vba 如何获取单元格的列字母（需要使其工作通过 Z 列，例如 AA、AB）
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How can i get the Column letters of a cell (need to make it work past the Z column such as AA,AB)
提问by Viladimir
Possible Duplicate:
VBA function to convert column number to letter?
可能的重复：
VBA 函数将列号转换为字母？
I want to have the column letters past the Z column, there are many techniques for getting it before Z column but after Z, nothing seems to work Is there a way to do so?
我想让列字母越过 Z 列，有很多技术可以在 Z 列之前获取它，但在 Z 之后，似乎没有任何效果 有没有办法做到这一点？
回答by Dick Kusleika
Another method:
另一种方法：
Public Function ColumnLettersFromRange(rInput As Range) As String
ColumnLettersFromRange = Split(rInput.Address, "$")(1)
End Function
回答by Peter Albert
This should do the job:
这应该可以完成这项工作：
Function ColumnName(rng As Range) As String
Dim s As String
s = rng.Address(False, False)
ColumnName = Left(s, Len(s) - Len(Format(rng.Row, "0")))
End Function
回答by NickSlash
Using worksheet functions to calculate the column letter isn't really a good plan. Using Peter Albert's VBA method is a much nicer way to do it!
使用工作表函数来计算列字母并不是一个好的计划。使用 Peter Albert 的 VBA 方法是一种更好的方法！
I had a go at making one using worksheet functions, just for fun :/
我尝试使用工作表函数制作一个，只是为了好玩：/
A-ZZ
A-ZZ
=IF(A1<27,CHAR(64+A1),IF(A1<703,CHAR(64+INT(A1/26))&CHAR(A1-INT(A1/26)+64),"TOO BIG!"))
A-XFD (doesn't work)
A-XFD（不起作用）
=IF(A1<27,CHAR(64+A1),IF(A1<703,CHAR(64+INT(A1/26))&CHAR(A1-(INT(A1/26)*26)+64),CHAR(64+INT(A1/676))&CHAR(64+(INT(A1-(INT(A1/676)*676))/26))&CHAR(64+INT(A1-((INT(A1-(INT(A1/676)*676))/26)*26)))))
The final one falls over trying to work out the 3rd character in the address, I just cant be bothered as using it is not a good plan!
最后一个在尝试计算地址中的第三个字符时失败了，我只是无法打扰，因为使用它不是一个好计划！
