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
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
其他回答
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)
}
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);
}
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
我最近问了同样的问题:
格式文件大小为MB, GB等。
虽然没有开箱即用的答案,但我可以接受这个解决方案:
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(在我的德语地区):
1 B
4,2 KB
41,4 MB
3,3 GB
我已经打开了一个问题,要求谷歌番石榴的这个功能。也许有人愿意支持它。