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

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

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 );

另一种皮肤的方法,没有任何类型的循环和负大小支持(对文件大小增量有意义):

public static class Format
{
    static string[] sizeSuffixes = {
        "B", "KB", "MB", "GB", "TB", "PB", "EB", "ZB", "YB" };

    public static string ByteSize(long size)
    {
        Debug.Assert(sizeSuffixes.Length > 0);

        const string formatTemplate = "{0}{1:0.#} {2}";

        if (size == 0)
        {
            return string.Format(formatTemplate, null, 0, sizeSuffixes[0]);
        }

        var absSize = Math.Abs((double)size);
        var fpPower = Math.Log(absSize, 1000);
        var intPower = (int)fpPower;
        var iUnit = intPower >= sizeSuffixes.Length
            ? sizeSuffixes.Length - 1
            : intPower;
        var normSize = absSize / Math.Pow(1000, iUnit);

        return string.Format(
            formatTemplate,
            size < 0 ? "-" : null, normSize, sizeSuffixes[iUnit]);
    }
}

下面是测试套件:

[TestFixture] public class ByteSize
{
    [TestCase(0, Result="0 B")]
    [TestCase(1, Result = "1 B")]
    [TestCase(1000, Result = "1 KB")]
    [TestCase(1500000, Result = "1.5 MB")]
    [TestCase(-1000, Result = "-1 KB")]
    [TestCase(int.MaxValue, Result = "2.1 GB")]
    [TestCase(int.MinValue, Result = "-2.1 GB")]
    [TestCase(long.MaxValue, Result = "9.2 EB")]
    [TestCase(long.MinValue, Result = "-9.2 EB")]
    public string Format_byte_size(long size)
    {
        return Format.ByteSize(size);
    }
}

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);

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]}";
}

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。