我发现了很多关于将原始字节信息转换为人类可读格式的信息，但我需要做相反的事情，即将字符串“1.6 GB”转换为长值1717990000.是否有内置/定义明确的方式，或者我几乎不得不自己动手？
[编辑]：这是我的第一次刺...
static class ByteFormat extends NumberFormat {
@Override
public StringBuffer format(double arg0, StringBuffer arg1, FieldPosition arg2) {
// TODO Auto-generated method stub
return null;
}
@Override
public StringBuffer format(long arg0, StringBuffer arg1, FieldPosition arg2) {
// TODO Auto-generated method stub
return null;
}
@Override
public Number parse(String arg0, ParsePosition arg1) {
return parse (arg0);
}
@Override
public Number parse(String arg0) {
int spaceNdx = arg0.indexOf(" ");
double ret = Double.parseDouble(arg0.substring(0, spaceNdx));
String unit = arg0.substring(spaceNdx + 1);
int factor = 0;
if (unit.equals("GB")) {
factor = 1073741824;
}
else if (unit.equals("MB")) {
factor = 1048576;
}
else if (unit.equals("KB")) {
factor = 1024;
}
return ret * factor;
}
}
答案 0 :(得分：4)
Andremoniy的答案的修订版本，正确区分公斤和kibi等。
private final static long KB_FACTOR = 1000;
private final static long KIB_FACTOR = 1024;
private final static long MB_FACTOR = 1000 * KB_FACTOR;
private final static long MIB_FACTOR = 1024 * KIB_FACTOR;
private final static long GB_FACTOR = 1000 * MB_FACTOR;
private final static long GIB_FACTOR = 1024 * MIB_FACTOR;
public static double parse(String arg0) {
int spaceNdx = arg0.indexOf(" ");
double ret = Double.parseDouble(arg0.substring(0, spaceNdx));
switch (arg0.substring(spaceNdx + 1)) {
case "GB":
return ret * GB_FACTOR;
case "GiB":
return ret * GIB_FACTOR;
case "MB":
return ret * MB_FACTOR;
case "MiB":
return ret * MIB_FACTOR;
case "KB":
return ret * KB_FACTOR;
case "KiB":
return ret * KIB_FACTOR;
}
return -1;
}
答案 1 :(得分：4)
一站式答案，解析为long：
public class SizeUtil {
public static String units = "BKMGTPEZY";
public static long parse(String arg0) {
int spaceNdx = arg0.indexOf(" ");
double ret = Double.parseDouble(arg0.substring(0, spaceNdx));
String unitString = arg0.substring(spaceNdx+1);
int unitChar = unitString.charAt(0);
int power = units.indexOf(unitChar);
boolean isSi = unitString.indexOf('i')!=-1;
int factor = 1024;
if (isSi)
{
factor = 1000;
}
return new Double(ret * Math.pow(factor, power)).longValue();
}
public static void main(String[] args) {
System.out.println(parse("300.00 GiB")); // requires a space
System.out.println(parse("300.00 GB"));
System.out.println(parse("300.00 B"));
System.out.println(parse("300 EB"));
}
}
答案 2 :(得分：2)
我从来没有听说过这样一个着名的库，它实现了这样的文本解析实用程序方法。但是你的解决方案似乎接近正确的实施。
我想在你的代码中纠正的唯一两件事是：
将方法Number parse(String arg0)定义为静态，因为它具有实用性
为每种类型的尺寸定义定义factor作为final static字段。
即。它会像这样：
private final static long KB_FACTOR = 1024;
private final static long MB_FACTOR = 1024 * KB_FACTOR;
private final static long GB_FACTOR = 1024 * MB_FACTOR;
public static double parse(String arg0) {
int spaceNdx = arg0.indexOf(" ");
double ret = Double.parseDouble(arg0.substring(0, spaceNdx));
switch (arg0.substring(spaceNdx + 1)) {
case "GB":
return ret * GB_FACTOR;
case "MB":
return ret * MB_FACTOR;
case "KB":
return ret * KB_FACTOR;
}
return -1;
}
答案 3 :(得分：2)
我知道这要晚得多，但我正在寻找一个兼顾SI prefix的类似功能。 所以我自己创建了一个，我认为它可能对其他人有用。
public static String units = "KMGTPE";
/**
* Converts from human readable to byte format
* @param number The number value of the amount to convert
* @param unit The unit: B, KB, MB, GB, TB, PB, EB
* @param si Si prefix
* @return byte value
*/
public static double parse(double number, String unit, boolean si)
{
String identifier = unit.substring(0, 1);
int index = units.indexOf(identifier);
//not already in bytes
if (index!=-1)
{
for (int i = 0; i <= index; i++)
number = number * (si ? 1000 : 1024);
}
return number;
}
我确信这也可以用于递归。打扰太简单......
