Comments on: Column Numbers To Letters http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/ Daily posts of Excel tips…and other stuff Mon, 06 Feb 2012 18:39:56 +0000 hourly 1 http://wordpress.org/?v=3.2.1 By: Bob Lorenz http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/#comment-63690 Bob Lorenz Mon, 16 May 2011 22:16:00 +0000 http://www.dailydoseofexcel.com/?p=567#comment-63690 <p>How about this one? I think it will work for any positive integer.</p> <div style="overflow: auto; white-space: nowrap;" class="codecolorer-container vb default"><div style="white-space: nowrap;" class="vb codecolorer"><span class="kw1">Function</span> ColumnNumberToLetter(<span class="kw1">ByVal</span> intRowNum <span class="kw1">As</span> <span class="kw1">Integer</span>) <span class="kw1">As</span> <span class="kw1">String</span><br> <span class="kw1">Dim</span> strTemp <span class="kw1">As</span> <span class="kw1">String</span><br> <br> <span class="kw1">Do</span> <span class="kw1">While</span> intRowNum > 0<br>     strTemp = Chr$((((intRowNum - 1) <span class="kw1">Mod</span> 26) + 1) + 64) & strTemp<br>     intRowNum = Int((intRowNum - 1) / 26)<br> <span class="kw1">Loop</span><br> ColumnNumberToLetter = strTemp<br> <span class="kw1">End</span> <span class="kw1">Function</span></div></div> How about this one? I think it will work for any positive integer.

Function ColumnNumberToLetter(ByVal intRowNum As Integer) As String
Dim strTemp As String

Do While intRowNum > 0
    strTemp = Chr$((((intRowNum – 1) Mod 26) + 1) + 64) & strTemp
    intRowNum = Int((intRowNum – 1) / 26)
Loop
ColumnNumberToLetter = strTemp
End Function

]]> By: Rick Rothstein (MVP - Excel) http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/#comment-52451 Rick Rothstein (MVP - Excel) Mon, 11 Oct 2010 20:22:00 +0000 http://www.dailydoseofexcel.com/?p=567#comment-52451 <p>@Jorgen,</p> <p>This is from the original blog article... "The challenge before you is to write a function that converts a column number to its letter equivalent assuming columns go to ZZZZ. That's about 450,000 columns - maybe more than I need." So, no, for that question, your code line would not work.</p> @Jorgen,

This is from the original blog article… “The challenge before you is to write a function that converts a column number to its letter equivalent assuming columns go to ZZZZ. That’s about 450,000 columns – maybe more than I need.” So, no, for that question, your code line would not work.

]]> By: Jorgen http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/#comment-52435 Jorgen Mon, 11 Oct 2010 16:12:00 +0000 http://www.dailydoseofexcel.com/?p=567#comment-52435 <p>Wouldn't this work?</p> <div style="overflow: auto; white-space: nowrap;" class="codecolorer-container vb default"><div style="white-space: nowrap;" class="vb codecolorer"> MsgBox Split(Selection.Address(1, 0), <span class="st0">"$"</span>)(0)</div></div> <p>//Jorgen</p> Wouldn’t this work?

 MsgBox Split(Selection.Address(1, 0), “$”)(0)

//Jorgen

]]> By: Rick Rothstein (MVP - Excel) http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/#comment-51177 Rick Rothstein (MVP - Excel) Fri, 24 Sep 2010 07:30:00 +0000 http://www.dailydoseofexcel.com/?p=567#comment-51177 <p>General Case Solution</p> <p>The following macro with accurately return the column number for a column letter input of up to 20 letters! So put the following into the indicated cells...</p> <p>A1: ZZZZZZZZZZZZZZZZZZZZ</p> <p>A2 =ColLetToNum(A1)</p> <p>and cell A2 will display the column number of 20725274851017785518433805270. Of course, this value had to be returned to the cell as a text string. Anyway, here is the function...</p> <div style="overflow: auto; white-space: nowrap;" class="codecolorer-container vb default"><div style="white-space: nowrap;" class="vb codecolorer"><span class="kw1">Function</span> ColLetToNum(<span class="kw1">ByVal</span> ColumnLetter <span class="kw1">As</span> <span class="kw1">String</span>) <span class="kw1">As</span> <span class="kw1">Variant</span><br>   Power = 1<br>   NoChars = <span class="kw1">String</span>(Len(ColumnLetter), <span class="st0">"@"</span>)<br>   L = Split(StrConv(Right(NoChars & UCase(ColumnLetter), Len(NoChars)), vbUnicode), Chr(0))<br>   <span class="kw1">For</span> X = 0 <span class="kw1">To</span> Len(NoChars) - 1<br>     ColLetToNum = ColLetToNum + Power * (Asc(L(Len(NoChars) - 1 - X)) - 64)<br>     Power = Power * <span class="kw1">CDec</span>(26)<br>   <span class="kw1">Next</span><br>   ColLetToNum = <span class="kw1">CStr</span>(ColLetToNum)<br> <span class="kw1">End</span> <span class="kw1">Function</span></div></div> General Case Solution

