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

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。

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";
}

更灵活和考虑最大pow尺寸列表 (升级后的l2aelba答案)

function formatBytes(bytes, decimals = 2, isBinary = false) {
      const sizes = ['Bytes', 'KB', 'MB', 'GB', 'TB']; // or ['B', 'KB', 'MB', 'GB', 'TB']
    
      if (!+bytes) {
        return `0 ${sizes[0]}`;
      }
    
      const inByte = isBinary ? 1024 : 1000;
      const dm = decimals < 0 ? 0 : decimals;
    
      const pow = Math.floor(Math.log(bytes) / Math.log(inByte));
      const maxPow = Math.min(pow, sizes.length - 1);
    
      return `${parseFloat((bytes / Math.pow(inByte, maxPow)).toFixed(dm))} ${
        sizes[maxPow]
      }`;
    }

这对我很有用。

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

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

    return bytes.toFixed(1) + ' ' + units[i]
}

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!