答案 4 :(得分：2)
Spring Framework在版本5.1上添加了一个DataSize类，该类允许将人类可读的数据大小解析为字节，并且还可以将其格式化为人类可读的形式。可以找到here。
如果使用Spring Framework，则可以升级到> = 5.1并使用此类。否则，您可以c / p它和相关类（同时遵守许可证）。
然后您可以使用它：
DataSize dataSize = DataSize.parse("16GB");
System.out.println(dataSize.toBytes());
将给出输出：
17179869184
但是，用于解析输入的模式
1GB，2GB，1638MB，但不能使用1.6GB）1GB但不能使用1 GB）我建议遵循兼容性/易于维护的约定。 但是，如果这不满足您的需要，则需要复制和编辑文件-这是一个很好的起点。
答案 5 :(得分：1)
另一个基于@gilbertpilz代码的选项。在这种情况下，使用正则表达式获取值和因数。它也不区分大小写。
private final static long KB_FACTOR = 1000;
private final static long KIB_FACTOR = 1024;
private final static long MB_FACTOR = 1000 * KB_FACTOR;
private final static long MIB_FACTOR = 1024 * KIB_FACTOR;
private final static long GB_FACTOR = 1000 * MB_FACTOR;
private final static long GIB_FACTOR = 1024 * MIB_FACTOR;
private long parse(String arg0) throws ParseException {
Pattern pattern = Pattern.compile("([0-9]+)(([KMG])I?B)");
Matcher match = pattern.matcher(arg0);
if( !match.matches() || match.groupCount()!=3)
throw new ParseException("Wrong format", 0);
long ret = Long.parseLong(match.group(0));
switch (match.group(2).toUpperCase()) {
case "GB":
return ret * GB_FACTOR;
case "GIB":
return ret * GIB_FACTOR;
case "MB":
return ret * MB_FACTOR;
case "MIB":
return ret * MIB_FACTOR;
case "KB":
return ret * KB_FACTOR;
case "KIB":
return ret * KIB_FACTOR;
}
throw new ParseException("Wrong format", 0);
}
答案 6 :(得分：0)
也可以使用以下方法并使之通用，而不依赖于空格字符进行解析。
感谢@RobAu提供上述提示。添加了一种新方法来获取字符串中第一个字母的索引，并更改了parse方法以基于该新方法来获取索引。我保留了原始的parse方法并添加了新的parseAny方法，因此可以比较结果。希望对别人有帮助。
还要感谢indexOf方法的答案-https://stackoverflow.com/a/11214786/6385674。
public class ConversionUtil {
public static String units = "BKMGTPEZY";
public static long parse(String arg0) {
int spaceNdx = arg0.indexOf(" ");
double ret = Double.parseDouble(arg0.substring(0, spaceNdx));
String unitString = arg0.substring(spaceNdx+1);
int unitChar = unitString.charAt(0);
int power = units.indexOf(unitChar);
boolean isSi = unitString.indexOf('i')!=-1;
int factor = 1024;
if (isSi)
{
factor = 1000;
}
return new Double(ret * Math.pow(factor, power)).longValue();
}
/** @return index of pattern in s or -1, if not found */
public static int indexOf(Pattern pattern, String s) {
Matcher matcher = pattern.matcher(s);
return matcher.find() ? matcher.start() : -1;
}
public static long parseAny(String arg0)
{
int index = indexOf(Pattern.compile("[A-Za-z]"), arg0);
double ret = Double.parseDouble(arg0.substring(0, index));
String unitString = arg0.substring(index);
int unitChar = unitString.charAt(0);
int power = units.indexOf(unitChar);
boolean isSi = unitString.indexOf('i')!=-1;
int factor = 1024;
if (isSi)
{
factor = 1000;
}
return new Double(ret * Math.pow(factor, power)).longValue();
}
public static void main(String[] args) {
System.out.println(parse("300.00 GiB")); // requires a space
System.out.println(parse("300.00 GB"));
System.out.println(parse("300.00 B"));
System.out.println(parse("300 EB"));
System.out.println(parseAny("300.00 GiB"));
System.out.println(parseAny("300M"));
}
}
答案 7 :(得分：0)
我写了一个文件大小的可读实用程序枚举类，希望对您有帮助！
/**
* The file size human readable utility class,
* provide mutual conversions from human readable size to byte size
*
* The similar function in stackoverflow, linked:
* https://stackoverflow.com/questions/3758606/how-to-convert-byte-size-into-human-readable-format-in-java?r=SearchResults
*
* Apache also provide similar function
* @see org.apache.commons.io.FileUtils#byteCountToDisplaySize(long)
*
* @author Ponfee
*/
public enum HumanReadables {
SI (1000, "B", "KB", "MB", "GB", "TB", "PB", "EB" /*, "ZB", "YB" */), //
BINARY(1024, "B", "KiB", "MiB", "GiB", "TiB", "PiB", "EiB"/*, "ZiB", "YiB"*/), //
;
private static final String FORMAT = "#,##0.##";
private static final Pattern PATTERN = Pattern.compile(".*[0-9]+.*");
private final int base;
private final String[] units;
private final long[] sizes;
HumanReadables(int base, String... units) {
this.base = base;
this.units = units;
