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
datasize至少在计算中可以满足这个需求。那么一个简单的装饰器就可以了。
其他回答
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
你可以使用StringUtils的TraditionalBinarPrefix:
public static String humanReadableInt(long number) {
return TraditionalBinaryPrefix.long2String(number, ””, 1);
}
创建接口:
public interface IUnits {
public String format(long size, String pattern);
public long getUnitSize();
}
创建StorageUnits类:
import java.text.DecimalFormat;
public class StorageUnits {
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;
enum Unit implements IUnits {
TERA_BYTE {
@Override
public String format(long size, String pattern) {
return format(size, getUnitSize(), "TB", pattern);
}
@Override
public long getUnitSize() {
return T;
}
@Override
public String toString() {
return "Terabytes";
}
},
GIGA_BYTE {
@Override
public String format(long size, String pattern) {
return format(size, getUnitSize(), "GB", pattern);
}
@Override
public long getUnitSize() {
return G;
}
@Override
public String toString() {
return "Gigabytes";
}
},
MEGA_BYTE {
@Override
public String format(long size, String pattern) {
return format(size, getUnitSize(), "MB", pattern);
}
@Override
public long getUnitSize() {
return M;
}
@Override
public String toString() {
return "Megabytes";
}
},
KILO_BYTE {
@Override
public String format(long size, String pattern) {
return format(size, getUnitSize(), "kB", pattern);
}
@Override
public long getUnitSize() {
return K;
}
@Override
public String toString() {
return "Kilobytes";
}
};
String format(long size, long base, String unit, String pattern) {
return new DecimalFormat(pattern).format(
Long.valueOf(size).doubleValue() /
Long.valueOf(base).doubleValue()
) + unit;
}
}
public static String format(long size, String pattern) {
for(Unit unit : Unit.values()) {
if(size >= unit.getUnitSize()) {
return unit.format(size, pattern);
}
}
return ("???(" + size + ")???");
}
public static String format(long size) {
return format(size, "#,##0.#");
}
}
叫它:
class Main {
public static void main(String... args) {
System.out.println(StorageUnits.format(21885));
System.out.println(StorageUnits.format(2188121545L));
}
}
输出:
21.4kB
2GB
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);
}
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);
}