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

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)}")

其他回答

这是一个Go版本。为了简单起见,我只包含了二进制输出情况。

func sizeOf(bytes int64) string {
    const unit = 1024
    if bytes < unit {
        return fmt.Sprintf("%d B", bytes)
    }

    fb := float64(bytes)
    exp := int(math.Log(fb) / math.Log(unit))
    pre := "KMGTPE"[exp-1]
    div := math.Pow(unit, float64(exp))
    return fmt.Sprintf("%.1f %ciB", fb / div, pre)
}

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。

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


private static final String[] Q = new String[]{"", "K", "M", "G", "T", "P", "E"};

public String getAsString(long bytes)
{
    for (int i = 6; i > 0; i--)
    {
        double step = Math.pow(1024, i);
        if (bytes > step) return String.format("%3.1f %s", bytes / step, Q[i]);
    }
    return Long.toString(bytes);
}
    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 "???";
    }