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

也许你可以使用下面的代码(在c#中):

long Kb = 1024;
long Mb = Kb * 1024;
long Gb = Mb * 1024;
long Tb = Gb * 1024;
long Pb = Tb * 1024;
long Eb = Pb * 1024;

if (size < Kb)  return size.ToString() + " byte";

if (size < Mb)  return (size / Kb).ToString("###.##") + " Kb.";
if (size < Gb)  return (size / Mb).ToString("###.##") + " Mb.";
if (size < Tb)  return (size / Gb).ToString("###.##") + " Gb.";
if (size < Pb)  return (size / Tb).ToString("###.##") + " Tb.";
if (size < Eb)  return (size / Pb).ToString("###.##") + " Pb.";
if (size >= Eb) return (size / Eb).ToString("###.##") + " Eb.";

return "invalid size";

实际上,兆字节已经足够人类阅读了。

long l = 1367343104l;
    
String s = String.format("%dm", l / 1024 / 1024);

1304米

我们可以完全避免使用缓慢的Math.pow()和Math.log()方法,而不会牺牲简单性,因为单位之间的因子(例如,B, KB, MB等)是1024,即2^10。Long类有一个方便的numberofleadingzero()方法,我们可以用它来告诉大小值落在哪个单元中。

重点:大小单位的距离为10位(1024 = 2^10),这意味着最高位的位置-换句话说,前导零的数量-相差10(字节= KB*1024, KB = MB*1024,等等)。

前导零数与大小单位的相关性:

# of leading 0's Size unit
>53 B (Bytes)
>43 KB
>33 MB
>23 GB
>13 TB
>3 PB
<=3 EB

最终代码:

public static String formatSize(long v) {
    if (v < 1024) return v + " B";
    int z = (63 - Long.numberOfLeadingZeros(v)) / 10;
    return String.format("%.1f %sB", (double)v / (1L << (z*10)), " KMGTPE".charAt(z));
}

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);
    }
}

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);
}