TA貢獻(xiàn)1794條經(jīng)驗(yàn) 獲得超8個(gè)贊
聽起來這應(yīng)該適合您：
function kFormatter(num) {
return Math.abs(num) > 999 ? Math.sign(num)*((Math.abs(num)/1000).toFixed(1)) + 'k' : Math.sign(num)*Math.abs(num)
}
console.log(kFormatter(1200)); // 1.2k
console.log(kFormatter(-1200)); // -1.2k
console.log(kFormatter(900)); // 900
console.log(kFormatter(-900)); // -900
TA貢獻(xiàn)2080條經(jīng)驗(yàn) 獲得超4個(gè)贊
這是一個(gè)避免所有if語句（使用的功能Math）的簡(jiǎn)單解決方案。
var SI_SYMBOL = ["", "k", "M", "G", "T", "P", "E"];
function abbreviateNumber(number){
// what tier? (determines SI symbol)
var tier = Math.log10(number) / 3 | 0;
// if zero, we don't need a suffix
if(tier == 0) return number;
// get suffix and determine scale
var suffix = SI_SYMBOL[tier];
var scale = Math.pow(10, tier * 3);
// scale the number
var scaled = number / scale;
// format number and add suffix
return scaled.toFixed(1) + suffix;
}
舉報(bào)