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

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];
    }
}

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;
}

使用位操作将是一个更好的解决方案。试试这个

function formatSizeUnits(bytes)
{
    if ( ( bytes >> 30 ) & 0x3FF )
        bytes = ( bytes >>> 30 ) + '.' + ( bytes & (3*0x3FF )) + 'GB' ;
    else if ( ( bytes >> 20 ) & 0x3FF )
        bytes = ( bytes >>> 20 ) + '.' + ( bytes & (2*0x3FF ) ) + 'MB' ;
    else if ( ( bytes >> 10 ) & 0x3FF )
        bytes = ( bytes >>> 10 ) + '.' + ( bytes & (0x3FF ) ) + 'KB' ;
    else if ( ( bytes >> 1 ) & 0x3FF )
        bytes = ( bytes >>> 1 ) + 'Bytes' ;
    else
        bytes = bytes + 'Byte' ;
    return bytes ;
}

function bytes2Size(byteVal){
    var units=["Bytes", "KB", "MB", "GB", "TB"];
    var kounter=0;
    var kb= 1024;
    var div=byteVal/1;
    while(div>=kb){
        kounter++;
        div= div/kb;
    }
    return div.toFixed(1) + " " + units[kounter];
}

我正在更新@ al冰岛的答案。由于小数点对于1,2位数的数字很重要，所以我舍入了第一个小数点并保留第一个小数点。对于3位数的数字，我舍入个位数，忽略所有小数点后的位置。

getMultiplers : function(bytes){
    var unit = 1000 ;
    if (bytes < unit) return bytes ;
    var exp = Math.floor(Math.log(bytes) / Math.log(unit));
    var pre = "kMGTPE".charAt(exp-1);
    var result = bytes / Math.pow(unit, exp);
    if(result/100 < 1)
        return (Math.round( result * 10 ) / 10) +pre;
    else
        return Math.round(result) + pre;
}