The following macro with accurately return the column number for a column letter input of up to 20 letters! So put the following into the indicated cells…

A1: ZZZZZZZZZZZZZZZZZZZZ

A2 =ColLetToNum(A1)

and cell A2 will display the column number of 20725274851017785518433805270. Of course, this value had to be returned to the cell as a text string. Anyway, here is the function…

Function ColLetToNum(ByVal ColumnLetter As String) As Variant
  Power = 1
  NoChars = String(Len(ColumnLetter), “@”)
  L = Split(StrConv(Right(NoChars & UCase(ColumnLetter), Len(NoChars)), vbUnicode), Chr(0))
  For X = 0 To Len(NoChars) – 1
    ColLetToNum = ColLetToNum + Power * (Asc(L(Len(NoChars) – 1 – X)) – 64)
    Power = Power * CDec(26)
  Next
  ColLetToNum = CStr(ColLetToNum)
End Function
]]> By: Rick Rothstein (MVP - Excel) http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/#comment-51170 Rick Rothstein (MVP - Excel) Fri, 24 Sep 2010 06:54:00 +0000 http://www.dailydoseofexcel.com/?p=567#comment-51170 <p>Damn! Damn! Damn! A couple of bad constants. Here is the correct letters (A to ZZZZ) to column number function...</p> <p>Function ColLetToNum(ByVal ColumnLetter As String) As Double<br> L = Split(StrConv(Right("@@@@" & UCase(ColumnLetter), 4), vbUnicode), Chr(0))<br> ColLetToNum = 17576# * Asc(L(0)) + 676# * Asc(L(1)) + 26# * Asc(L(2)) + Asc(L(3)) - 1169856#<br> End Function</p> Damn! Damn! Damn! A couple of bad constants. Here is the correct letters (A to ZZZZ) to column number function…

Function ColLetToNum(ByVal ColumnLetter As String) As Double
L = Split(StrConv(Right(“@@@@” & UCase(ColumnLetter), 4), vbUnicode), Chr(0))
ColLetToNum = 17576# * Asc(L(0)) + 676# * Asc(L(1)) + 26# * Asc(L(2)) + Asc(L(3)) – 1169856#
End Function

]]> By: Rick Rothstein (MVP - Excel) http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/#comment-51169 Rick Rothstein (MVP - Excel) Fri, 24 Sep 2010 06:40:00 +0000 http://www.dailydoseofexcel.com/?p=567#comment-51169 <p>Okay, for the original BLOG challenge to convert column letters from A up to ZZZZ to their numerical column number, this function should do that...</p> <div style="overflow: auto; white-space: nowrap;" class="codecolorer-container vb default"><div style="white-space: nowrap;" class="vb codecolorer"><span class="kw1">Function</span> ColLetToNum(<span class="kw1">ByVal</span> ColumnLetter <span class="kw1">As</span> <span class="kw1">String</span>) <span class="kw1">As</span> <span class="kw1">Double</span><br>   L = Split(StrConv(Right(<span class="st0">"@@@@"</span> & UCase(ColumnLetter), 4), vbUnicode), Chr(0))<br>   ColLetToNum = 17576# * Asc(L(0)) + 626# * Asc(L(1)) + 26# * Asc(L(2)) + Asc(L(3)) - 1166656#<br> <span class="kw1">End</span> <span class="kw1">Function</span></div></div> Okay, for the original BLOG challenge to convert column letters from A up to ZZZZ to their numerical column number, this function should do that…

Function ColLetToNum(ByVal ColumnLetter As String) As Double
  L = Split(StrConv(Right(“@@@@” & UCase(ColumnLetter), 4), vbUnicode), Chr(0))
  ColLetToNum = 17576# * Asc(L(0)) + 626# * Asc(L(1)) + 26# * Asc(L(2)) + Asc(L(3)) – 1166656#
End Function
]]> By: Rick Rothstein (MVP - Excel) http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/#comment-51157 Rick Rothstein (MVP - Excel) Fri, 24 Sep 2010 04:35:00 +0000 http://www.dailydoseofexcel.com/?p=567#comment-51157 <p>Never mind... I just re-read the challenge and my code doesn't really address it. Sorry.</p> Never mind… I just re-read the challenge and my code doesn’t really address it. Sorry.

