我如何获得一个人类可读的文件大小字节缩写使用。net ?
例子: 输入7,326,629,显示6.98 MB
我如何获得一个人类可读的文件大小字节缩写使用。net ?
例子: 输入7,326,629,显示6.98 MB
当前回答
1-liner(加上前缀常量)
const String prefixes = " KMGTPEY";
/// <summary> Returns the human-readable file size for an arbitrary, 64-bit file size. </summary>
public static String HumanSize(UInt64 bytes)
=> Enumerable
.Range(0, prefixes.Length)
.Where(i => bytes < 1024U<<(i*10))
.Select(i => $"{(bytes>>(10*i-10))/1024:0.###} {prefixes[i]}B")
.First();
或者,如果你想减少LINQ对象的分配,使用相同的for循环变量:
/// <summary>
/// Returns the human-readable file size for an arbitrary, 64-bit file size.
/// </summary>
public static String HumanSize(UInt64 bytes)
{
const String prefixes = " KMGTPEY";
for (var i = 0; i < prefixes.Length; i++)
if (bytes < 1024U<<(i*10))
return $"{(bytes>>(10*i-10))/1024:0.###} {prefixes[i]}B";
throw new ArgumentOutOfRangeException(nameof(bytes));
}
其他回答
如果你试图匹配Windows资源管理器的详细信息视图中显示的大小,这是你想要的代码:
[DllImport("shlwapi.dll", CharSet = CharSet.Unicode)]
private static extern long StrFormatKBSize(
long qdw,
[MarshalAs(UnmanagedType.LPTStr)] StringBuilder pszBuf,
int cchBuf);
public static string BytesToString(long byteCount)
{
var sb = new StringBuilder(32);
StrFormatKBSize(byteCount, sb, sb.Capacity);
return sb.ToString();
}
这不仅会与资源管理器完全匹配,而且还会为您提供翻译后的字符串,并匹配Windows版本的差异(例如在Win10中,K = 1000 vs.之前的版本K = 1024)。
string[] suffixes = { "B", "KB", "MB", "GB", "TB", "PB", "EB", "ZB", "YB" };
int s = 0;
long size = fileInfo.Length;
while (size >= 1024)
{
s++;
size /= 1024;
}
string humanReadable = String.Format("{0} {1}", size, suffixes[s]);
int size = new FileInfo( filePath ).Length / 1024;
string humanKBSize = string.Format( "{0} KB", size );
string humanMBSize = string.Format( "{0} MB", size / 1024 );
string humanGBSize = string.Format( "{0} GB", size / 1024 / 1024 );
那么递归呢:
private static string ReturnSize(double size, string sizeLabel)
{
if (size > 1024)
{
if (sizeLabel.Length == 0)
return ReturnSize(size / 1024, "KB");
else if (sizeLabel == "KB")
return ReturnSize(size / 1024, "MB");
else if (sizeLabel == "MB")
return ReturnSize(size / 1024, "GB");
else if (sizeLabel == "GB")
return ReturnSize(size / 1024, "TB");
else
return ReturnSize(size / 1024, "PB");
}
else
{
if (sizeLabel.Length > 0)
return string.Concat(size.ToString("0.00"), sizeLabel);
else
return string.Concat(size.ToString("0.00"), "Bytes");
}
}
然后你称之为:
return ReturnSize(size, string.Empty);
比如@NET3的解决方案。使用shift而不是除法来测试字节的范围,因为除法占用更多的CPU成本。
private static readonly string[] UNITS = new string[] { "B", "KB", "MB", "GB", "TB", "PB", "EB" };
public static string FormatSize(ulong bytes)
{
int c = 0;
for (c = 0; c < UNITS.Length; c++)
{
ulong m = (ulong)1 << ((c + 1) * 10);
if (bytes < m)
break;
}
double n = bytes / (double)((ulong)1 << (c * 10));
return string.Format("{0:0.##} {1}", n, UNITS[c]);
}