我如何获得一个人类可读的文件大小字节缩写使用。net ?
例子: 输入7,326,629,显示6.98 MB
我如何获得一个人类可读的文件大小字节缩写使用。net ?
例子: 输入7,326,629,显示6.98 MB
当前回答
所有溶液的混合:-)
/// <summary>
/// Converts a numeric value into a string that represents the number expressed as a size value in bytes,
/// kilobytes, megabytes, or gigabytes, depending on the size.
/// </summary>
/// <param name="fileSize">The numeric value to be converted.</param>
/// <returns>The converted string.</returns>
public static string FormatByteSize(double fileSize)
{
FileSizeUnit unit = FileSizeUnit.B;
while (fileSize >= 1024 && unit < FileSizeUnit.YB)
{
fileSize = fileSize / 1024;
unit++;
}
return string.Format("{0:0.##} {1}", fileSize, unit);
}
/// <summary>
/// Converts a numeric value into a string that represents the number expressed as a size value in bytes,
/// kilobytes, megabytes, or gigabytes, depending on the size.
/// </summary>
/// <param name="fileInfo"></param>
/// <returns>The converted string.</returns>
public static string FormatByteSize(FileInfo fileInfo)
{
return FormatByteSize(fileInfo.Length);
}
}
public enum FileSizeUnit : byte
{
B,
KB,
MB,
GB,
TB,
PB,
EB,
ZB,
YB
}
其他回答
int size = new FileInfo( filePath ).Length / 1024;
string humanKBSize = string.Format( "{0} KB", size );
string humanMBSize = string.Format( "{0} MB", size / 1024 );
string humanGBSize = string.Format( "{0} GB", size / 1024 / 1024 );
string[] suffixes = { "B", "KB", "MB", "GB", "TB", "PB", "EB", "ZB", "YB" };
int s = 0;
long size = fileInfo.Length;
while (size >= 1024)
{
s++;
size /= 1024;
}
string humanReadable = String.Format("{0} {1}", size, suffixes[s]);
那么递归呢:
private static string ReturnSize(double size, string sizeLabel)
{
if (size > 1024)
{
if (sizeLabel.Length == 0)
return ReturnSize(size / 1024, "KB");
else if (sizeLabel == "KB")
return ReturnSize(size / 1024, "MB");
else if (sizeLabel == "MB")
return ReturnSize(size / 1024, "GB");
else if (sizeLabel == "GB")
return ReturnSize(size / 1024, "TB");
else
return ReturnSize(size / 1024, "PB");
}
else
{
if (sizeLabel.Length > 0)
return string.Concat(size.ToString("0.00"), sizeLabel);
else
return string.Concat(size.ToString("0.00"), "Bytes");
}
}
然后你称之为:
return ReturnSize(size, string.Empty);
比如@NET3的解决方案。使用shift而不是除法来测试字节的范围,因为除法占用更多的CPU成本。
private static readonly string[] UNITS = new string[] { "B", "KB", "MB", "GB", "TB", "PB", "EB" };
public static string FormatSize(ulong bytes)
{
int c = 0;
for (c = 0; c < UNITS.Length; c++)
{
ulong m = (ulong)1 << ((c + 1) * 10);
if (bytes < m)
break;
}
double n = bytes / (double)((ulong)1 << (c * 10));
return string.Format("{0:0.##} {1}", n, UNITS[c]);
}
下面是@ deepe1的BigInteger版本的答案,它绕过了long的大小限制(因此支持yottabyte和理论上的任何后面的限制):
public static string ToBytesString(this BigInteger byteCount, string format = "N3")
{
string[] suf = { "B", "KiB", "MiB", "GiB", "TiB", "PiB", "EiB", "YiB" };
if (byteCount.IsZero)
{
return $"{0.0.ToString(format)} {suf[0]}";
}
var abs = BigInteger.Abs(byteCount);
var place = Convert.ToInt32(Math.Floor(BigInteger.Log(abs, 1024)));
var pow = Math.Pow(1024, place);
// since we need to do this with integer math, get the quotient and remainder
var quotient = BigInteger.DivRem(abs, new BigInteger(pow), out var remainder);
// convert the remainder to a ratio and add both back together as doubles
var num = byteCount.Sign * (Math.Floor((double)quotient) + ((double)remainder / pow));
return $"{num.ToString(format)} {suf[place]}";
}