String
Dim second_Column As String
Dim third_Column As String
first_Column = Range("B2").Text
second_Column = Range("B3").Text
third_Column = Range("B4").Text
Here the cells B2 - B4 are where the user inputs the values. Generally I want to be able to not have the B3 and B4 anymore. I feel like the Offset(0,1) might be able to help somehow but so far I've been unable to implement it correctly.
Thank you,
Jesse Smothermon
Here are two functions that will help you dealing with columns > "Z". They convert the textual form of a column to a column index (as a Long value) and vice versa:
Function ColTextToInt(ByVal col As String) As Long
Dim c1 As String, c2 As String
col = UCase(col) 'Make sure we are dealing with "A", not with "a"
If Len(col) = 1 Then 'if "A" to "Z" is given, there is just one letter to decode
ColTextToInt = Asc(col) - Asc("A") + 1
ElseIf Len(col) = 2 Then
c1 = Left(col, 1) ' two letter columns: split to left and right letter
c2 = Right(col, 1)
' calculate the column indexes from both letters
ColTextToInt = (Asc(c1) - Asc("A") + 1) * 26 + (Asc(c2) - Asc("A") + 1)
Else
ColTextToInt = 0
End If
End Function
Function ColIntToText(ByVal col As Long) As String
Dim i1 As Long, i2 As Long
i1 = (col - 1) \ 26 ' col - 1 =i1*26+i2 : this calculates i1 and i2 from col
i2 = (col - 1) Mod 26
ColIntToText = Chr(Asc("A") + i2) ' if i1 is 0, this is the column from "A" to "Z"
If i1 > 0 Then 'in this case, i1 represents the first letter of the two-letter columns
ColIntToText = Chr(Asc("A") + i1 - 1) & ColIntToText ' add the first letter to the result
End If
End Function
Now your problem can be solved easily, for example
newColumn = ColIntToText(ColTextToInt(oldColumn)+1)
EDITED accordingly to the remark of mwolfe02:
Of course, if you are not interested in the column names, but just want to get a range object of a specific cell in a given row right beneath a column given by the user, this code is "overkill". In this case, a simple
Dim r as Range
Dim row as long, oldColumn as String
' ... init row and oldColumn here ...
Set r = mysheet.Range(oldColumn & row).Offset(0,1)
' now use r to manipulate the cell right to the original cell
will do it.
You were on the right track with Offset. Here is a test function that shows a couple different approaches to take with it:
Sub test()
Dim first_Column As String
Dim second_Column As String
Dim third_Column As String
Dim r As Range
first_Column = Range("B2").Text
second_Column = Range("B2").Offset(1, 0).Text
third_Column = Range("B2").Offset(2, 0).Text
Debug.Print first_Column, second_Column, third_Column
Set r = Range("B2")
first_Column = r.Text
Set r = r.Offset(1, 0)
second_Column = r.Text
Set r = r.Offset(1, 0)
third_Column = r.Text
Debug.Print first_Column, second_Column, third_Column
End Sub
UPDATE: After re-reading your question I realize you were trying to do offsets based on a user-entered column letter. #rskar's answer will shift the column letter, but it will be a lot easier to work with the column number in code. For example:
Sub test()
Dim first_Col As Integer, second_Col As Integer
first_Col = Cells(, Range("B2").Text).Column
second_Col = first_Col + 1
Cells.Columns(first_Col).Font.Bold = True
Cells.Columns(second_Col).Font.Italic = True
End Sub
There are a few syntactical problems with #rskar's answer. However, it was helpful in producing a function that grabs a column "letter", based on an input column "letter" and a desired offset to the right:
Public Function GetNextCol(TheCol As String, OffsetRight As Integer) As String
Dim TempCol1 As String
Dim TempCol2 As String
TempCol1 = Range(TheCol & "1").Address
TempCol2 = Range(TempCol1).Offset(0, OffsetRight).Address(0, 0, xlA1)
GetNextCol = Left(TempCol2, Len(TempCol2) - 1)
End Function
In light of the comments of others (and they all raised valid points), here is a much better solution to the problem, using Offset and Address:
Dim first_Column As String
Dim second_Column As String
Dim p As Integer
first_Column = Range("B2").Text
second_Column = _
Range(first_Column + ":" + first_Column).Offset(0, 1).Address(0, 0, xlA1)
p = InStr(second_Column, ":")
second_Column = Left(second_Column, p - 1)
The above should work for any valid column name, "Z" and "AA" etc. included.
Make use of the Asc() and Chr() functions in VBA, like so:
Dim first_Column As String
Dim second_Column As String
first_Column = Range("B2").Text
second_Column = Chr(Asc(first_Column) + 1)
The Asc(s) function returns the ASCII code (in integer, usually between 0 and 255) of the first character of a string "s".
The Chr(c) function returns a string containing the character which corresponds to the given code "c".
Upper case letters (A thru Z) are ASCII codes 65 thru 90. Just google ASCII for more detail.
NOTE: The above code will be fine so long as the first_Column is between "A" and "Y"; for columns "AA" etc., it will take a little more work, but Asc() and Chr() will still be the ticket to coding for that.