下面是一个快速，简单和可读的代码片段来实现这一点:

/**
 * Converts byte size to human readable strings (also declares useful constants)
 *
 * @see <a href="https://en.wikipedia.org/wiki/File_size">File size</a>
 */
@SuppressWarnings("SpellCheckingInspection")
public class HumanReadableSize {
    public static final double
            KILO = 1000L, // 1000 power 1 (10 power 3)
            KIBI = 1024L, // 1024 power 1 (2 power 10)
            MEGA = KILO * KILO, // 1000 power 2 (10 power 6)
            MEBI = KIBI * KIBI, // 1024 power 2 (2 power 20)
            GIGA = MEGA * KILO, // 1000 power 3 (10 power 9)
            GIBI = MEBI * KIBI, // 1024 power 3 (2 power 30)
            TERA = GIGA * KILO, // 1000 power 4 (10 power 12)
            TEBI = GIBI * KIBI, // 1024 power 4 (2 power 40)
            PETA = TERA * KILO, // 1000 power 5 (10 power 15)
            PEBI = TEBI * KIBI, // 1024 power 5 (2 power 50)
            EXA = PETA * KILO, // 1000 power 6 (10 power 18)
            EXBI = PEBI * KIBI; // 1024 power 6 (2 power 60)

    private static final DecimalFormat df = new DecimalFormat("#.##");

    public static String binaryBased(long size) {
        if (size < 0) {
            throw new IllegalArgumentException("Argument cannot be negative");
        } else if (size < KIBI) {
            return df.format(size).concat("B");
        } else if (size < MEBI) {
            return df.format(size / KIBI).concat("KiB");
        } else if (size < GIBI) {
            return df.format(size / MEBI).concat("MiB");
        } else if (size < TEBI) {
            return df.format(size / GIBI).concat("GiB");
        } else if (size < PEBI) {
            return df.format(size / TEBI).concat("TiB");
        } else if (size < EXBI) {
            return df.format(size / PEBI).concat("PiB");
        } else {
            return df.format(size / EXBI).concat("EiB");
        }
    }

    public static String decimalBased(long size) {
        if (size < 0) {
            throw new IllegalArgumentException("Argument cannot be negative");
        } else if (size < KILO) {
            return df.format(size).concat("B");
        } else if (size < MEGA) {
            return df.format(size / KILO).concat("KB");
        } else if (size < GIGA) {
            return df.format(size / MEGA).concat("MB");
        } else if (size < TERA) {
            return df.format(size / GIGA).concat("GB");
        } else if (size < PETA) {
            return df.format(size / TERA).concat("TB");
        } else if (size < EXA) {
            return df.format(size / PETA).concat("PB");
        } else {
            return df.format(size / EXA).concat("EB");
        }
    }
}

注意:

上面的代码冗长而简单。 它不使用循环(循环应该只在您不知道在编译期间需要迭代多少次时使用) 它不会进行不必要的库调用(StringBuilder, Math等) 上面的代码是快速的，使用非常少的内存。基于在我个人的入门级云计算机上运行的基准测试，它是最快的(在这些情况下性能并不重要，但仍然如此) 以上代码是一个很好的答案的修改版本

我最近问了同样的问题:

格式文件大小为MB, GB等。

虽然没有开箱即用的答案，但我可以接受这个解决方案:

private static final long K = 1024;
private static final long M = K * K;
private static final long G = M * K;
private static final long T = G * K;

public static String convertToStringRepresentation(final long value){
    final long[] dividers = new long[] { T, G, M, K, 1 };
    final String[] units = new String[] { "TB", "GB", "MB", "KB", "B" };
    if(value < 1)
        throw new IllegalArgumentException("Invalid file size: " + value);
    String result = null;
    for(int i = 0; i < dividers.length; i++){
        final long divider = dividers[i];
        if(value >= divider){
            result = format(value, divider, units[i]);
            break;
        }
    }
    return result;
}

private static String format(final long value,
    final long divider,
    final String unit){
    final double result =
        divider > 1 ? (double) value / (double) divider : (double) value;
    return new DecimalFormat("#,##0.#").format(result) + " " + unit;
}

