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

下面是一个自动确定单位的简明答案。

public static string ToBytesCount(this long bytes)
{
    int unit = 1024;
    string unitStr = "B";
    if (bytes < unit)
    {
        return string.Format("{0} {1}", bytes, unitStr);
    }
    int exp = (int)(Math.Log(bytes) / Math.Log(unit));
    return string.Format("{0:##.##} {1}{2}", bytes / Math.Pow(unit, exp), "KMGTPEZY"[exp - 1], unitStr);
}

“b”代表bit,“b”代表Byte,“KMGTPEZY”分别代表kilo, mega, giga, tera, peta, exa, zetta和yotta

可以将ISO/IEC80000纳入考虑范围:

public static string ToBytesCount(this long bytes, bool isISO = true)
{
    int unit = isISO ? 1024 : 1000;
    string unitStr = "B";
    if (bytes < unit)
    {
        return string.Format("{0} {1}", bytes, unitStr);
    }
    int exp = (int)(Math.Log(bytes) / Math.Log(unit));
    return string.Format("{0:##.##} {1}{2}{3}", bytes / Math.Pow(unit, exp), "KMGTPEZY"[exp - 1], isISO ? "i" : "", unitStr);
}

其他回答

如果你试图匹配Windows资源管理器的详细信息视图中显示的大小,这是你想要的代码:

[DllImport("shlwapi.dll", CharSet = CharSet.Unicode)]
private static extern long StrFormatKBSize(
    long qdw,
    [MarshalAs(UnmanagedType.LPTStr)] StringBuilder pszBuf,
    int cchBuf);

public static string BytesToString(long byteCount)
{
    var sb = new StringBuilder(32);
    StrFormatKBSize(byteCount, sb, sb.Capacity);
    return sb.ToString();
}

这不仅会与资源管理器完全匹配,而且还会为您提供翻译后的字符串,并匹配Windows版本的差异(例如在Win10中,K = 1000 vs.之前的版本K = 1024)。

这可能不是最有效或最优化的方法,但如果您不熟悉对数数学,它更容易阅读,并且对于大多数情况来说应该足够快。

string[] sizes = { "B", "KB", "MB", "GB", "TB" };
double len = new FileInfo(filename).Length;
int order = 0;
while (len >= 1024 && order < sizes.Length - 1) {
    order++;
    len = len/1024;
}

// Adjust the format string to your preferences. For example "{0:0.#}{1}" would
// show a single decimal place, and no space.
string result = String.Format("{0:0.##} {1}", len, sizes[order]);

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);
    }
}
[DllImport ( "Shlwapi.dll", CharSet = CharSet.Auto )]
public static extern long StrFormatByteSize ( 
        long fileSize
        , [MarshalAs ( UnmanagedType.LPTStr )] StringBuilder buffer
        , int bufferSize );


/// <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="filelength">The numeric value to be converted.</param>
/// <returns>the converted string</returns>
public static string StrFormatByteSize (long filesize) {
     StringBuilder sb = new StringBuilder( 11 );
     StrFormatByteSize( filesize, sb, sb.Capacity );
     return sb.ToString();
}

来自:http://www.pinvoke.net/default.aspx/shlwapi/StrFormatByteSize.html

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]);
}