如何在 Java 中将字节大小转换为人类可读的格式？

有趣的事实：这里发布的原始代码段是 Stack Overflow 上有史以来被复制最多的 Java 代码段，但它存在缺陷。它已修复，但它变得混乱。本文全文： 有史以来被复制最多的 Stack Overflow 片段存在缺陷！

来源：Formatting byte size to human readable format | Programming.Guide

国际单位制 (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 Ki = 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)

---

如果您的项目可以依赖于 org.apache.commons.io，那么 FileUtils.byteCountToDisplaySize(long size) 会起作用。

JavaDoc for this method

---

使用 Android 内置类

对于 Android，有一个类，Formatter。只需一行代码，您就完成了。

android.text.format.Formatter.formatShortFileSize(activityContext, bytes);

它类似于 formatFileSize()，但试图生成更短的数字（显示更少的小数）。

android.text.format.Formatter.formatFileSize(activityContext, bytes);

它将内容大小格式化为字节、千字节、兆字节等形式。

---

我们可以完全避免使用慢速 Math.pow() 和 Math.log() 方法而不牺牲简单性，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 numberOfLeadingZeros() 方法，我们可以使用它来判断大小值属于哪个单位。

关键点：大小单位的距离为 10 位 (1024 = 2^10)，这意味着最高一位的位置——或者换句话说，前导零的数量——相差 10（字节 = KB*1024，KB = MB *1024 等）。

前导零的数量与大小单位之间的相关性：

# of leading 0's   Size unit
-------------------------------
>53                B (Bytes)
>43                KB
>33                MB
>23                GB
>13                TB
>3                 PB
<=2                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));
}

---

我最近问了同样的问题：

Format file size as MB, GB, etc.

虽然没有开箱即用的答案，但我可以接受以下解决方案：

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;

public static String convertToStringRepresentation(final long value){
    final long[] dividers = new long[] { T, G, M, K, 1 };
    final String[] units = new String[] { "TB", "GB", "MB", "KB", "B" };
    if(value < 1)
        throw new IllegalArgumentException("Invalid file size: " + value);
    String result = null;
    for(int i = 0; i < dividers.length; i++){
        final long divider = dividers[i];
        if(value >= divider){
            result = format(value, divider, units[i]);
            break;
        }
    }
    return result;
}

private static String format(final long value,
    final long divider,
    final String unit){
    final double result =
        divider > 1 ? (double) value / (double) divider : (double) value;
    return new DecimalFormat("#,##0.#").format(result) + " " + unit;
}

测试代码：

public static void main(final String[] args){
    final long[] l = new long[] { 1l, 4343l, 43434334l, 3563543743l };
    for(final long ll : l){
        System.out.println(convertToStringRepresentation(ll));
    }
}

输出（在我的德语语言环境中）：

1 B
4,2 KB
41,4 MB
3,3 GB

我打开了一个 issue requesting this functionality for Google Guava。也许有人会愿意支持它。

---

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";
    }
}

---

这是 aioobe's answer 的修改版本。

变化：

语言环境参数，因为某些语言使用 .和其他，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);
}

---

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);
}

---

Byte Units 允许您这样做：

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));

我编写了另一个名为 storage-units 的库，它允许您这样做：

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);

如果您想强制某个单位，请执行以下操作：

String formattedUnit9 = StorageUnits.formatAsKibibyte(input2);
String formattedUnit10 = StorageUnits.formatAsCommonMegabyte(input2);

Assert.assertEquals("1024.00 KiB", formattedUnit9);
Assert.assertEquals("1.00 MB", formattedUnit10);

---

如果您使用 Android，则只需使用 android.text.format.Formatter.formatFileSize()。优点是它易于使用，并且取决于语言环境以很好地向用户展示它。缺点是它不处理 EB，而且只用于公制单位（每 Kilo 为 1000 字节，无法将其用作 1024 字节）。

或者，这是基于 this popular post 的解决方案：

interface BytesFormatter {
    /**called when the type of the result to format is Long. Example: 123KB
     * @param unitPowerIndex the unit-power we need to format to. Examples: 0 is bytes, 1 is kb, 2 is mb, etc...
     * available units and their order: B,K,M,G,T,P,E
     * @param isMetric true if each kilo==1000, false if kilo==1024
     * */
    fun onFormatLong(valueToFormat: Long, unitPowerIndex: Int, isMetric: Boolean): String

