3 回答
TA贡献1806条经验 获得超5个赞
对于整数:
var x=12345678;
x=x.toString();
var lastThree = x.substring(x.length-3);
var otherNumbers = x.substring(0,x.length-3);
if(otherNumbers != '')
lastThree = ',' + lastThree;
var res = otherNumbers.replace(/\B(?=(\d{2})+(?!\d))/g, ",") + lastThree;
alert(res);
对于浮点数:
var x=12345652457.557;
x=x.toString();
var afterPoint = '';
if(x.indexOf('.') > 0)
afterPoint = x.substring(x.indexOf('.'),x.length);
x = Math.floor(x);
x=x.toString();
var lastThree = x.substring(x.length-3);
var otherNumbers = x.substring(0,x.length-3);
if(otherNumbers != '')
lastThree = ',' + lastThree;
var res = otherNumbers.replace(/\B(?=(\d{2})+(?!\d))/g, ",") + lastThree + afterPoint;
alert(res);
TA贡献1834条经验 获得超8个赞
我迟到了,但我想这会有所帮助的:)
您可以使用Number.prototype.toLocaleString()
句法
numObj.toLocaleString([locales [, options]])
var number = 123456.789;
// India uses thousands/lakh/crore separators
document.getElementById('result').innerHTML = number.toLocaleString('en-IN');
// → 1,23,456.789
document.getElementById('result1').innerHTML = number.toLocaleString('en-IN', {
maximumFractionDigits: 2,
style: 'currency',
currency: 'INR'
});
// → Rs.123,456.79
<div id="result"></div>
<div id="result1"></div>
TA贡献1895条经验 获得超7个赞
对于整数,不需要其他操作。
这将匹配末尾的每个数字,其后具有1个或多个双位数字模式,并将其替换为自身+“,”:
"125465778".replace(/(\d)(?=(\d\d)+$)/g, "$1,");
-> "1,25,46,57,78"
但是,由于我们希望末尾有3个,因此让我们通过在输入的匹配末尾之前添加额外的“ \ d”来明确声明这一点:
"125465778".replace(/(\d)(?=(\d\d)+\d$)/g, "$1,");
-> "12,54,65,778"
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