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;
}

这不是关于将字节转换为其他单位，但它有助于正确显示当前地区的数字和单位:

bytes.toLocaleString(undefined, {
  style: 'unit',
  unit: 'gigabyte',
})

更多选择和细节可以在这里找到:https://v8.dev/features/intl-numberformat#units

function formatBytes(bytes) {
    var marker = 1024; // Change to 1000 if required
    var decimal = 3; // Change as required
    var kiloBytes = marker; // One Kilobyte is 1024 bytes
    var megaBytes = marker * marker; // One MB is 1024 KB
    var gigaBytes = marker * marker * marker; // One GB is 1024 MB
    var teraBytes = marker * marker * marker * marker; // One TB is 1024 GB

    // return bytes if less than a KB
    if(bytes < kiloBytes) return bytes + " Bytes";
    // return KB if less than a MB
    else if(bytes < megaBytes) return(bytes / kiloBytes).toFixed(decimal) + " KB";
    // return MB if less than a GB
    else if(bytes < gigaBytes) return(bytes / megaBytes).toFixed(decimal) + " MB";
    // return GB if less than a TB
    else return(bytes / gigaBytes).toFixed(decimal) + " GB";
}

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。

这是一个坚实的有效的方法来转换字节。你唯一需要做的就是安装mathjs库进行精确的计算。复制粘贴即可。

import { multiply, divide, round } from "mathjs";

class Size {
  constructor(value, unit) {
    this.value = value;
    this.unit = unit.toUpperCase();
  }
}

async function byteToSize(bytes) {
  const B = 1;
  const KB = multiply(B, 1024);
  const MB = multiply(KB, 1024);
  const GB = multiply(MB, 1024);
  const TB = multiply(GB, 1024);
  const PB = multiply(TB, 1024);

  if (bytes <= KB) {
    // @returns BYTE

    const result = round(divide(bytes, B));
    const unit = `B`;

    return new Size(result, unit);
  }

  if (bytes <= MB) {
    // @returns KILOBYTE

    const result = round(divide(bytes, KB));
    const unit = `KB`;

    return new Size(result, unit);
  }

  if (bytes <= GB) {
    // @returns MEGABYTE

    const result = round(divide(bytes, MB));
    const unit = `MB`;

    return new Size(result, unit);
  }

  if (bytes <= TB) {
    // @returns GIGABYTE

    const result = round(divide(bytes, GB));
    const unit = `GB`;

    return new Size(result, unit);
  }

  if (bytes <= PB) {
    // @returns TERABYTE

    const result = divide(bytes, TB).toFixed(2);
    const unit = `TB`;

    return new Size(result, unit);
  }

  if (bytes >= PB) {
    // @returns PETABYTE

    const result = divide(bytes, PB).toFixed(2);
    const unit = `PB`;

    return new Size(result, unit);
  }
}

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];
}