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

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

我通常是这样做的:

public static String getFileSize(double size) {
    return _getFileSize(size,0,1024);
}

public static String _getFileSize(double size, int i, double base) {
    String units = " KMGTP";
    String unit = (i>0)?(""+units.charAt(i)).toUpperCase()+"i":"";
    if(size<base)
        return size +" "+unit.trim()+"B";
    else {
        size = Math.floor(size/base);
        return _getFileSize(size,++i,base);
    }
}

你可以使用StringUtils的TraditionalBinarPrefix:

public static String humanReadableInt(long number) {
    return TraditionalBinaryPrefix.long2String(number, ””, 1);
}

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。