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
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)
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。
下面是从aioobe转换到Kotlin的转换:
/**
* https://stackoverflow.com/a/3758880/1006741
*/
fun Long.humanReadableByteCountBinary(): String {
val b = when (this) {
Long.MIN_VALUE -> Long.MAX_VALUE
else -> abs(this)
}
return when {
b < 1024L -> "$this B"
b <= 0xfffccccccccccccL shr 40 -> "%.1f KiB".format(Locale.UK, this / 1024.0)
b <= 0xfffccccccccccccL shr 30 -> "%.1f MiB".format(Locale.UK, this / 1048576.0)
b <= 0xfffccccccccccccL shr 20 -> "%.1f GiB".format(Locale.UK, this / 1.073741824E9)
b <= 0xfffccccccccccccL shr 10 -> "%.1f TiB".format(Locale.UK, this / 1.099511627776E12)
b <= 0xfffccccccccccccL -> "%.1f PiB".format(Locale.UK, (this shr 10) / 1.099511627776E12)
else -> "%.1f EiB".format(Locale.UK, (this shr 20) / 1.099511627776E12)
}
}
使用下面的函数来获得确切的信息。它是基于atm_cashwithdraw概念生成的。
getFullMemoryUnit(): Total: [123 MB], Max: [1 GB, 773 MB, 512 KB], Free: [120 MB, 409 KB, 304 Bytes]
public static String getFullMemoryUnit(long unit) {
long BYTE = 1024, KB = BYTE, MB = KB * KB, GB = MB * KB, TB = GB * KB;
long KILO_BYTE, MEGA_BYTE = 0, GIGA_BYTE = 0, TERA_BYTE = 0;
unit = Math.abs(unit);
StringBuffer buffer = new StringBuffer();
if ( unit / TB > 0 ) {
TERA_BYTE = (int) (unit / TB);
buffer.append(TERA_BYTE+" TB");
unit -= TERA_BYTE * TB;
}
if ( unit / GB > 0 ) {
GIGA_BYTE = (int) (unit / GB);
if (TERA_BYTE != 0) buffer.append(", ");
buffer.append(GIGA_BYTE+" GB");
unit %= GB;
}
if ( unit / MB > 0 ) {
MEGA_BYTE = (int) (unit / MB);
if (GIGA_BYTE != 0) buffer.append(", ");
buffer.append(MEGA_BYTE+" MB");
unit %= MB;
}
if ( unit / KB > 0 ) {
KILO_BYTE = (int) (unit / KB);
if (MEGA_BYTE != 0) buffer.append(", ");
buffer.append(KILO_BYTE+" KB");
unit %= KB;
}
if ( unit > 0 ) buffer.append(", "+unit+" Bytes");
return buffer.toString();
}
我刚刚修改了facebookarchive-StringUtils的代码以获得以下格式。与使用apache.hadoop-StringUtils时得到的格式相同
getMemoryUnit(): Total: [123.0 MB], Max: [1.8 GB], Free: [120.4 MB]
public static String getMemoryUnit(long bytes) {
DecimalFormat oneDecimal = new DecimalFormat("0.0");
float BYTE = 1024.0f, KB = BYTE, MB = KB * KB, GB = MB * KB, TB = GB * KB;
long absNumber = Math.abs(bytes);
double result = bytes;
String suffix = " Bytes";
if (absNumber < MB) {
result = bytes / KB;
suffix = " KB";
} else if (absNumber < GB) {
result = bytes / MB;
suffix = " MB";
} else if (absNumber < TB) {
result = bytes / GB;
suffix = " GB";
}
return oneDecimal.format(result) + suffix;
}
以上方法的使用示例:
public static void main(String[] args) {
Runtime runtime = Runtime.getRuntime();
int availableProcessors = runtime.availableProcessors();
long heapSize = Runtime.getRuntime().totalMemory();
long heapMaxSize = Runtime.getRuntime().maxMemory();
long heapFreeSize = Runtime.getRuntime().freeMemory();
System.out.format("Total: [%s], Max: [%s], Free: [%s]\n", heapSize, heapMaxSize, heapFreeSize);
System.out.format("getMemoryUnit(): Total: [%s], Max: [%s], Free: [%s]\n",
getMemoryUnit(heapSize), getMemoryUnit(heapMaxSize), getMemoryUnit(heapFreeSize));
System.out.format("getFullMemoryUnit(): Total: [%s], Max: [%s], Free: [%s]\n",
getFullMemoryUnit(heapSize), getFullMemoryUnit(heapMaxSize), getFullMemoryUnit(heapFreeSize));
}
字节来获取上面的格式
Total: [128974848], Max: [1884815360], Free: [126248240]
为了以人类可读的格式显示时间，请使用函数millisToShortDHMS(长持续时间)。
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
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