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
static string[] ExcelColumnAlphabetIdentifiers = new string[] { "", "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" };
public static string ExcelColumnAlphabetIdentifier( int ColumnNumber)
{
StringBuilder sb = new StringBuilder();
int remainder = ColumnNumber;
do
{
sb.Append(ExcelColumnAlphabetIdentifiers[remainder % 26]);
remainder = remainder / 26;
}
while (remainder > 0);
return sb.ToString();
}
其他回答
抱歉,这是Python而不是c#,但至少结果是正确的:
def excel_column_number_to_name(column_number):
output = ""
index = column_number-1
while index >= 0:
character = chr((index%26)+ord('A'))
output = output + character
index = index/26 - 1
return output[::-1]
for i in xrange(1, 1024):
print "%4d : %s" % (i, excel_column_number_to_name(i))
通过这些测试用例:
列号:494286 => ABCDZ 列号:27 => 列号:52 => AZ
已经有30多个解决方案了,但这里是我的一行c#解决方案……
public string IntToExcelColumn(int i)
{
return ((i<16926? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i<2730? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i<26? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + ((char)((i%26)+65)));
}
我是这样做的:
private string GetExcelColumnName(int columnNumber)
{
string columnName = "";
while (columnNumber > 0)
{
int modulo = (columnNumber - 1) % 26;
columnName = Convert.ToChar('A' + modulo) + columnName;
columnNumber = (columnNumber - modulo) / 26;
}
return columnName;
}
这个片段适用于A到ZZ列名
string columnName = columnNumber > 26 ? Convert.ToChar(64 + (columnNumber / 26)).ToString() + Convert.ToChar(64 + (columnNumber % 26)) : Convert.ToChar(64 + columnNumber).ToString();
如果有人需要在没有VBA的Excel中做到这一点,这里有一种方法:
=SUBSTITUTE(ADDRESS(1;colNum;4);"1";"")
其中colNum是列号
在VBA中:
Function GetColumnName(colNum As Integer) As String
Dim d As Integer
Dim m As Integer
Dim name As String
d = colNum
name = ""
Do While (d > 0)
m = (d - 1) Mod 26
name = Chr(65 + m) + name
d = Int((d - m) / 26)
Loop
GetColumnName = name
End Function