]]> By: Rick Rothstein (MVP - Excel) http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/#comment-51155 Rick Rothstein (MVP - Excel) Fri, 24 Sep 2010 04:31:00 +0000 http://www.dailydoseofexcel.com/?p=567#comment-51155 <p>I guess even simpler would be this...</p> <p>Function ColLetToNum(ByVal ColumnLetter As String) As Double<br> ColLetToNum = Columns(ColumnLetter).Column<br> End Function</p> I guess even simpler would be this…

Function ColLetToNum(ByVal ColumnLetter As String) As Double
ColLetToNum = Columns(ColumnLetter).Column
End Function

]]> By: Rick Rothstein (MVP - Excel) http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/#comment-51154 Rick Rothstein (MVP - Excel) Fri, 24 Sep 2010 04:30:00 +0000 http://www.dailydoseofexcel.com/?p=567#comment-51154 <p>I know this is an old thread; but, unless I missed it, it doesn't look like anyone ever posted this extremely simple function...</p> <p>Function ColLetToNum(ByVal ColumnLetter As String) As Double<br> ColLetToNum = Range(ColumnLetter & "1?).Column<br> End Function</p> I know this is an old thread; but, unless I missed it, it doesn’t look like anyone ever posted this extremely simple function…

Function ColLetToNum(ByVal ColumnLetter As String) As Double
ColLetToNum = Range(ColumnLetter & “1?).Column
End Function

]]> By: Sean http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/#comment-49549 Sean Mon, 30 Aug 2010 07:46:00 +0000 http://www.dailydoseofexcel.com/?p=567#comment-49549 <p>@ incognito. You would have to define a user defined variable type that will be bigger than double. This will allow for enough accuracy to use this solution from above</p> <p>Function ColNumToLet(ByVal ColumnNumber As Double) As String<br> If ColumnNumber 3.267215265265262E+15 Then<br> ColNumToLet = "Error"<br> Else<br> ColNumToLet = ""<br> If ColumnNumber 26<br> If myMod(ColumnNumber, 26) = 0 Then<br> ColNumToLet = "Z" & ColNumToLet<br> ColumnNumber = ColumnNumber - 1<br> Else<br> ColNumToLet = Chr(myMod(ColumnNumber, 26) + 64) & ColNumToLet<br> End If<br> ColumnNumber = WorksheetFunction.RoundDown(ColumnNumber / 26, 0)<br> Loop<br> If Not ColumnNumber = 0 Then ColNumToLet = Chr(ColumnNumber + 64) & ColNumToLet<br> End If<br> End If<br> End Function</p> <p>Function myMod(Numerator As Double, Denominator As Double) As Double<br> myMod = Numerator - (WorksheetFunction.RoundDown(Numerator / Denominator, 0) * Denominator)<br> End Function</p> @ incognito. You would have to define a user defined variable type that will be bigger than double. This will allow for enough accuracy to use this solution from above

Function ColNumToLet(ByVal ColumnNumber As Double) As String
If ColumnNumber 3.267215265265262E+15 Then
ColNumToLet = “Error”
Else
ColNumToLet = “”
If ColumnNumber 26
If myMod(ColumnNumber, 26) = 0 Then
ColNumToLet = “Z” & ColNumToLet
ColumnNumber = ColumnNumber – 1
Else
ColNumToLet = Chr(myMod(ColumnNumber, 26) + 64) & ColNumToLet
End If
ColumnNumber = WorksheetFunction.RoundDown(ColumnNumber / 26, 0)
Loop
If Not ColumnNumber = 0 Then ColNumToLet = Chr(ColumnNumber + 64) & ColNumToLet
End If
End If
End Function

Function myMod(Numerator As Double, Denominator As Double) As Double
myMod = Numerator – (WorksheetFunction.RoundDown(Numerator / Denominator, 0) * Denominator)
End Function

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