function num(n) {
if (n >= 1000 && n < 10000) return (n/1000).toFixed(3) + "K";
if (n >= 10000 && n < 100000) return (n/1000).toFixed(1) + "K";
if (n >= 100000 && n < 1000000) return (n/1000).toFixed(0) + "K";
if (n >= 1000000 && n < 10000000) return (n/1000000).toFixed(3) + "M";
if (n >= 10000000 && n < 100000000) return (n/1000000).toFixed(1) + "M";
if (n >= 100000000 && n < 1000000000) return (n/1000000).toFixed(0) + "M";
if (n >= 1000000000 && n < 10000000000) return (n/1000000000).toFixed(3) + "B";
if (n >= 10000000000 && n < 100000000000) return (n/1000000000).toFixed(1) + "B";
if (n >= 100000000000 && n < 1000000000000) return (n/1000000000).toFixed(0) + "B";
if (n >= 1000000000000 && n < 10000000000000) return (n/1000000000000).toFixed(3) + "T";
if (n >= 10000000000000 && n < 100000000000000) return (n/1000000000000).toFixed(1) + "T";
if (n >= 100000000000000 && n < 1000000000000000) return (n/1000000000000).toFixed(0) + "T";
return n;
}
因为在某些时候我可能会向上发展到数百人的力量,是否有更简单的方法来做到这一点?
解决方法:
function formatNumber(number) {
var i = 0; units = [ "", "K", "M", "B", "T" ]; // etc
while (number > 1000) {
number /= 1000;
i += 1;
}
return Math.floor(number * 1000) / 1000 + units[i];
}
formatNumber(1234567); // 1.234M
formatNumber(1230567); // 1.23M
对于非常大的数字,这可能会更快:
function formatNumber(number) {
var i; units = [ "", "K", "M", "B", "T" ]; // etc
i = Math.round(Math.log(number) / Math.log(10) / 3);
number /= Math.pow(10, i * 3);
return Math.floor(number * 1000) / 1000 + units[i];
}
formatNumber(1234567); // 1.234M
formatNumber(1230567); // 1.23M