如何在Java中将字节大小转换为人类可读的格式?

比如1024应该变成“1 Kb”，1024*1024应该变成“1 Mb”。

我有点厌倦了为每个项目写这个实用方法。在Apache Commons中有这样的静态方法吗?

当前回答

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)

2010-09-21 09:22:21

其他回答

创建接口:

public interface IUnits {
    public String format(long size, String pattern);
    public long getUnitSize();
}

创建StorageUnits类:

import java.text.DecimalFormat;

public class StorageUnits {

    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;

    enum Unit implements IUnits {

        TERA_BYTE {
            @Override
            public String format(long size, String pattern) {
                return format(size, getUnitSize(), "TB", pattern);
            }
            @Override
            public long getUnitSize() {
                return T;
            }
            @Override
            public String toString() {
                return "Terabytes";
            }
        },
        GIGA_BYTE {
            @Override
            public String format(long size, String pattern) {
                return format(size, getUnitSize(), "GB", pattern);
            }
            @Override
            public long getUnitSize() {
                return G;
            }
            @Override
            public String toString() {
                return "Gigabytes";
            }
        },
        MEGA_BYTE {
            @Override
            public String format(long size, String pattern) {
                return format(size, getUnitSize(), "MB", pattern);
            }
            @Override
            public long getUnitSize() {
                return M;
            }
            @Override
            public String toString() {
                return "Megabytes";
            }
        },
        KILO_BYTE {
            @Override
            public String format(long size, String pattern) {
                return format(size, getUnitSize(), "kB", pattern);
            }
            @Override
            public long getUnitSize() {
                return K;
            }
            @Override
            public String toString() {
                return "Kilobytes";
            }

        };

        String format(long size, long base, String unit, String pattern) {
            return new DecimalFormat(pattern).format(
                           Long.valueOf(size).doubleValue() /
                           Long.valueOf(base).doubleValue()
            ) + unit;
        }
    }

    public static String format(long size, String pattern) {
        for(Unit unit : Unit.values()) {
            if(size >= unit.getUnitSize()) {
                return unit.format(size, pattern);
            }
        }
        return ("???(" + size + ")???");
    }

    public static String format(long size) {
        return format(size, "#,##0.#");
    }
}

叫它:

class Main {
    public static void main(String... args) {
        System.out.println(StorageUnits.format(21885));
        System.out.println(StorageUnits.format(2188121545L));
    }
}

输出:

21.4kB
2GB

2017-07-29 19:37:05

下面是上面Java正确的共识答案的c# .NET等价版本 (下面还有一个代码更短的例子):

    public static String BytesNumberToHumanReadableString(long bytes, bool SI1000orBinary1024)
    {
        int unit = SI1000orBinary1024 ? 1000 : 1024;
        if (bytes < unit)
            return bytes + " B";

        int exp = (int)(Math.Log(bytes) / Math.Log(unit));
        String pre = (SI1000orBinary1024 ? "kMGTPE" : "KMGTPE")[(exp - 1)] + (SI1000orBinary1024 ? "" : "i");
        return String.Format("{0:F1} {1}B", bytes / Math.Pow(unit, exp), pre);
    }

从技术上讲，如果我们坚持使用国际单位制，这个程序适用于任何常规的数字使用。专家们还给出了许多不错的答案。假设您正在对gridview上的数字进行数据绑定，有必要从它们中查看性能优化例程。

PS:这个帖子是因为当我在做一个c#项目时，这个问题/答案出现在谷歌搜索的顶部。

2016-03-31 04:14:15

Kotlin版本通过扩展属性

如果您正在使用Kotlin，那么通过这些扩展名属性格式化文件大小非常容易。它是无循环的，完全基于纯数学。

HumanizeUtils.kt

import java.io.File
import kotlin.math.log2
import kotlin.math.pow

/**
 * @author aminography
 */

val File.formatSize: String
    get() = length().formatAsFileSize

val Int.formatAsFileSize: String
    get() = toLong().formatAsFileSize

val Long.formatAsFileSize: String
    get() = log2(if (this != 0L) toDouble() else 1.0).toInt().div(10).let {
        val precision = when (it) {
            0 -> 0; 1 -> 1; else -> 2
        }
        val prefix = arrayOf("", "K", "M", "G", "T", "P", "E", "Z", "Y")
        String.format("%.${precision}f ${prefix[it]}B", toDouble() / 2.0.pow(it * 10.0))
    }

用法:

println("0:          " + 0.formatAsFileSize)
println("170:        " + 170.formatAsFileSize)
println("14356:      " + 14356.formatAsFileSize)
println("968542985:  " + 968542985.formatAsFileSize)
println("8729842496: " + 8729842496.formatAsFileSize)

println("file: " + file.formatSize)

结果:

0:          0 B
170:        170 B
14356:      14.0 KB
968542985:  923.67 MB
8729842496: 8.13 GB

file: 6.15 MB

2020-08-13 06:27:32

我通常是这样做的:

public static String getFileSize(double size) {
    return _getFileSize(size,0,1024);
}

public static String _getFileSize(double size, int i, double base) {
    String units = " KMGTP";
    String unit = (i>0)?(""+units.charAt(i)).toUpperCase()+"i":"";
    if(size<base)
        return size +" "+unit.trim()+"B";
    else {
        size = Math.floor(size/base);
        return _getFileSize(size,++i,base);
    }
}

2020-10-02 07:56:32

这是aioobe答案的修改版本。

变化:

Locale参数，因为有些语言使用。其他的，作为小数点。人类可读的代码

private static final String[] SI_UNITS = { "B", "kB", "MB", "GB", "TB", "PB", "EB" };
private static final String[] BINARY_UNITS = { "B", "KiB", "MiB", "GiB", "TiB", "PiB", "EiB" };

public static String humanReadableByteCount(final long bytes, final boolean useSIUnits, final Locale locale)
{
    final String[] units = useSIUnits ? SI_UNITS : BINARY_UNITS;
    final int base = useSIUnits ? 1000 : 1024;

    // When using the smallest unit no decimal point is needed, because it's the exact number.
    if (bytes < base) {
        return bytes + " " + units[0];
    }

    final int exponent = (int) (Math.log(bytes) / Math.log(base));
    final String unit = units[exponent];
    return String.format(locale, "%.1f %s", bytes / Math.pow(base, exponent), unit);
}

2015-11-17 09:18:06

如何在Java中将字节大小转换为人类可读的格式?

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