我该如何在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”,而1024*1024应该变成“1 Mb”。
我已经厌烦了每个项目都要写这个实用程序方法。在Apache Commons中有没有静态方法可以完成这个任务?
我该如何在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”,而1024*1024应该变成“1 Mb”。
我已经厌烦了每个项目都要写这个实用程序方法。在Apache Commons中有没有静态方法可以完成这个任务?
完整故事请看这篇文章:Stack Overflow 历史上被复制最多的代码片段存在缺陷!
来源:如何将字节大小格式化为易于阅读的格式 | Programming.Guide
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());
}
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());
}
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)
du
命令的行为,例如。) - aioobelet bytes = 110592
NSByteCountFormatter.stringFromByteCount(Int64(bytes), countStyle: NSByteCountFormatterCountStyle.File)
将产生"111 KB"的结果。 - duthenFileUtils.byteCountToDisplaySize(long size)
可以解决您的问题,如果您的项目可以依赖于 org.apache.commons.io
。
com.zendesk.util
)的任何人,您应该有以下API作为选项:FileUtils.humanReadableFileSize(bytesSize)
。 - Thiengo我们可以完全避免使用缓慢的 Math.pow()
和 Math.log()
方法而不牺牲简洁性,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 numberOfLeadingZeros()
方法,我们可以使用它来确定大小值属于哪个单位。
关键点:大小单位距离 10 位比特(1024 = 2^10),意味着最高有效比特的位置,或者说前导零的数量,相差 10 位(字节 = KB * 1024,KB = MB * 1024 等)。
前导零数量和大小单位之间的对应关系:
前导零数量 | 大小单位 |
---|---|
>53 | B (字节) |
>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));
}
Math.scalb(v, z * -10)
代替(double)v / (1L << (z*10))
。 - Holger最近我也问了同样的问题:
虽然没有现成的答案,但是我可以接受以下解决方案:
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));
}
}
输出结果(在我的德语环境下):
1 B
4,2 KB
41,4 MB
3,3 GB
我已经为Google Guava开启了一个功能请求问题。也许有人愿意支持它。
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";
}
}
这是aioobe的回答的修改版。
更改内容:
Locale
参数,因为一些语言使用.
而其他语言使用,
作为小数点。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);
}
如果您正在使用Kotlin,使用这些扩展属性对文件大小进行格式化非常容易。它是无循环的,并完全基于纯数学。
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(coerceAtLeast(1).toDouble()).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))
}
println("0: " + 0.formatAsFileSize)
println("170: " + 170.formatAsFileSize)
println("14356: " + 14356.formatAsFileSize)
println("968542985: " + 968542985.formatAsFileSize)
println("8729842496: " + 8729842496.formatAsFileSize)
println("file: " + file.formatSize)
0: 0 B
170: 170 B
14356: 14.0 KB
968542985: 923.67 MB
8729842496: 8.13 GB
file: 6.15 MB
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);
}
Q
的长度有关吗? - Peter Mortensen另外,这里有一个基于这个流行帖子的解决方案:
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)}")
Long.MAX_VALUE
,它将不会显示正确的值。你怎么看? - android developerByte Units可以让你像这样做:
long input1 = 1024;
long input2 = 1024 * 1024;
Assert.assertEquals("1 KiB", BinaryByteUnit.format(input1));
Assert.assertEquals("1 MiB", BinaryByteUnit.format(input2));
Assert.assertEquals("1.024 KB", DecimalByteUnit.format(input1, "#.0"));
Assert.assertEquals("1.049 MB", DecimalByteUnit.format(input2, "#.000"));
NumberFormat format = new DecimalFormat("#.#");
Assert.assertEquals("1 KiB", BinaryByteUnit.format(input1, format));
Assert.assertEquals("1 MiB", BinaryByteUnit.format(input2, format));
我编写了另一个名为 storage-units 的库,它可以让你像这样做:
String formattedUnit1 = StorageUnits.formatAsCommonUnit(input1, "#");
String formattedUnit2 = StorageUnits.formatAsCommonUnit(input2, "#");
String formattedUnit3 = StorageUnits.formatAsBinaryUnit(input1);
String formattedUnit4 = StorageUnits.formatAsBinaryUnit(input2);
String formattedUnit5 = StorageUnits.formatAsDecimalUnit(input1, "#.00", Locale.GERMAN);
String formattedUnit6 = StorageUnits.formatAsDecimalUnit(input2, "#.00", Locale.GERMAN);
String formattedUnit7 = StorageUnits.formatAsBinaryUnit(input1, format);
String formattedUnit8 = StorageUnits.formatAsBinaryUnit(input2, format);
Assert.assertEquals("1 kB", formattedUnit1);
Assert.assertEquals("1 MB", formattedUnit2);
Assert.assertEquals("1.00 KiB", formattedUnit3);
Assert.assertEquals("1.00 MiB", formattedUnit4);
Assert.assertEquals("1,02 kB", formattedUnit5);
Assert.assertEquals("1,05 MB", formattedUnit6);
Assert.assertEquals("1 KiB", formattedUnit7);
Assert.assertEquals("1 MiB", formattedUnit8);
如需强制使用某个单位,请按照以下步骤操作:
String formattedUnit9 = StorageUnits.formatAsKibibyte(input2);
String formattedUnit10 = StorageUnits.formatAsCommonMegabyte(input2);
Assert.assertEquals("1024.00 KiB", formattedUnit9);
Assert.assertEquals("1.00 MB", formattedUnit10);