我想格式化我的数字，总是显示2小数点后，四舍五入适用的地方。
例子:
number display
------ -------
1 1.00
1.341 1.34
1.345 1.35
我一直在用这个:
parseFloat(num).toFixed(2);
但是它把1显示为1，而不是1.00。
当前回答
Number(1).toFixed(2); // 1.00
Number(1.341).toFixed(2); // 1.34
Number(1.345).toFixed(2); // 1.34 NOTE: See andy's comment below.
Number(1.3450001).toFixed(2); // 1.35
. getelementbyid(“line1”)。innerHTML = Number(1).toFixed(2); . getelementbyid(么)。innerHTML = Number(1.341).toFixed(2); . getelementbyid(“line3”)。innerHTML = Number(1.345).toFixed(2); . getelementbyid(“line4”)。innerHTML = Number(1.3450001).toFixed(2); < span id = " line1 " > < / span > < br / > < span id = "么" > < / span > < br / > < span id = " line3 " > < / span > < br / > < span id = " line4 " > < / span >
其他回答
你在找地板吗?
变量数 = 1.42482; var num2 = 1; var fnum = Math.floor（num）.toFixed（2）; var fnum2 = Math.floor（num2）.toFixed（2）; console.log（fnum + “ 和 ” + fnum2）;两个值均为 1.00
我建议你用
new Intl.NumberFormat('de-DE', { minimumFractionDigits: 2, maximumFractionDigits: 2 }).format(num)
这样，您还将拥有指定国家的本地格式，并且它将确保显示精确的2个小数(无论num是1还是1.12345，它将分别显示1.00和1.12)
在这个例子中，我使用德语本地化，因为我想我的数字显示与千分隔符，所以这将是一些输出:
1 => 1,00
1.12 => 1,12
1.1234 => 1,12
1234 => 1.234,00
1234.1234 => 1.234,12
var quantity = 12;
var import1 = 12.55;
var total = quantity * import1;
var answer = parseFloat(total).toFixed(2);
document.write(answer);
parseInt(number * 100) / 100;为我工作。
RegExp -替代方法
在输入时，你有字符串(因为你使用解析)，所以我们可以通过只使用字符串操作和整数计算得到结果
let toFix2 = (n) = > n.replace (d / \(?) -(+)。\ \ \ \ d (d + 1) / (_), s, i、d (r) = {> 让k= (+r[0]>=5)+ +d - (r==5 && s=='-'); 返回s +(+i+(k>99)) + "。"(k +(> 99%) ? ? 9: 00”(k≥0 " k " + k)); }) / /测试 console.log toFix2(“1”)); console.log (toFix2 1.341”()); console.log (toFix2 1.345”()); console.log (toFix2 1.005”());
解释
s is sign, i is integer part, d are first two digits after dot, r are other digits (we use r[0] value to calc rounding) k contains information about last two digits (represented as integer number) if r[0] is >=5 then we add 1 to d - but in case when we have minus number (s=='-') and r is exact equal to 5 then in this case we substract 1 (for compatibility reasons - in same way Math.round works for minus numbers e.g Math.round(-1.5)==-1) after that if last two digits k are greater than 99 then we add one to integer part i