    /**called when the type of the result to format is Double. Example: 1.23KB
     * @param unitPowerIndex the unit-power we need to format to. Examples: 0 is bytes, 1 is kb, 2 is mb, etc...
     * available units and their order: B,K,M,G,T,P,E
     * @param isMetric true if each kilo==1000, false if kilo==1024
     * */
    fun onFormatDouble(valueToFormat: Double, unitPowerIndex: Int, isMetric: Boolean): String
}

/**
 * formats the bytes to a human readable format, by providing the values to format later in the unit that we've found best to fit it
 *
 * @param isMetric true if each kilo==1000, false if kilo==1024
 * */
fun bytesIntoHumanReadable(
    @IntRange(from = 0L) bytesToFormat: Long, bytesFormatter: BytesFormatter,
    isMetric: Boolean = true
): String {
    val units = if (isMetric) 1000L else 1024L
    if (bytesToFormat < units)
        return bytesFormatter.onFormatLong(bytesToFormat, 0, isMetric)
    var bytesLeft = bytesToFormat
    var unitPowerIndex = 0
    while (unitPowerIndex < 6) {
        val newBytesLeft = bytesLeft / units
        if (newBytesLeft < units) {
            val byteLeftAsDouble = bytesLeft.toDouble() / units
            val needToShowAsInteger =
                byteLeftAsDouble == (bytesLeft / units).toDouble()
            ++unitPowerIndex
            if (needToShowAsInteger) {
                bytesLeft = newBytesLeft
                break
            }
            return bytesFormatter.onFormatDouble(byteLeftAsDouble, unitPowerIndex, isMetric)
        }
        bytesLeft = newBytesLeft
        ++unitPowerIndex
    }
    return bytesFormatter.onFormatLong(bytesLeft, unitPowerIndex, isMetric)
}

Sample usage:

// val valueToTest = 2_000L
// val valueToTest = 2_000_000L
// val valueToTest = 2_000_000_000L
// val valueToTest = 9_000_000_000_000_000_000L
// val valueToTest = 9_200_000_000_000_000_000L
val bytesToFormat = Random.nextLong(Long.MAX_VALUE)
val bytesFormatter = object : BytesFormatter {
    val numberFormat = NumberFormat.getNumberInstance(Locale.ROOT).also {
        it.maximumFractionDigits = 2
        it.minimumFractionDigits = 0
    }

    private fun formatByUnit(formattedNumber: String, threePowerIndex: Int, isMetric: Boolean): String {
        val sb = StringBuilder(formattedNumber.length + 4)
        sb.append(formattedNumber)
        val unitsToUse = "B${if (isMetric) "k" else "K"}MGTPE"
        sb.append(unitsToUse[threePowerIndex])
        if (threePowerIndex > 0)
            if (isMetric) sb.append('B') else sb.append("iB")
        return sb.toString()
    }

    override fun onFormatLong(valueToFormat: Long, unitPowerIndex: Int, isMetric: Boolean): String {
        return formatByUnit(String.format("%,d", valueToFormat), unitPowerIndex, isMetric)
    }

    override fun onFormatDouble(valueToFormat: Double, unitPowerIndex: Int, isMetric: Boolean): String {
        //alternative for using numberFormat :
        //val formattedNumber = String.format("%,.2f", valueToFormat).let { initialFormattedString ->
        //    if (initialFormattedString.contains('.'))
        //        return@let initialFormattedString.dropLastWhile { it == '0' }
        //    else return@let initialFormattedString
        //}
        return formatByUnit(numberFormat.format(valueToFormat), unitPowerIndex, isMetric)
    }
}
Log.d("AppLog", "formatting of $bytesToFormat bytes (${String.format("%,d", bytesToFormat)})")
Log.d("AppLog", bytesIntoHumanReadable(bytesToFormat, bytesFormatter))
Log.d("AppLog", "Android:${android.text.format.Formatter.formatFileSize(this, bytesToFormat)}")

---

• Kotlin 版本通过扩展属性

如果您使用的是 kotlin，则很容易通过这些扩展属性来格式化文件大小。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

