假设我有一个值15.7784514，我想把它显示为15.77，没有舍入。
var num = parseFloat(15.7784514);
document.write(num.toFixed(1)+"<br />");
document.write(num.toFixed(2)+"<br />");
document.write(num.toFixed(3)+"<br />");
document.write(num.toFixed(10));
结果:
15.8
15.78
15.778
15.7784514000
如何显示15.77?
当前回答
我选择写这个来手动删除剩余的字符串，这样我就不必处理数字带来的数学问题:
num = num.toString(); //If it's not already a String
num = num.slice(0, (num.indexOf("."))+3); //With 3 exposing the hundredths place
Number(num); //If you need it back as a Number
这将得到“15.77”，其中num = 15.7784514;
其他回答
已经有一些合适的答案与正则表达式和算术计算，你也可以试试这个
function myFunction() {
var str = 12.234556;
str = str.toString().split('.');
var res = str[1].slice(0, 2);
document.getElementById("demo").innerHTML = str[0]+'.'+res;
}
// output: 12.23
将数字转换为字符串，匹配到小数点后第二位:
功能召唤(形式) 原型。值，圆 var with2Decimals toString () = num。竞赛(- ^ \ d + (? d: \。{0.2的)? /)[0 - 9] rounded。价值2决定 的 <形式onsubmit=“报复性calc(this)”> 原始号码:<输入方式/> <br /> ' Rounded number: < name name=" Rounded" type="文本" placeholder="readonly" readonly> < / form >
与toString不同，toFixed方法在某些情况下会失败，所以要非常小心。
function limitDecimalsWithoutRounding(val, decimals){
let parts = val.toString().split(".");
return parseFloat(parts[0] + "." + parts[1].substring(0, decimals));
}
var num = parseFloat(15.7784514);
var new_num = limitDecimalsWithoutRounding(num, 2);
基于David D的回答:
function NumberFormat(num,n) {
var num = (arguments[0] != null) ? arguments[0] : 0;
var n = (arguments[1] != null) ? arguments[1] : 2;
if(num > 0){
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length > 1) {
if(n > 0){
var temp = numarr[0] + ".";
for(var i = 0; i < n; i++){
if(i < numarr[1].length){
temp += numarr[1].charAt(i);
}
}
num = Number(temp);
}
}
}
}
return Number(num);
}
console.log('NumberFormat(123.85,2)',NumberFormat(123.85,2));
console.log('NumberFormat(123.851,2)',NumberFormat(123.851,2));
console.log('NumberFormat(0.85,2)',NumberFormat(0.85,2));
console.log('NumberFormat(0.851,2)',NumberFormat(0.851,2));
console.log('NumberFormat(0.85156,2)',NumberFormat(0.85156,2));
console.log('NumberFormat(0.85156,4)',NumberFormat(0.85156,4));
console.log('NumberFormat(0.85156,8)',NumberFormat(0.85156,8));
console.log('NumberFormat(".85156",2)',NumberFormat(".85156",2));
console.log('NumberFormat("0.85156",2)',NumberFormat("0.85156",2));
console.log('NumberFormat("1005.85156",2)',NumberFormat("1005.85156",2));
console.log('NumberFormat("0",2)',NumberFormat("0",2));
console.log('NumberFormat("",2)',NumberFormat("",2));
console.log('NumberFormat(85156,8)',NumberFormat(85156,8));
console.log('NumberFormat("85156",2)',NumberFormat("85156",2));
console.log('NumberFormat("85156.",2)',NumberFormat("85156.",2));
// NumberFormat(123.85,2) 123.85
// NumberFormat(123.851,2) 123.85
// NumberFormat(0.85,2) 0.85
// NumberFormat(0.851,2) 0.85
// NumberFormat(0.85156,2) 0.85
// NumberFormat(0.85156,4) 0.8515
// NumberFormat(0.85156,8) 0.85156
// NumberFormat(".85156",2) 0.85
// NumberFormat("0.85156",2) 0.85
// NumberFormat("1005.85156",2) 1005.85
// NumberFormat("0",2) 0
// NumberFormat("",2) 0
// NumberFormat(85156,8) 85156
// NumberFormat("85156",2) 85156
// NumberFormat("85156.",2) 85156
我也面临着同样的问题，并决定在TS中使用字符串操作。
如果没有足够的小数，它将返回原始值
const getDecimalsWithoutRounding = (value: number, numberOfDecimals: number) => {
const stringValue: string = value?.toString();
const dotIdx: number = stringValue?.indexOf('.');
if (dotIdx) {
return parseFloat(stringValue.slice(0, dotIdx + numberOfDecimals + 1));
} else {
return value;
}
};
console.log(getDecimalsWithoutRounding(3.34589, 2)); /// 3.34
console.log(getDecimalsWithoutRounding(null, 2)); ///null
console.log(getDecimalsWithoutRounding(55.123456789, 5)); /// 55.12345
console.log(getDecimalsWithoutRounding(10, 2)); /// 10
console.log(getDecimalsWithoutRounding(10.6, 5)); /// 10.6
