我得到这段代码通过PHP隐蔽大小字节。

现在我想使用JavaScript将这些大小转换为人类可读的大小。我尝试将这段代码转换为JavaScript,看起来像这样:

function formatSizeUnits(bytes){
  if      (bytes >= 1073741824) { bytes = (bytes / 1073741824).toFixed(2) + " GB"; }
  else if (bytes >= 1048576)    { bytes = (bytes / 1048576).toFixed(2) + " MB"; }
  else if (bytes >= 1024)       { bytes = (bytes / 1024).toFixed(2) + " KB"; }
  else if (bytes > 1)           { bytes = bytes + " bytes"; }
  else if (bytes == 1)          { bytes = bytes + " byte"; }
  else                          { bytes = "0 bytes"; }
  return bytes;
}

这是正确的做法吗?有没有更简单的方法?


当前回答

我有一个问题的元数据从服务器返回不同大小的单位。我使用了@Alicejim response,并试图让它更通用。在这里分享代码,也许会帮助到一些人。

enum SizeUnits {
   Bytes = 'Bytes',
   KB = 'KB',
   MB = 'MB',
   GB = 'GB',
   TB = 'TB',
   PB = 'PB',
   EB = 'EB',
   ZB = 'ZB',
   YB = 'YB'
}
function convertSizeUnites(size: number, sourceSizeUnits: SizeUnits, targetSizeUnits: SizeUnits) {
    const i = Object.keys(SizeUnits).indexOf(sourceSizeUnits);
    const sizeInBytes = size * Math.pow(1024, i);
    const j = Object.keys(SizeUnits).indexOf(targetSizeUnits);
    return sizeInBytes / Math.pow(1024, j);
}
function formatSize(size: number, measureUnit: SizeUnits, decimals = 2) {
    if (size === 0) return '0 Bytes';
    const sizeInBytes = convertSizeUnites(size, measureUnit, SizeUnits.Bytes);
    const dm = decimals < 0 ? 0 : decimals;
    const i = Math.floor(Math.log(sizeInBytes) / Math.log(1024));
    return parseFloat((sizeInBytes / Math.pow(1024, i)).toFixed(dm)) + ' ' + 
    Object.keys(SizeUnits)[i];
}

其他回答

函数by睾丸(字节){ const sizes = ['Bytes', 'KB', 'MB', 'GB', 'TB']; If (bytes === 0)返回'n/a'; const i = Math.min(Math.floor(Math.log(bytes) / Math.log(1024))),大小。长度- 1); 如果(i === 0)返回' ${bytes} ${sizes[i]} '; 返回“${(字节/(1024 * *我)).toFixed(1)} ${大小[我]}'; } console.log (bytesToSize (400000)) console.log (bytesToSize (5000000))

这个解决方案建立在以前的解决方案的基础上,但同时考虑了公制和二进制单位:

function formatBytes(bytes, decimals, binaryUnits) {
    if(bytes == 0) {
        return '0 Bytes';
    }
    var unitMultiple = (binaryUnits) ? 1024 : 1000; 
    var unitNames = (unitMultiple === 1024) ? // 1000 bytes in 1 Kilobyte (KB) or 1024 bytes for the binary version (KiB)
        ['Bytes', 'KiB', 'MiB', 'GiB', 'TiB', 'PiB', 'EiB', 'ZiB', 'YiB']: 
        ['Bytes', 'KB', 'MB', 'GB', 'TB', 'PB', 'EB', 'ZB', 'YB'];
    var unitChanges = Math.floor(Math.log(bytes) / Math.log(unitMultiple));
    return parseFloat((bytes / Math.pow(unitMultiple, unitChanges)).toFixed(decimals || 0)) + ' ' + unitNames[unitChanges];
}

例子:

formatBytes(293489203947847, 1);    // 293.5 TB
formatBytes(1234, 0);   // 1 KB
formatBytes(4534634523453678343456, 2); // 4.53 ZB
formatBytes(4534634523453678343456, 2, true));  // 3.84 ZiB
formatBytes(4566744, 1);    // 4.6 MB
formatBytes(534, 0);    // 534 Bytes
formatBytes(273403407, 0);  // 273 MB

我有一个问题的元数据从服务器返回不同大小的单位。我使用了@Alicejim response,并试图让它更通用。在这里分享代码,也许会帮助到一些人。

enum SizeUnits {
   Bytes = 'Bytes',
   KB = 'KB',
   MB = 'MB',
   GB = 'GB',
   TB = 'TB',
   PB = 'PB',
   EB = 'EB',
   ZB = 'ZB',
   YB = 'YB'
}
function convertSizeUnites(size: number, sourceSizeUnits: SizeUnits, targetSizeUnits: SizeUnits) {
    const i = Object.keys(SizeUnits).indexOf(sourceSizeUnits);
    const sizeInBytes = size * Math.pow(1024, i);
    const j = Object.keys(SizeUnits).indexOf(targetSizeUnits);
    return sizeInBytes / Math.pow(1024, j);
}
function formatSize(size: number, measureUnit: SizeUnits, decimals = 2) {
    if (size === 0) return '0 Bytes';
    const sizeInBytes = convertSizeUnites(size, measureUnit, SizeUnits.Bytes);
    const dm = decimals < 0 ? 0 : decimals;
    const i = Math.floor(Math.log(sizeInBytes) / Math.log(1024));
    return parseFloat((sizeInBytes / Math.pow(1024, i)).toFixed(dm)) + ' ' + 
    Object.keys(SizeUnits)[i];
}

这里有一句话:

val => ['Bytes','Kb','Mb','Gb','Tb'][Math.floor(Math.log2(val)/10)]

甚至:

v => 'BKMGT'[~~(Math.log2(v)/10)]

与数:

function shortenBytes(n) {
    const k = n > 0 ? Math.floor((Math.log2(n)/10)) : 0;
    const rank = (k > 0 ? 'KMGT'[k - 1] : '') + 'b';
    const count = Math.floor(n / Math.pow(1024, k));
    return count + rank;
}

我用react和typescript就能做到。

export const FormatBytes = (bytes: number) => {
  const units = ['b', 'kb', 'mb', 'gb', 'tb'];

  let i = 0;

  for (i; bytes >= 1024 && i < 4; i++) {
    bytes /= 1024;
  }

  return `${bytes.toFixed(2)} ${units[i]}`;
};