我得到这段代码通过PHP隐蔽大小字节。

现在我想使用JavaScript将这些大小转换为人类可读的大小。我尝试将这段代码转换为JavaScript,看起来像这样:

function formatSizeUnits(bytes){
  if      (bytes >= 1073741824) { bytes = (bytes / 1073741824).toFixed(2) + " GB"; }
  else if (bytes >= 1048576)    { bytes = (bytes / 1048576).toFixed(2) + " MB"; }
  else if (bytes >= 1024)       { bytes = (bytes / 1024).toFixed(2) + " KB"; }
  else if (bytes > 1)           { bytes = bytes + " bytes"; }
  else if (bytes == 1)          { bytes = bytes + " byte"; }
  else                          { bytes = "0 bytes"; }
  return bytes;
}

这是正确的做法吗?有没有更简单的方法?


当前回答

const byteConversion = (bytes: number, decimals = 2) => {
if (bytes === 0) return '0 B';

const kiloByte = 1000;
const decimal = decimals < 0 ? 0 : decimals;
const sizes = ['B', 'KB', 'MB', 'GB', 'TB', 'PB', 'EB', 'ZB', 'YB'];

const i: number = Math.floor(Math.log(bytes) / Math.log(kiloByte));

return `${parseFloat((bytes / kiloByte ** i).toFixed(decimal))} ${sizes[i]}`;

};

其他回答

由此可见:(来源)


Unminified and es6’ed:(社区)

function formatBytes(bytes, decimals = 2) { if (!+bytes) return '0 Bytes' const k = 1024 const dm = decimals < 0 ? 0 : decimals const sizes = ['Bytes', 'KB', 'MB', 'GB', 'TB', 'PB', 'EB', 'ZB', 'YB'] const i = Math.floor(Math.log(bytes) / Math.log(k)) return `${parseFloat((bytes / Math.pow(k, i)).toFixed(dm))} ${sizes[i]}` } // Demo code document.body.innerHTML += `<input type="text" oninput="document.querySelector('p').innerHTML=formatBytes(this.value)" value="1000"><p>1000 Bytes</p>`

简化版(由StackOverflow社区提供,由JSCompress提供)

function formatBytes(a,b=2){if(!+a)return"0 Bytes";const c=0>b?0:b,d=Math.floor(Math.log(a)/Math.log(1024));return`${parseFloat((a/Math.pow(1024,d)).toFixed(c))} ${["Bytes","KB","MB","GB","TB","PB","EB","ZB","YB"][d]}`}

用法:

// formatBytes(bytes, decimals)

formatBytes(1024)       // 1 KB
formatBytes('1024')     // 1 KB
formatBytes(1234)       // 1.21 KB
formatBytes(1234, 3)    // 1.205 KB
formatBytes(0)          // 0 Bytes
formatBytes('0')        // 0 Bytes

PS:更改k = 1000或大小=["…如你所愿(比特或字节)

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

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 ;
}

一行程序

const b2s = t = > {let’e = Math .对数(t) / 10 | 0; return (t / 1024 * * (e = e < = 0 ? 0 toFixed: e))(3) +“BKMGP”[e]}; console . log (b2s (0)); console . log (b2s (123)); console . log (b2s (123123)); console . log (b2s (123123123)); console . log (b2s (123123123123)); console . log (b2s (123123123123123));

我使用递归和分配水平变量为适当的单位。

函数getReadableByte(count, decimal=0, level=0) { 让unitList =[“字节”,“知识库”,“m”,“g”、“肺结核”,“PT”); if (count >= 1024.0 && (level+1 < unitList.length)) { 返回getReadableByte(count/1024, decimal, ++level) } 返回' ${小数?(count).toFixed(decimal): Math.round(count)}${unitList[level]} ' } 2) console.log (getReadableByte (120)