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
我的观点是:
千字节的前缀是kB(小写K) 由于这些函数用于表示目的,因此应该提供区域性,例如:CurrentCulture,“{0时。##}{1}",文件大小,单位); 根据上下文,千字节可以是1000字节或1024字节。MB、GB等也是如此。
其他回答
我使用下面的Long扩展方法将其转换为人类可读的大小字符串。这个方法是在Stack Overflow上发布的相同问题的Java解决方案的c#实现。
/// <summary>
/// Convert a byte count into a human readable size string.
/// </summary>
/// <param name="bytes">The byte count.</param>
/// <param name="si">Whether or not to use SI units.</param>
/// <returns>A human readable size string.</returns>
public static string ToHumanReadableByteCount(
this long bytes
, bool si
)
{
var unit = si
? 1000
: 1024;
if (bytes < unit)
{
return $"{bytes} B";
}
var exp = (int) (Math.Log(bytes) / Math.Log(unit));
return $"{bytes / Math.Pow(unit, exp):F2} " +
$"{(si ? "kMGTPE" : "KMGTPE")[exp - 1] + (si ? string.Empty : "i")}B";
}
下面是一个Log10的方法:
using System;
class Program {
static string NumberFormat(double n) {
var n2 = (int)Math.Log10(n) / 3;
var n3 = n / Math.Pow(1e3, n2);
return String.Format("{0:f3}", n3) + new[]{"", " k", " M", " G"}[n2];
}
static void Main() {
var s = NumberFormat(9012345678);
Console.WriteLine(s == "9.012 G");
}
}
https://learn.microsoft.com/dotnet/api/system.math.log10
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
}
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]);
又多了一种方法,不管怎样。我喜欢上面提到的@humbads优化解决方案,所以复制了原理,但我实现了一点不同。
我认为它是否应该是一个扩展方法是有争议的(因为不是所有的long都必须是字节大小),但我喜欢它们,当我下次需要它时,我可以在某个地方找到它!
关于单位,我想我从来没有说过“Kibibyte”或“Mebibyte”,虽然我对这种强制而非进化的标准持怀疑态度,但我认为从长远来看,这将避免混淆。
public static class LongExtensions
{
private static readonly long[] numberOfBytesInUnit;
private static readonly Func<long, string>[] bytesToUnitConverters;
static LongExtensions()
{
numberOfBytesInUnit = new long[6]
{
1L << 10, // Bytes in a Kibibyte
1L << 20, // Bytes in a Mebibyte
1L << 30, // Bytes in a Gibibyte
1L << 40, // Bytes in a Tebibyte
1L << 50, // Bytes in a Pebibyte
1L << 60 // Bytes in a Exbibyte
};
// Shift the long (integer) down to 1024 times its number of units, convert to a double (real number),
// then divide to get the final number of units (units will be in the range 1 to 1023.999)
Func<long, int, string> FormatAsProportionOfUnit = (bytes, shift) => (((double)(bytes >> shift)) / 1024).ToString("0.###");
bytesToUnitConverters = new Func<long,string>[7]
{
bytes => bytes.ToString() + " B",
bytes => FormatAsProportionOfUnit(bytes, 0) + " KiB",
bytes => FormatAsProportionOfUnit(bytes, 10) + " MiB",
bytes => FormatAsProportionOfUnit(bytes, 20) + " GiB",
bytes => FormatAsProportionOfUnit(bytes, 30) + " TiB",
bytes => FormatAsProportionOfUnit(bytes, 40) + " PiB",
bytes => FormatAsProportionOfUnit(bytes, 50) + " EiB",
};
}
public static string ToReadableByteSizeString(this long bytes)
{
if (bytes < 0)
return "-" + Math.Abs(bytes).ToReadableByteSizeString();
int counter = 0;
while (counter < numberOfBytesInUnit.Length)
{
if (bytes < numberOfBytesInUnit[counter])
return bytesToUnitConverters[counter](bytes);
counter++;
}
return bytesToUnitConverters[counter](bytes);
}
}