测试代码:

public static void main(final String[] args){
    final long[] l = new long[] { 1l, 4343l, 43434334l, 3563543743l };
    for(final long ll : l){
        System.out.println(convertToStringRepresentation(ll));
    }
}

输出(在我的德语地区):

1 B
4,2 KB
41,4 MB
3,3 GB

我已经打开了一个问题，要求谷歌番石榴的这个功能。也许有人愿意支持它。

String[] fileSizeUnits = {"bytes", "KB", "MB", "GB", "TB", "PB", "EB", "ZB", "YB"};

public String calculateProperFileSize(double bytes){
    String sizeToReturn = "";
    int index = 0;
    for(index = 0; index < fileSizeUnits.length; index++){
        if(bytes < 1024){
            break;
        }
        bytes = bytes / 1024;
    }

    System.out.println("File size in proper format: " + bytes + " " + fileSizeUnits[index]);
    sizeToReturn = String.valueOf(bytes) + " " + fileSizeUnits[index];
    return sizeToReturn;
}

只需添加更多的文件单元(如果有任何缺失)，你将看到单元大小达到该单元(如果你的文件有那么长):

也许你可以使用下面的代码(在c#中):

long Kb = 1024;
long Mb = Kb * 1024;
long Gb = Mb * 1024;
long Tb = Gb * 1024;
long Pb = Tb * 1024;
long Eb = Pb * 1024;

if (size < Kb)  return size.ToString() + " byte";

if (size < Mb)  return (size / Kb).ToString("###.##") + " Kb.";
if (size < Gb)  return (size / Mb).ToString("###.##") + " Mb.";
if (size < Tb)  return (size / Gb).ToString("###.##") + " Gb.";
if (size < Pb)  return (size / Tb).ToString("###.##") + " Tb.";
if (size < Eb)  return (size / Pb).ToString("###.##") + " Pb.";
if (size >= Eb) return (size / Eb).ToString("###.##") + " Eb.";

return "invalid size";

有趣的事实:这里发布的原始代码片段是Stack Overflow上被复制最多的Java代码片段，它是有缺陷的。它被修好了，但却变得一团糟。 本文的完整故事:有史以来复制最多的堆栈溢出代码片段是有缺陷的!

来源:格式化字节大小到人类可读的格式|编程。指南

SI（1 k = 1，000）

public static String humanReadableByteCountSI(long bytes) {
    if (-1000 < bytes && bytes < 1000) {
        return bytes + " B";
    }
    CharacterIterator ci = new StringCharacterIterator("kMGTPE");
    while (bytes <= -999_950 || bytes >= 999_950) {
        bytes /= 1000;
        ci.next();
    }
    return String.format("%.1f %cB", bytes / 1000.0, ci.current());
}

二进制(1's = 1,024)

public static String humanReadableByteCountBin(long bytes) {
    long absB = bytes == Long.MIN_VALUE ? Long.MAX_VALUE : Math.abs(bytes);
    if (absB < 1024) {
        return bytes + " B";
    }
    long value = absB;
    CharacterIterator ci = new StringCharacterIterator("KMGTPE");
    for (int i = 40; i >= 0 && absB > 0xfffccccccccccccL >> i; i -= 10) {
        value >>= 10;
        ci.next();
    }
    value *= Long.signum(bytes);
    return String.format("%.1f %ciB", value / 1024.0, ci.current());
}

示例输出:

                             SI     BINARY

                  0:        0 B        0 B
                 27:       27 B       27 B
                999:      999 B      999 B
               1000:     1.0 kB     1000 B
               1023:     1.0 kB     1023 B
               1024:     1.0 kB    1.0 KiB
               1728:     1.7 kB    1.7 KiB
             110592:   110.6 kB  108.0 KiB
            7077888:     7.1 MB    6.8 MiB
          452984832:   453.0 MB  432.0 MiB
        28991029248:    29.0 GB   27.0 GiB
      1855425871872:     1.9 TB    1.7 TiB
9223372036854775807:     9.2 EB    8.0 EiB   (Long.MAX_VALUE)

FileUtils。如果你的项目依赖于org.apache.commons.io, byteCountToDisplaySize(长尺寸)可以工作。

此方法的JavaDoc