如何在c#中将数值转换为Excel列名,而不使用直接从Excel中获取值的自动化。
Excel 2007的范围可能是1到16384,这是它支持的列数。结果值应以excel列名的形式出现,例如A、AA、AAA等。
如何在c#中将数值转换为Excel列名,而不使用直接从Excel中获取值的自动化。
Excel 2007的范围可能是1到16384,这是它支持的列数。结果值应以excel列名的形式出现,例如A、AA、AAA等。
当前回答
下面是一个Actionscript版本:
private var columnNumbers:Array = ['A', 'B', 'C', 'D', 'E', 'F' , 'G', 'H', 'I', 'J', 'K' ,'L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];
private function getExcelColumnName(columnNumber:int) : String{
var dividend:int = columnNumber;
var columnName:String = "";
var modulo:int;
while (dividend > 0)
{
modulo = (dividend - 1) % 26;
columnName = columnNumbers[modulo] + columnName;
dividend = int((dividend - modulo) / 26);
}
return columnName;
}
其他回答
在Delphi (Pascal)中:
function GetExcelColumnName(columnNumber: integer): string;
var
dividend, modulo: integer;
begin
Result := '';
dividend := columnNumber;
while dividend > 0 do begin
modulo := (dividend - 1) mod 26;
Result := Chr(65 + modulo) + Result;
dividend := (dividend - modulo) div 26;
end;
end;
抱歉,这是Python而不是c#,但至少结果是正确的:
def ColIdxToXlName(idx):
if idx < 1:
raise ValueError("Index is too small")
result = ""
while True:
if idx > 26:
idx, r = divmod(idx - 1, 26)
result = chr(r + ord('A')) + result
else:
return chr(idx + ord('A') - 1) + result
for i in xrange(1, 1024):
print "%4d : %s" % (i, ColIdxToXlName(i))
精炼原始的解决方案(在c#中):
public static class ExcelHelper
{
private static Dictionary<UInt16, String> l_DictionaryOfColumns;
public static ExcelHelper() {
l_DictionaryOfColumns = new Dictionary<ushort, string>(256);
}
public static String GetExcelColumnName(UInt16 l_Column)
{
UInt16 l_ColumnCopy = l_Column;
String l_Chars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
String l_rVal = "";
UInt16 l_Char;
if (l_DictionaryOfColumns.ContainsKey(l_Column) == true)
{
l_rVal = l_DictionaryOfColumns[l_Column];
}
else
{
while (l_ColumnCopy > 26)
{
l_Char = l_ColumnCopy % 26;
if (l_Char == 0)
l_Char = 26;
l_ColumnCopy = (l_ColumnCopy - l_Char) / 26;
l_rVal = l_Chars[l_Char] + l_rVal;
}
if (l_ColumnCopy != 0)
l_rVal = l_Chars[l_ColumnCopy] + l_rVal;
l_DictionaryOfColumns.ContainsKey(l_Column) = l_rVal;
}
return l_rVal;
}
}
我今天必须做这个工作,我的实现使用递归:
private static string GetColumnLetter(string colNumber)
{
if (string.IsNullOrEmpty(colNumber))
{
throw new ArgumentNullException(colNumber);
}
string colName = String.Empty;
try
{
var colNum = Convert.ToInt32(colNumber);
var mod = colNum % 26;
var div = Math.Floor((double)(colNum)/26);
colName = ((div > 0) ? GetColumnLetter((div - 1).ToString()) : String.Empty) + Convert.ToChar(mod + 65);
}
finally
{
colName = colName == String.Empty ? "A" : colName;
}
return colName;
}
该方法将数字视为字符串,而以“0”开头的数字(A = 0)
我想在我使用的静态类中加入,用于在col index和col Label之间进行交互。我对ColumnLabel方法使用了修改后的可接受答案
public static class Extensions
{
public static string ColumnLabel(this int col)
{
var dividend = col;
var columnLabel = string.Empty;
int modulo;
while (dividend > 0)
{
modulo = (dividend - 1) % 26;
columnLabel = Convert.ToChar(65 + modulo).ToString() + columnLabel;
dividend = (int)((dividend - modulo) / 26);
}
return columnLabel;
}
public static int ColumnIndex(this string colLabel)
{
// "AD" (1 * 26^1) + (4 * 26^0) ...
var colIndex = 0;
for(int ind = 0, pow = colLabel.Count()-1; ind < colLabel.Count(); ++ind, --pow)
{
var cVal = Convert.ToInt32(colLabel[ind]) - 64; //col A is index 1
colIndex += cVal * ((int)Math.Pow(26, pow));
}
return colIndex;
}
}
用这个…
30.ColumnLabel(); // "AD"
"AD".ColumnIndex(); // 30