如何舍入浮点数，使小数点后只有两位尾随数字? [英] How can I round a float such that there are only two trailing digits after the decimal point?

问题描述

我有一个浮点数，例如137.57667565656，但我想四舍五入，使得小数点后只有两个尾随数字，例如新的浮点数将是137.58.

I have a float number like 137.57667565656 but I would like to round it such that there are only two trailing digits after the decimal point like the new float number will be 137.58.

到目前为止，我已经尝试过:

I tried this so far:

(Math.round((value*100)/100)).toFixed(2).toString();

但不幸的是，它将我的值四舍五入为137.00.它将小数位添加为零，为什么呢?

But unfortunately, it rounds my value to 137.00. It adds the decimals places as zeroes, why?

如何实现以上目标?

推荐答案

您期望什么?

(value*100)/100

仅返回 value 的原始值，因此

Math.round((value*100)/100))

等同于:

Math.round(value)

您将拥有:

Math.round(value).toFixed(2).toString();

因此 value 会四舍五入为整数， toFixed 将添加两个小数位并返回一个字符串，因此 toString 部分是多余的.如果您希望将 value 舍入到小数点后四位，则:

so value is rounded to an integer, toFixed will add two decimal places and return a string so the toString part is redundant. If you wish to round value to four decimal places, then:

value.toFixed(4)

将完成这项工作:

var x = 137.57667565656;

console.log(x.toFixed(4)); // 137.5767

如果要将其四舍五入，但将其显示为4，则:

If you want to round it to 2 places but present it as 4, then:

Number(x.toFixed(2)).toFixed(4)  // 137.5800

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