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

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);
}

其他回答

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);
    }
}

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);
}

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

下面是上面Java正确的共识答案的c# .NET等价版本 (下面还有一个代码更短的例子):

    public static String BytesNumberToHumanReadableString(long bytes, bool SI1000orBinary1024)
    {
        int unit = SI1000orBinary1024 ? 1000 : 1024;
        if (bytes < unit)
            return bytes + " B";

        int exp = (int)(Math.Log(bytes) / Math.Log(unit));
        String pre = (SI1000orBinary1024 ? "kMGTPE" : "KMGTPE")[(exp - 1)] + (SI1000orBinary1024 ? "" : "i");
        return String.Format("{0:F1} {1}B", bytes / Math.Pow(unit, exp), pre);
    }

从技术上讲,如果我们坚持使用国际单位制,这个程序适用于任何常规的数字使用。专家们还给出了许多不错的答案。假设您正在对gridview上的数字进行数据绑定,有必要从它们中查看性能优化例程。

PS:这个帖子是因为当我在做一个c#项目时,这个问题/答案出现在谷歌搜索的顶部。

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