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);
  }
}

当与字节相关时，有两种真实的方法来表示大小，它们是SI单位(10^3)或IEC单位(2^10)。也有JEDEC，但他们的方法是模糊的和令人困惑的。我注意到其他示例有错误，例如使用KB而不是KB来表示千字节，因此我决定编写一个函数，使用当前接受的度量单位的范围来解决这些情况。

在结尾有一个格式化的地方，这将使数字看起来更好一点(至少在我看来)，如果它不适合你的目的，请随意删除这个格式。

享受。

// pBytes: the size in bytes to be converted.
// pUnits: 'si'|'iec' si units means the order of magnitude is 10^3, iec uses 2^10

function prettyNumber(pBytes, pUnits) {
    // Handle some special cases
    if(pBytes == 0) return '0 Bytes';
    if(pBytes == 1) return '1 Byte';
    if(pBytes == -1) return '-1 Byte';

    var bytes = Math.abs(pBytes)
    if(pUnits && pUnits.toLowerCase() && pUnits.toLowerCase() == 'si') {
        // SI units use the Metric representation based on 10^3 as a order of magnitude
        var orderOfMagnitude = Math.pow(10, 3);
        var abbreviations = ['Bytes', 'kB', 'MB', 'GB', 'TB', 'PB', 'EB', 'ZB', 'YB'];
    } else {
        // IEC units use 2^10 as an order of magnitude
        var orderOfMagnitude = Math.pow(2, 10);
        var abbreviations = ['Bytes', 'KiB', 'MiB', 'GiB', 'TiB', 'PiB', 'EiB', 'ZiB', 'YiB'];
    }
    var i = Math.floor(Math.log(bytes) / Math.log(orderOfMagnitude));
    var result = (bytes / Math.pow(orderOfMagnitude, i));

    // This will get the sign right
    if(pBytes < 0) {
        result *= -1;
    }

    // This bit here is purely for show. it drops the percision on numbers greater than 100 before the units.
    // it also always shows the full number of bytes if bytes is the unit.
    if(result >= 99.995 || i==0) {
        return result.toFixed(0) + ' ' + abbreviations[i];
    } else {
        return result.toFixed(2) + ' ' + abbreviations[i];
    }
}

@ al冰岛jm也给出了同样的答案，但以一种“更说教”的方式。谢谢!= D

function formatBytes(numBytes, decPlaces) {
    /* Adjust the number of bytes informed for the most appropriate metric according
    to its value.

    Args:
        numBytes (number): The number of bytes (integer);
        decPlaces (Optional[number])): The number of decimal places (integer). If
            it is "undefined" the value "2" will be adopted.

    Returns:
        string: The number adjusted together with the most appropriate metric. */

    if (numBytes === 0) {
        return "0 Bytes";
    }

    // NOTE: 1 KB is equal to 1024 Bytes. By Questor
    // [Ref(s).: https://en.wikipedia.org/wiki/Kilobyte ]
    var oneKByte = 1024;

    // NOTE: Treats if the "decPlaces" is "undefined". If it is "undefined" the value
    // "2" will be adopted. By Questor
    if (decPlaces === undefined || decPlaces === "") {
        decPlaces = 2;
    }

    var byteMtrcs = ["Bytes", "KB", "MB", "GB", "TB", "PB", "EB", "ZB", "YB"];
    // Byte Metrics

    // NOTE: Defines the factor for the number of bytes and the metric. By Questor
    var mtrcNumbFactor = Math.floor(Math.log(numBytes) / Math.log(oneKByte));
    // Metrics Number Factor

    return parseFloat((numBytes / Math.pow(oneKByte, mtrcNumbFactor)).
            toFixed(decPlaces)) + " " + byteMtrcs[mtrcNumbFactor];
}

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。

我让算法快了7倍(就像忍者一样):

function rafaelFormatBytes(number){ if(number == null || number === undefined || number <= 0) { return '0 Bytes'; } var scaleCounter = 0; var scaleInitials = [' Bytes',' KB',' MB',' GB',' TB',' PB',' EB',' ZB',' YB']; while (number >= 1024 && scaleCounter < scaleInitials.length - 1){ number /= 1024; scaleCounter++; } if(scaleCounter >= scaleInitials.length) scaleCounter = scaleInitials.length - 1; var compactNumber = number.toFixed(2) .replace(/\.?0+$/,'') .replace(/\B(?=(\d{3})+(?!\d))/g, ","); compactNumber += scaleInitials[scaleCounter]; return compactNumber.trim(); } var testNum = 0; var startTime, endTime; function start() { startTime = new Date(); }; function end() { endTime = new Date(); var timeDiff = endTime - startTime; //in ms // strip the ms timeDiff /= 1000; // get seconds var seconds = Math.round(timeDiff, 5); console.log(timeDiff + " seconds"); } function formatBytes(a,b=2,k=1024){with(Math){let d=floor(log(a)/log(k));return 0==a?"0 Bytes":parseFloat((a/pow(k,d)).toFixed(max(0,b)))+" "+["Bytes","KB","MB","GB","TB","PB","EB","ZB","YB"][d]}} console.log(formatBytes(1000000000000000000000000000)); console.log(rafaelFormatBytes(1000000000000000000000000000)); start(); for(i=0; i<100000; i++){ formatBytes(1000000000000000); } end(); start(); for(i=0; i<100000; i++){ rafaelFormatBytes(1000000000000000); } end();

．.． 输出:

827.18 YB
827.18 YB
0.293 seconds
0.039 seconds

耶稣基督!

只是稍微修改了@zayarTun答案的代码，以包括一个额外的参数，表示结果中的小数数(如果小数为零，则不需要显示像15.00 KB这样的结果，而是15 KB就足够了，这就是为什么我在parseFloat()中包装结果值)

  bytesForHuman(bytes, decimals = 2) {
    let units = ['B', 'KB', 'MB', 'GB', 'TB', 'PB']

    let i = 0
    
    for (i; bytes > 1024; i++) {
        bytes /= 1024;
    }

    return parseFloat(bytes.toFixed(decimals)) + ' ' + units[i]
  }