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

下面是从aioobe转换到Kotlin的转换:

/**
 * https://stackoverflow.com/a/3758880/1006741
 */
fun Long.humanReadableByteCountBinary(): String {
    val b = when (this) {
        Long.MIN_VALUE -> Long.MAX_VALUE
        else -> abs(this)
    }
    return when {
        b < 1024L -> "$this B"
        b <= 0xfffccccccccccccL shr 40 -> "%.1f KiB".format(Locale.UK, this / 1024.0)
        b <= 0xfffccccccccccccL shr 30 -> "%.1f MiB".format(Locale.UK, this / 1048576.0)
        b <= 0xfffccccccccccccL shr 20 -> "%.1f GiB".format(Locale.UK, this / 1.073741824E9)
        b <= 0xfffccccccccccccL shr 10 -> "%.1f TiB".format(Locale.UK, this / 1.099511627776E12)
        b <= 0xfffccccccccccccL -> "%.1f PiB".format(Locale.UK, (this shr 10) / 1.099511627776E12)
        else -> "%.1f EiB".format(Locale.UK, (this shr 20) / 1.099511627776E12)
    }
}

其他回答

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)}")

filename=filedilg.getSelectedFile().getAbsolutePath();
File file=new File(filename);

String disp=FileUtils.byteCountToDisplaySize(file.length());
System.out.println("THE FILE PATH IS "+file+"THIS File SIZE IS IN MB "+disp);

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。

我们可以完全避免使用缓慢的Math.pow()和Math.log()方法,而不会牺牲简单性,因为单位之间的因子(例如,B, KB, MB等)是1024,即2^10。Long类有一个方便的numberofleadingzero()方法,我们可以用它来告诉大小值落在哪个单元中。

重点:大小单位的距离为10位(1024 = 2^10),这意味着最高位的位置-换句话说,前导零的数量-相差10(字节= KB*1024, KB = MB*1024,等等)。

前导零数与大小单位的相关性:

# of leading 0's Size unit
>53 B (Bytes)
>43 KB
>33 MB
>23 GB
>13 TB
>3 PB
<=3 EB

最终代码:

public static String formatSize(long v) {
    if (v < 1024) return v + " B";
    int z = (63 - Long.numberOfLeadingZeros(v)) / 10;
    return String.format("%.1f %sB", (double)v / (1L << (z*10)), " KMGTPE".charAt(z));
}

这是另一个简洁的解决方案,没有循环,但具有区域敏感格式和正确的二进制前缀:

import java.util.Locale;

public final class Bytes {

  private Bytes() {
  }

  public static String format(long value, Locale locale) {
    if (value < 1024) {
      return value + " B";
    }
    int z = (63 - Long.numberOfLeadingZeros(value)) / 10;
    return String.format(locale, "%.1f %siB", (double) value / (1L << (z * 10)), " KMGTPE".charAt(z));
  }
}

测试:

Locale locale = Locale.getDefault()
System.out.println(Bytes.format(1L, locale))
System.out.println(Bytes.format(2L * 1024, locale))
System.out.println(Bytes.format(3L * 1024 * 1024, locale))
System.out.println(Bytes.format(4L * 1024 * 1024 * 1024, locale))
System.out.println(Bytes.format(5L * 1024 * 1024 * 1024 * 1024, locale))
System.out.println(Bytes.format(6L * 1024 * 1024 * 1024 * 1024 * 1024, locale))
System.out.println(Bytes.format(Long.MAX_VALUE, locale))

输出:

1 B
2.0 KiB
3.0 MiB
4.0 GiB
5.0 GiB
6.0 PiB
8.0 EiB