为了求长度，你必须把数字变成字符串

var num = 123;

alert((num + "").length);

or

alert(num.toString().length);

var x = 1234567;
String(x).length;

它比. tostring()(在接受的答案中)短。

为了求长度，你必须把数字变成字符串

var num = 123;

alert((num + "").length);

or

alert(num.toString().length);

首先将其转换为字符串:

var mynumber = 123;
alert((""+mynumber).length);

添加一个空字符串将隐式地导致mynumber变成一个字符串。

I'm perplex about converting into a string the given number because such an algorithm won't be robust and will be prone to errors: it will show all its limitations especially in case it has to evaluate very long numbers. In fact before converting the long number into a string it will "collapse" into its exponential notation equivalent (example: 1.2345e4). This notation will be converted into a string and this resulting string will be evaluated for returning its length. All of this will give a wrong result. So I suggest not to use that approach.

看看下面的代码，并运行代码片段来比较不同的行为:

let num = 116234567891011121415113441236542134465236441625344625344625623456723423523429798771121411511034412365421344652364416253446253446254461253446221314623879235441623683749283441136232514654296853446323214617456789101112141511344122354416236837492834411362325146542968534463232146172368374928344113623251465429685; let lenFromMath; let lenFromString; // The suggested way: lenFromMath = Math.ceil(Math.log10(num + 1)); // this works in fact returns 309 // The discouraged way: lenFromString = String(num).split("").length; // this doesn't work in fact returns 23 /*It is also possible to modify the prototype of the primitive "Number" (but some programmer might suggest this is not a good practice). But this is will also work:*/ Number.prototype.lenght = () => {return Math.ceil(Math.log10(num + 1));} lenFromPrototype = num.lenght(); console.log({lenFromMath, lenFromPrototype, lenFromString});

我想纠正@Neal的答案，这对整数来说很好，但在前一种情况下，数字1将返回0的长度。

function Longueur(numberlen)
{
    var length = 0, i; //define `i` with `var` as not to clutter the global scope
    numberlen = parseInt(numberlen);
    for(i = numberlen; i >= 1; i)
    {
        ++length;
        i = Math.floor(i/10);
    }
    return length;
}