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)
}
其他回答
字节单位允许你这样做:
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));
我写了另一个叫做storage-units的库,它允许你这样做:
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);
如果你想强制某个单位,可以这样做:
String formattedUnit9 = StorageUnits.formatAsKibibyte(input2);
String formattedUnit10 = StorageUnits.formatAsCommonMegabyte(input2);
Assert.assertEquals("1024.00 KiB", formattedUnit9);
Assert.assertEquals("1.00 MB", formattedUnit10);
Kotlin爱好者可以使用这个扩展:
fun Long.readableFormat(): String {
if (this <= 0 ) return "0"
val units = arrayOf("B", "kB", "MB", "GB", "TB")
val digitGroups = (log10(this.toDouble()) / log10(1024.0)).toInt()
return DecimalFormat("#,##0.#").format(this / 1024.0.pow(digitGroups.toDouble())).toString() + " " + units[digitGroups]
}
现在使用
val size : Long = 90836457
val readbleString = size.readableFormat()
另一种方法
val Long.formatSize : String
get() {
if (this <= 0) return "0"
val units = arrayOf("B", "kB", "MB", "GB", "TB")
val digitGroups = (log10(this.toDouble()) / log10(1024.0)).toInt()
return DecimalFormat("#,##0.#").format(this / 1024.0.pow(digitGroups.toDouble())).toString() + " " + units[digitGroups]
}
现在使用
val size : Long = 90836457
val readbleString = size.formatSize
实际上,兆字节已经足够人类阅读了。
long l = 1367343104l;
String s = String.format("%dm", l / 1024 / 1024);
1304米
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
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)