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

    public static String floatForm (double d)
    {
       return new DecimalFormat("#.##").format(d);
    }


    public static String bytesToHuman (long size)
    {
        long Kb = 1  * 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 floatForm(        size     ) + " byte";
        if (size >= Kb && size < Mb)    return floatForm((double)size / Kb) + " Kb";
        if (size >= Mb && size < Gb)    return floatForm((double)size / Mb) + " Mb";
        if (size >= Gb && size < Tb)    return floatForm((double)size / Gb) + " Gb";
        if (size >= Tb && size < Pb)    return floatForm((double)size / Tb) + " Tb";
        if (size >= Pb && size < Eb)    return floatForm((double)size / Pb) + " Pb";
        if (size >= Eb)                 return floatForm((double)size / Eb) + " Eb";

        return "???";
    }

其他回答

如果在Android上,你可以简单地调用Android .text. format . formatter的一个静态方法。

https://developer.android.com/reference/android/text/format/Formatter

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

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)

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

Kotlin爱好者可以使用这个扩展:

fun Long.readableFormat(): String {
    if (this <= 0 ) return "0"
    val units = arrayOf("B", "kB", "MB", "GB", "TB")
    val digitGroups = (log10(this.toDouble()) / log10(1024.0)).toInt()
    return DecimalFormat("#,##0.#").format(this / 1024.0.pow(digitGroups.toDouble())).toString() + " " + units[digitGroups]
}

现在使用

val size : Long = 90836457
val readbleString = size.readableFormat()

另一种方法

val Long.formatSize : String
    get() {
        if (this <= 0) return "0"
        val units = arrayOf("B", "kB", "MB", "GB", "TB")
        val digitGroups = (log10(this.toDouble()) / log10(1024.0)).toInt()
        return DecimalFormat("#,##0.#").format(this / 1024.0.pow(digitGroups.toDouble())).toString() + " " + units[digitGroups]
    }

现在使用

val size : Long = 90836457
val readbleString = size.formatSize