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?
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
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);
}
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:${android.text.format.Formatter.formatFileSize(this, bytesToFormat)}")
下面是从aioobe转换到Kotlin的转换:
/**
* https://stackoverflow.com/a/3758880/1006741
*/
fun Long.humanReadableByteCountBinary(): String {
val b = when (this) {
Long.MIN_VALUE -> Long.MAX_VALUE
else -> abs(this)
}
return when {
b < 1024L -> "$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)
}
}
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));
}
我使用了一个比公认答案稍作修改的方法:
public static String formatFileSize(long bytes) {
if (bytes <= 0)
return "";
if (bytes < 1000)
return bytes + " B";
CharacterIterator ci = new StringCharacterIterator("kMGTPE");
while (bytes >= 99_999) {
bytes /= 1000;
ci.next();
}
return String.format(Locale.getDefault(), "%.1f %cB", bytes / 1000.0, ci.current());
}
因为我想看到另一个输出:
SI
0: <--------- instead of 0 B
27: 27 B
999: 999 B
1000: 1.0 kB
1023: 1.0 kB
1024: 1.0 kB
1728: 1.7 kB
110592: 0.1 MB <--------- instead of 110.6 kB
7077888: 7.1 MB
452984832: 0.5 GB <--------- instead of 453.0 MB
28991029248: 29.0 GB
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