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

我只是想分享我的想法。我遇到了这个问题,所以我的解决方案是这样的。这将把低单位转换为高单位,反之亦然,只需提供参数toUnit和fromUnit

export function fileSizeConverter(size: number, fromUnit: string, toUnit: string ): number | string {
  const units: string[] = ['B', 'KB', 'MB', 'GB', 'TB'];
  const from = units.indexOf(fromUnit.toUpperCase());
  const to = units.indexOf(toUnit.toUpperCase());
  const BASE_SIZE = 1024;
  let result: number | string = 0;

  if (from < 0 || to < 0 ) { return result = 'Error: Incorrect units'; }

  result = from < to ? size / (BASE_SIZE ** to) : size * (BASE_SIZE ** from);

  return result.toFixed(2);
}

我从这里得到了灵感

其他回答

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;
}

根据al冰岛m的答案,我在小数点后去掉了0:

function formatBytes(bytes, decimals) {
    if(bytes== 0)
    {
        return "0 Byte";
    }
    var k = 1024; //Or 1 kilo = 1000
    var sizes = ["Bytes", "KB", "MB", "GB", "TB", "PB"];
    var i = Math.floor(Math.log(bytes) / Math.log(k));
    return parseFloat((bytes / Math.pow(k, i)).toFixed(decimals)) + " " + sizes[i];
}

一行程序

const b2s = t = > {let’e = Math .对数(t) / 10 | 0; return (t / 1024 * * (e = e < = 0 ? 0 toFixed: e))(3) +“BKMGP”[e]}; console . log (b2s (0)); console . log (b2s (123)); console . log (b2s (123123)); console . log (b2s (123123123)); console . log (b2s (123123123123)); console . log (b2s (123123123123123));

这是目前排名最高的答案的后续。

边界情况

我发现了一个边缘情况:非常少量的字节!具体来说,当字节数在-1和1之间(独占)时。

例如,考虑0.25字节。在这种情况下,Math.floor(Math.log(0.25) / Math.log(1024))将返回-1。由于-1不是一个有效的索引,formatBytes(0.25)将返回类似“0.25 undefined”的值。

下面是一个使用Wolfram Alpha的边缘情况的示例。

Fix

我通过添加Math来解决这个问题。马克斯(0,…):

数学。max(0, Math.floor(Math.log(bytes) / Math.log(1024))

数学。Max(0,…)确保索引值始终至少为0。

这是一个字节应该如何显示给人类:

function bytesToHuman(bytes, decimals = 2) {
  // https://en.wikipedia.org/wiki/Orders_of_magnitude_(data)
  const units = ["bytes", "KiB", "MiB", "GiB", "TiB", "PiB", "EiB"]; // etc

  let i = 0;
  let h = 0;

  let c = 1 / 1023; // change it to 1024 and see the diff

  for (; h < c && i < units.length; i++) {
    if ((h = Math.pow(1024, i) / bytes) >= c) {
      break;
    }
  }

  // remove toFixed and let `locale` controls formatting
  return (1 / h).toFixed(decimals).toLocaleString() + " " + units[i];
}

// test
for (let i = 0; i < 9; i++) {
  let val = i * Math.pow(10, i);
  console.log(val.toLocaleString() + " bytes is the same as " + bytesToHuman(val));

}

// let's fool around
console.log(bytesToHuman(1023));
console.log(bytesToHuman(1024));
console.log(bytesToHuman(1025));