this.sizes = new long[this.units.length];
this.sizes[0] = 1;
for (int i = 1; i < this.sizes.length; i++) {
this.sizes[i] = this.sizes[i - 1] * this.base; // Maths.pow(this.base, i);
}
}
/**
* Returns a string of bytes count human readable size
*
* @param size the size
* @return human readable size
*/
public strictfp String human(long size) {
if (size == 0) {
return "0" + this.units[0];
}
String signed = "";
if (size < 0) {
signed = "-";
size = size == Long.MIN_VALUE ? Long.MAX_VALUE : -size;
}
/*int unit = (int) Maths.log(size, this.base);
return signed + format(size / Math.pow(this.base, unit)) + " " + this.units[unit];*/
int unit = find(size);
return new StringBuilder(13) // 13 max length like as "-1,023.45 GiB"
.append(signed)
.append(formatter().format(size / (double) this.sizes[unit]))
.append(" ")
.append(this.units[unit])
.toString();
}
public strictfp long parse(String size) {
return parse(size, false);
}
/**
* Parse the readable byte count, allowed suffix units: "1", "1B", "1MB", "1MiB", "1M"
*
* @param size the size
* @param strict the strict, if BINARY then verify whether contains "i"
* @return a long value bytes count
*/
public strictfp long parse(String size, boolean strict) {
if (size == null || size.isEmpty()) {
return 0L;
}
if (!PATTERN.matcher(size).matches()) {
throw new IllegalArgumentException("Invalid format [" + size + "]");
}
String str = size = size.trim();
long factor = this.sizes[0];
switch (str.charAt(0)) {
case '+': str = str.substring(1); break;
case '-': str = str.substring(1); factor = -1L; break;
}
int end = 0, lastPos = str.length() - 1;
// last character isn't a digit
char c = str.charAt(lastPos - end);
if (c == 'i') {
// last pos cannot end with "i"
throw new IllegalArgumentException("Invalid format [" + size + "], cannot end with \"i\".");
}
if (c == 'B') {
end++;
c = str.charAt(lastPos - end);
boolean flag = isBlank(c);
while (isBlank(c) && end < lastPos) {
end++;
c = str.charAt(lastPos - end);
}
// if "B" head has space char, then the first head non space char must be a digit
if (flag && !Character.isDigit(c)) {
throw new IllegalArgumentException("Invalid format [" + size + "]: \"" + c + "\".");
}
}
if (!Character.isDigit(c)) {
// if not a digit character, then assume is a unit character
if (c == 'i') {
if (this == SI) {
// SI cannot contains "i"
throw new IllegalArgumentException("Invalid SI format [" + size + "], cannot contains \"i\".");
}
end++;
c = str.charAt(lastPos - end);
} else {
if (this == BINARY && strict) {
// if strict, then BINARY must contains "i"
throw new IllegalArgumentException("Invalid BINARY format [" + size + "], miss character \"i\".");
}
}
switch (c) {
case 'K': factor *= this.sizes[1]; break;
case 'M': factor *= this.sizes[2]; break;
case 'G': factor *= this.sizes[3]; break;
case 'T': factor *= this.sizes[4]; break;
case 'P': factor *= this.sizes[5]; break;
case 'E': factor *= this.sizes[6]; break;
/*
case 'Z': factor *= this.bytes[7]; break;
case 'Y': factor *= this.bytes[8]; break;
*/
default: throw new IllegalArgumentException("Invalid format [" + size + "]: \"" + c + "\".");
}
do {
end++;
c = str.charAt(lastPos - end);
} while (isBlank(c) && end < lastPos);
}
str = str.substring(0, str.length() - end);
try {
return (long) (factor * formatter().parse(str).doubleValue());
} catch (NumberFormatException | ParseException e) {
throw new IllegalArgumentException("Failed to parse [" + size + "]: \"" + str + "\".");
}
}
public int base() {
return this.base;
}
public String[] units() {
return Arrays.copyOf(this.units, this.units.length);
}
public long[] sizes() {
return Arrays.copyOf(this.sizes, this.sizes.length);
}
private int find(long bytes) {
int n = this.sizes.length;
for (int i = 1; i < n; i++) {
if (bytes < this.sizes[i]) {
return i - 1;
}
}
return n - 1;
}
private DecimalFormat formatter() {
return new DecimalFormat(FORMAT);
}
private boolean isBlank(char c) {
return c == ' ' || c == '\t';
}
}