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

@ 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];
}

var大小=[“字节”,“知识库”,“m”,“g”,“结核”,“铅”、“海尔哥哥”,“ZB”,“YB”); 函数formatBytes(字节，小数){ For (var I = 0, r = bytes, b = 1024;R > b;I ++) r /= b; 返回' ${parseFloat(r.toFixed(decimal))} ${SIZES[i]} '; ｝

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];
    }
}

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 ;
}

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))