---

    public static String floatForm (double d)
    {
       return new DecimalFormat("#.##").format(d);
    }


    public static String bytesToHuman (long size)
    {
        long Kb = 1  * 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 floatForm(        size     ) + " byte";
        if (size >= Kb && size < Mb)    return floatForm((double)size / Kb) + " Kb";
        if (size >= Mb && size < Gb)    return floatForm((double)size / Mb) + " Mb";
        if (size >= Gb && size < Tb)    return floatForm((double)size / Gb) + " Gb";
        if (size >= Tb && size < Pb)    return floatForm((double)size / Tb) + " Tb";
        if (size >= Pb && size < Eb)    return floatForm((double)size / Pb) + " Pb";
        if (size >= Eb)                 return floatForm((double)size / Eb) + " Eb";

        return "???";
    }

---

org.springframework.util.unit.DataSize 至少在计算中可以满足这个要求。然后一个简单的装饰器就可以了。

---

现在有一个包含单位格式的可用库。我将它添加到 triava 库中，因为唯一的其他现有库似乎是用于 Android 的。

它可以在 3 种不同的系统（SI、IEC、JEDEC）和各种输出选项中以任意精度格式化数字。以下是 triava unit tests 中的一些代码示例：

UnitFormatter.formatAsUnit(1126, UnitSystem.SI, "B");
// = "1.13kB"
UnitFormatter.formatAsUnit(2094, UnitSystem.IEC, "B");
// = "2.04KiB"

打印精确的公斤、兆值（此处为 W = 瓦特）：

UnitFormatter.formatAsUnits(12_000_678, UnitSystem.SI, "W", ", ");
// = "12MW, 678W"

您可以传递 DecimalFormat 来自定义输出：

UnitFormatter.formatAsUnit(2085, UnitSystem.IEC, "B", new DecimalFormat("0.0000"));
// = "2.0361KiB"

对于公斤或兆值的任意操作，您可以将它们拆分为组件：

UnitComponent uc = new  UnitComponent(123_345_567_789L, UnitSystem.SI);
int kilos = uc.kilo(); // 567
int gigas = uc.giga(); // 123

---

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

---

另一个没有循环但具有区域设置敏感格式和正确二进制前缀的简洁解决方案：

import java.util.Locale;

public final class Bytes {

  private Bytes() {
  }

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

测试：

Locale locale = Locale.getDefault()
System.out.println(Bytes.format(1L, locale))
System.out.println(Bytes.format(2L * 1024, locale))
System.out.println(Bytes.format(3L * 1024 * 1024, locale))
System.out.println(Bytes.format(4L * 1024 * 1024 * 1024, locale))
System.out.println(Bytes.format(5L * 1024 * 1024 * 1024 * 1024, locale))
System.out.println(Bytes.format(6L * 1024 * 1024 * 1024 * 1024 * 1024, locale))
System.out.println(Bytes.format(Long.MAX_VALUE, locale))

输出：

1 B
2.0 KiB
3.0 MiB
4.0 GiB
5.0 GiB
6.0 PiB
8.0 EiB

---

您可以使用 StringUtils 的 TraditionalBinarPrefix：

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

---

这是一个 Go 版本。为简单起见，我只包含了二进制输出的情况。

func sizeOf(bytes int64) string {
    const unit = 1024
    if bytes < unit {
        return fmt.Sprintf("%d B", bytes)
    }

    fb := float64(bytes)
    exp := int(math.Log(fb) / math.Log(unit))
    pre := "KMGTPE"[exp-1]
    div := math.Pow(unit, float64(exp))
    return fmt.Sprintf("%.1f %ciB", fb / div, pre)
}

---

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;
}

只需添加更多文件单元（如果缺少），您将看到单元大小达到该单元（如果您的文件有那么长）：

---

我使用的方法比接受的答案稍有修改：

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  

---

对于 kotlin 爱好者，请使用此扩展

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 

---

这是上面 Java 正确共识答案的 C# .NET 等效项（下面还有另一个代码较短）：

    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 SI 单位，则此例程适用于任何常规使用的数字。专家还有许多其他很好的答案。假设您正在对网格视图上的数字进行数据绑定，值得从它们中检查性能优化例程。

PS：这是因为这个问题/答案在我做 C# 项目时在谷歌搜索中出现在首位。

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也许您可以使用此代码（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";

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这是转换为 Kotlin 的 the conversion from aioobe：

/**
 * 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)
    }
}

---

public String humanReadable(long size) {
    long limit = 10 * 1024;
    long limit2 = limit * 2 - 1;
    String negative = "";
    if(size < 0) {
        negative = "-";
        size = Math.abs(size);
    }

    if(size < limit) {
        return String.format("%s%s bytes", negative, size);
    } else {
        size = Math.round((double) size / 1024);
        if (size < limit2) {
            return String.format("%s%s kB", negative, size);
        } else {
            size = Math.round((double)size / 1024);
            if (size < limit2) {
                return String.format("%s%s MB", negative, size);
            } else {
                size = Math.round((double)size / 1024);
                if (size < limit2) {
                    return String.format("%s%s GB", negative, size);
                } else {
                    size = Math.round((double)size / 1024);
                        return String.format("%s%s TB", negative, size);
                }
            }
        }
    }
}

