number=number*26+Asc(Mid(temp,2,1))-Asc("A")+1'Third letter IfLen(letter)>=3Then number=number*26+Asc(Mid(temp,3,1))-Asc("A")+1IfLen(letter)>=4Then LetterToNum="Letter more than three"Exit Function End If End If
In Excel sheet, type ‘=Col_Letter_To_Number(“AA”)’ & press Enter key. The function will return 27 as its column number. Same way, this function can also be used in any vba macro function.
a column number to its corresponding column letter in Excel with VBA code using 3 methods: converting a specific column number to its column letter, converting a user-input column number to its letter, and creating a UDF (User-Defined Function) to convert a column number to its letter.Method...
点击后代码会运行,程序中我给出的是”AG”列,要求求出这列的列数值,结果如下: 代码见程序文件:VBA_ConvertColumnNumberToLetter.xlsm
Use this VBA function to convert a column number to a letter. This is useful if you need to build a range and you want to do so with the A1 notation.
Get vba code & Excel formula to get Column letter from column number, in here. For example, If you pass ‘1’ to the function in this code, it will return as “A”. Similarly “B” for 2, “C” for 3 & so on. Lets first see the Excel vba code & then its equivalent formula...
rowNumber = rowNumber \ 26 Loop RowToLetter = letter End Function 此函数接受一个行号作为输入，返回对应的字母表示。例如，当输入1时，返回"A"；当输入26时，返回"Z"；当输入27时，返回"AA"。通过这种方式，你可以在VBA中灵活地使用字母来表示单元格的行号，为你的代码增加更多的可读性和灵活...
For i = 1 To myRng.Cells.Count myStr = myRng.Cells(i) letter = UCase(myStr) outputRng.Cells(i) = outputRng.Cells(i) & letter Next i End Sub Next, the code uses a For Next loop to iterate over each cell in myRng. For each cell, it converts the value to uppercase using ...
Errorhandler ColLetter = Left(Cells(1, ColNumber).Address(0, 0), 1 - (ColNumber > 26)) Exit Function Errorhandler: MsgBox "Error encountered, please re-enter " End Function '### '2.函数作用:返回列标2 '### Function ColIntToLetter(intCol As Integer) As String '' Dim intPart As ...
Function ColIntToLetter(intCol As Integer) As String '' Dim intPart As Integer Dim intRemainder As Integer If intCol > 255 Or intCol <= 0 Then MsgBox ("The Wrong Column Number: " & CStr(intCol)) Exit Function End If intPart = intCol \ 26 intRemainder = intCol Mod 26 If intPart...