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
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
Function outColLetterFromNumber(iCol as Integer) as String
sAddr = Cells(1, iCol).Address
aSplit = Split(sAddr, "$")
outColLetterFromNumber = aSplit(1)
End Function
Function ConvertToLetter(iCol As Integer) As String
Dim iAlpha As Integer
Dim iRemainder As Integer
iAlpha = Int(iCol / 27)
iRemainder = iCol - (iAlpha * 26)
If iAlpha > 0 Then
ConvertToLetter = Chr(iAlpha + 64)
End If
If iRemainder > 0 Then
ConvertToLetter = ConvertToLetter & Chr(iRemainder + 64)
End If
End Function
Public Function ColumnLetter(Column As Integer) As String
If Column < 1 Then Exit Function
ColumnLetter = ColumnLetter(Int((Column - 1) / 26)) & Chr(((Column - 1) Mod 26) + Asc("A"))
End Function
我遇到的问题是基于 MATCH ()函数的输出滚动到特定的列。我没有将列号并行转换为列号字母,而是选择临时将引用样式从 A1切换到 R1C1。这样我就可以直接滚动到列号,而不必使用 VBA 函数。要在两个引用样式之间轻松切换,可以使用以下 VBA 代码:
Sub toggle_reference_style()
If Application.ReferenceStyle = xlR1C1 Then
Application.ReferenceStyle = xlA1
Else
Application.ReferenceStyle = xlR1C1
End If
End Sub
Sub GiveAddress()
Dim Chara As String
Chara = ""
Dim Num As Integer
Dim ColNum As Long
ColNum = InputBox("Input the column number")
Do
If ColNum < 27 Then
Chara = Chr(ColNum + 64) & Chara
Exit Do
Else
Num = ColNum / 26
If (Num * 26) > ColNum Then Num = Num - 1
If (Num * 26) = ColNum Then Num = ((ColNum - 1) / 26) - 1
Chara = Chr((ColNum - (26 * Num)) + 64) & Chara
ColNum = Num
End If
Loop
MsgBox "Address is '" & Chara & "'."
End Sub
Public Function ColumnLetter(Optional ColumnNumber As Long = 0) As String
If ColumnNumber = 0 Then
ColumnLetter = Split(Application.Caller.Address(True, False, xlA1), "$")(0)
Else
ColumnLetter = Split(Cells(1, ColumnNumber).Address(True, False, xlA1), "$")(0)
End If
End Function
Private Sub RefEdit1_Change()
Me.Label1.Caption = NOtoLETTER(RefEdit1.Value) ' you may assign to a variable var=....'
End Sub
Function NOtoLETTER(REFedit)
Dim First As Long, Second As Long
First = InStr(REFedit, "$") 'first poz of $
Second = InStr(First + 1, REFedit, "$") 'second poz of $
NOtoLETTER = Mid(REFedit, First + 1, Second - First - 1) 'extract COLUMN LETTER
End Function
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
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;
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