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使用以下函数获取准确信息。它是根据 ATM_CashWithdrawl 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。使用 apache.hadoop-StringUtils 时将获得相同的格式

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;
}

上述方法的示例用法：

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));
}

得到上述格式的字节

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(long duration)。

---

试试 JSR 363。它的单元扩展模块，如 Unicode CLDR（在 GitHub: uom-systems 中）为您完成所有这些工作。

您可以使用每个实现中包含的 MetricPrefix 或 BinaryPrefix（与上面的一些示例类似），如果您在印度或附近国家生活和工作，IndianPrefix（也在 uom-systems 的公共模块中） ) 也允许您使用和格式化“Crore Bytes”或“Lakh Bytes”。

---

我一般都是这样的，你觉得呢？

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);
    }
}

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下面是一个快速、简单且易读的代码片段来实现这一点：

/**
 * Converts byte size to human readable strings (also declares useful constants)
 *
 * @see <a href="https://en.wikipedia.org/wiki/File_size">File size</a>
 */
@SuppressWarnings("SpellCheckingInspection")
public class HumanReadableSize {
    public static final double
            KILO = 1000L, // 1000 power 1 (10 power 3)
            KIBI = 1024L, // 1024 power 1 (2 power 10)
            MEGA = KILO * KILO, // 1000 power 2 (10 power 6)
            MEBI = KIBI * KIBI, // 1024 power 2 (2 power 20)
            GIGA = MEGA * KILO, // 1000 power 3 (10 power 9)
            GIBI = MEBI * KIBI, // 1024 power 3 (2 power 30)
            TERA = GIGA * KILO, // 1000 power 4 (10 power 12)
            TEBI = GIBI * KIBI, // 1024 power 4 (2 power 40)
            PETA = TERA * KILO, // 1000 power 5 (10 power 15)
            PEBI = TEBI * KIBI, // 1024 power 5 (2 power 50)
            EXA = PETA * KILO, // 1000 power 6 (10 power 18)
            EXBI = PEBI * KIBI; // 1024 power 6 (2 power 60)

    private static final DecimalFormat df = new DecimalFormat("#.##");

    public static String binaryBased(long size) {
        if (size < 0) {
            throw new IllegalArgumentException("Argument cannot be negative");
        } else if (size < KIBI) {
            return df.format(size).concat("B");
        } else if (size < MEBI) {
            return df.format(size / KIBI).concat("KiB");
        } else if (size < GIBI) {
            return df.format(size / MEBI).concat("MiB");
        } else if (size < TEBI) {
            return df.format(size / GIBI).concat("GiB");
        } else if (size < PEBI) {
            return df.format(size / TEBI).concat("TiB");
        } else if (size < EXBI) {
            return df.format(size / PEBI).concat("PiB");
        } else {
            return df.format(size / EXBI).concat("EiB");
        }
    }

    public static String decimalBased(long size) {
        if (size < 0) {
            throw new IllegalArgumentException("Argument cannot be negative");
        } else if (size < KILO) {
            return df.format(size).concat("B");
        } else if (size < MEGA) {
            return df.format(size / KILO).concat("KB");
        } else if (size < GIGA) {
            return df.format(size / MEGA).concat("MB");
        } else if (size < TERA) {
            return df.format(size / GIGA).concat("GB");
        } else if (size < PETA) {
            return df.format(size / TERA).concat("TB");
        } else if (size < EXA) {
            return df.format(size / PETA).concat("PB");
        } else {
            return df.format(size / EXA).concat("EB");
        }
    }
}

笔记：

上面的代码冗长而直接。它不使用循环（仅当您不知道在编译期间需要迭代多少次时才应使用循环）它不会进行不必要的库调用（StringBuilder、Math 等） 上面的代码速度很快并且使用的内存非常少.基于在我的个人入门级云机器上运行的基准测试，它是最快的（在这些情况下性能并不重要，但仍然如此）上面的代码是一个很好的答案的修改版本

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filename=filedilg.getSelectedFile().getAbsolutePath();
File file=new File(filename);

String disp=FileUtils.byteCountToDisplaySize(file.length());
System.out.println("THE FILE PATH IS "+file+"THIS File SIZE IS IN MB "+disp);

---