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
这个答案是用javaScript写的:
function getCharFromNumber(columnNumber){
var dividend = columnNumber;
var columnName = "";
var modulo;
while (dividend > 0)
{
modulo = (dividend - 1) % 26;
columnName = String.fromCharCode(65 + modulo).toString() + columnName;
dividend = parseInt((dividend - modulo) / 26);
}
return columnName;
}
其他回答
int nCol = 127;
string sChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol >= 26)
{
int nChar = nCol % 26;
nCol = (nCol - nChar) / 26;
// You could do some trick with using nChar as offset from 'A', but I am lazy to do it right now.
sCol = sChars[nChar] + sCol;
}
sCol = sChars[nCol] + sCol;
更新:Peter的评论是正确的。这就是我在浏览器中编写代码的结果。:-)我的解决方案是不编译,它是最左边的字母,它是在反向顺序构建字符串-现在都固定了。
除了bug之外,该算法基本上是将一个数字从10进制转换为26进制。
更新2:Joel Coehoorn是对的-上面的代码将返回AB为27。如果它是一个以26为底的实数,AA就等于A, Z之后的下一个数字就是BA。
int nCol = 127;
string sChars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol > 26)
{
int nChar = nCol % 26;
if (nChar == 0)
nChar = 26;
nCol = (nCol - nChar) / 26;
sCol = sChars[nChar] + sCol;
}
if (nCol != 0)
sCol = sChars[nCol] + sCol;
递归很简单。
public static string GetStandardExcelColumnName(int columnNumberOneBased)
{
int baseValue = Convert.ToInt32('A');
int columnNumberZeroBased = columnNumberOneBased - 1;
string ret = "";
if (columnNumberOneBased > 26)
{
ret = GetStandardExcelColumnName(columnNumberZeroBased / 26) ;
}
return ret + Convert.ToChar(baseValue + (columnNumberZeroBased % 26) );
}
有点晚了,但这里是我使用的代码(c#):
private static readonly string _Alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
public static int ColumnNameParse(string value)
{
// assumes value.Length is [1,3]
// assumes value is uppercase
var digits = value.PadLeft(3).Select(x => _Alphabet.IndexOf(x));
return digits.Aggregate(0, (current, index) => (current * 26) + (index + 1));
}
Objective-C实现:
-(NSString*)getColumnName:(int)n {
NSString *name = @"";
while (n>0) {
n--;
char c = (char)('A' + n%26);
name = [NSString stringWithFormat:@"%c%@",c,name];
n = n/26;
}
return name;
}
迅速实现:
func getColumnName(n:Int)->String{
var columnName = ""
var index = n
while index>0 {
index--
let char = Character(UnicodeScalar(65 + index%26))
columnName = "\(char)\(columnName)"
index = index / 26
}
return columnName
}
答案是基于:https://stackoverflow.com/a/4532562/2231118
这是我用python编写的解决方案
import math
num = 3500
row_number = str(math.ceil(num / 702))
letters = ''
num = num - 702 * math.floor(num / 702)
while num:
mod = (num - 1) % 26
letters += chr(mod + 65)
num = (num - 1) // 26
result = row_number + ("".join(reversed(letters)))
print(result)