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?
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+$/,'')
.replace(/\B(?=(\d{3})+(?!\d))/g, ",");
compactNumber += scaleInitials[scaleCounter];
return compactNumber.trim();
}
var testNum = 0;
var startTime, endTime;
function start() {
startTime = new Date();
};
function end() {
endTime = new Date();
var timeDiff = endTime - startTime; //in ms
// strip the ms
timeDiff /= 1000;
// get seconds
var seconds = Math.round(timeDiff, 5);
console.log(timeDiff + " seconds");
}
function formatBytes(a,b=2,k=1024){with(Math){let d=floor(log(a)/log(k));return 0==a?"0 Bytes":parseFloat((a/pow(k,d)).toFixed(max(0,b)))+" "+["Bytes","KB","MB","GB","TB","PB","EB","ZB","YB"][d]}}
console.log(formatBytes(1000000000000000000000000000));
console.log(rafaelFormatBytes(1000000000000000000000000000));
start();
for(i=0; i<100000; i++){
formatBytes(1000000000000000);
}
end();
start();
for(i=0; i<100000; i++){
rafaelFormatBytes(1000000000000000);
}
end();
．.． 输出:
827.18 YB
827.18 YB
0.293 seconds
0.039 seconds
耶稣基督!
我让算法快了7倍(就像忍者一样):
function rafaelFormatBytes(number){
if(number == null || number === undefined || number <= 0) {
return '0 Bytes';
}
var scaleCounter = 0;
var scaleInitials = [' Bytes',' KB',' MB',' GB',' TB',' PB',' EB',' ZB',' YB'];
while (number >= 1024 && scaleCounter < scaleInitials.length - 1){
number /= 1024;
scaleCounter++;
}
if(scaleCounter >= scaleInitials.length) scaleCounter = scaleInitials.length - 1;
var compactNumber = number.toFixed(2)
.replace(/\.?0+$/,'')
.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!
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。
