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
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
其他回答
String[] fileSizeUnits = {"bytes", "KB", "MB", "GB", "TB", "PB", "EB", "ZB", "YB"};
public String calculateProperFileSize(double bytes){
String sizeToReturn = "";
int index = 0;
for(index = 0; index < fileSizeUnits.length; index++){
if(bytes < 1024){
break;
}
bytes = bytes / 1024;
}
System.out.println("File size in proper format: " + bytes + " " + fileSizeUnits[index]);
sizeToReturn = String.valueOf(bytes) + " " + fileSizeUnits[index];
return sizeToReturn;
}
只需添加更多的文件单元(如果有任何缺失),你将看到单元大小达到该单元(如果你的文件有那么长):
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";
}
}
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
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);
}
你可以使用StringUtils的TraditionalBinarPrefix:
public static String humanReadableInt(long number) {
return TraditionalBinaryPrefix.long2String(number, ””, 1);
}