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 String humanReadable(long size) {
    long limit = 10 * 1024;
    long limit2 = limit * 2 - 1;
    String negative = "";
    if(size < 0) {
        negative = "-";
        size = Math.abs(size);
    }

    if(size < limit) {
        return String.format("%s%s bytes", negative, size);
    } else {
        size = Math.round((double) size / 1024);
        if (size < limit2) {
            return String.format("%s%s kB", negative, size);
        } else {
            size = Math.round((double)size / 1024);
            if (size < limit2) {
                return String.format("%s%s MB", negative, size);
            } else {
                size = Math.round((double)size / 1024);
                if (size < limit2) {
                    return String.format("%s%s GB", negative, size);
                } else {
                    size = Math.round((double)size / 1024);
                        return String.format("%s%s TB", negative, size);
                }
            }
        }
    }
}

其他回答

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);
}
private String bytesIntoHumanReadable(long bytes) {
    long kilobyte = 1024;
    long megabyte = kilobyte * 1024;
    long gigabyte = megabyte * 1024;
    long terabyte = gigabyte * 1024;

    if ((bytes >= 0) && (bytes < kilobyte)) {
        return bytes + " B";

    } else if ((bytes >= kilobyte) && (bytes < megabyte)) {
        return (bytes / kilobyte) + " KB";

    } else if ((bytes >= megabyte) && (bytes < gigabyte)) {
        return (bytes / megabyte) + " MB";

    } else if ((bytes >= gigabyte) && (bytes < terabyte)) {
        return (bytes / gigabyte) + " GB";

    } else if (bytes >= terabyte) {
        return (bytes / terabyte) + " TB";

    } else {
        return bytes + " Bytes";
    }
}

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)

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。

如果你使用Android,你可以简单地使用Android .text.format. formatter . formatfilesize()。它的优点是易于使用,并且它取决于区域设置,以便为用户更好地显示它。缺点是它不处理EB,而且它只用于公制单位(每个Kilo是1000字节,不能作为1024字节使用)。

或者,这里有一个基于这篇热门文章的解决方案:


interface BytesFormatter {
    /**called when the type of the result to format is Long. Example: 123KB
     * @param unitPowerIndex the unit-power we need to format to. Examples: 0 is bytes, 1 is kb, 2 is mb, etc...
     * available units and their order: B,K,M,G,T,P,E
     * @param isMetric true if each kilo==1000, false if kilo==1024
     * */
    fun onFormatLong(valueToFormat: Long, unitPowerIndex: Int, isMetric: Boolean): String

    /**called when the type of the result to format is Double. Example: 1.23KB
     * @param unitPowerIndex the unit-power we need to format to. Examples: 0 is bytes, 1 is kb, 2 is mb, etc...
     * available units and their order: B,K,M,G,T,P,E
     * @param isMetric true if each kilo==1000, false if kilo==1024
     * */
    fun onFormatDouble(valueToFormat: Double, unitPowerIndex: Int, isMetric: Boolean): String
}

/**
 * formats the bytes to a human readable format, by providing the values to format later in the unit that we've found best to fit it
 *
 * @param isMetric true if each kilo==1000, false if kilo==1024
 * */
fun bytesIntoHumanReadable(
    @IntRange(from = 0L) bytesToFormat: Long, bytesFormatter: BytesFormatter,
    isMetric: Boolean = true
): String {
    val units = if (isMetric) 1000L else 1024L
    if (bytesToFormat < units)
        return bytesFormatter.onFormatLong(bytesToFormat, 0, isMetric)
    var bytesLeft = bytesToFormat
    var unitPowerIndex = 0
    while (unitPowerIndex < 6) {
        val newBytesLeft = bytesLeft / units
        if (newBytesLeft < units) {
            val byteLeftAsDouble = bytesLeft.toDouble() / units
            val needToShowAsInteger =
                byteLeftAsDouble == (bytesLeft / units).toDouble()
            ++unitPowerIndex
            if (needToShowAsInteger) {
                bytesLeft = newBytesLeft
                break
            }
            return bytesFormatter.onFormatDouble(byteLeftAsDouble, unitPowerIndex, isMetric)
        }
        bytesLeft = newBytesLeft
        ++unitPowerIndex
    }
    return bytesFormatter.onFormatLong(bytesLeft, unitPowerIndex, isMetric)
}

Sample usage:

// val valueToTest = 2_000L
// val valueToTest = 2_000_000L
// val valueToTest = 2_000_000_000L
// val valueToTest = 9_000_000_000_000_000_000L
// val valueToTest = 9_200_000_000_000_000_000L
val bytesToFormat = Random.nextLong(Long.MAX_VALUE)
val bytesFormatter = object : BytesFormatter {
    val numberFormat = NumberFormat.getNumberInstance(Locale.ROOT).also {
        it.maximumFractionDigits = 2
        it.minimumFractionDigits = 0
    }

    private fun formatByUnit(formattedNumber: String, threePowerIndex: Int, isMetric: Boolean): String {
        val sb = StringBuilder(formattedNumber.length + 4)
        sb.append(formattedNumber)
        val unitsToUse = "B${if (isMetric) "k" else "K"}MGTPE"
        sb.append(unitsToUse[threePowerIndex])
        if (threePowerIndex > 0)
            if (isMetric) sb.append('B') else sb.append("iB")
        return sb.toString()
    }

    override fun onFormatLong(valueToFormat: Long, unitPowerIndex: Int, isMetric: Boolean): String {
        return formatByUnit(String.format("%,d", valueToFormat), unitPowerIndex, isMetric)
    }

    override fun onFormatDouble(valueToFormat: Double, unitPowerIndex: Int, isMetric: Boolean): String {
        //alternative for using numberFormat :
        //val formattedNumber = String.format("%,.2f", valueToFormat).let { initialFormattedString ->
        //    if (initialFormattedString.contains('.'))
        //        return@let initialFormattedString.dropLastWhile { it == '0' }
        //    else return@let initialFormattedString
        //}
        return formatByUnit(numberFormat.format(valueToFormat), unitPowerIndex, isMetric)
    }
}
Log.d("AppLog", "formatting of $bytesToFormat bytes (${String.format("%,d", bytesToFormat)})")
Log.d("AppLog", bytesIntoHumanReadable(bytesToFormat, bytesFormatter))
Log.d("AppLog", "Android:${android.text.format.Formatter.formatFileSize(this, bytesToFormat)}")