是的，你需要转换为字符串，以找到长度。例如

var x=100;// type of x is number
var x=100+"";// now the type of x is string
document.write(x.length);//which would output 3.

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;
}

在一次测试中，我也被问到类似的问题。

找到一个数字的长度而不转换为字符串

const numbers = [1, 10, 100, 12, 123, -1, -10, -100, -12, -123, 0, -0]

const numberLength = number => {

  let length = 0
  let n = Math.abs(number)

  do {
    n /=  10
    length++
  } while (n >= 1)

  return length
}

console.log(numbers.map(numberLength)) // [ 1, 2, 3, 2, 3, 1, 2, 3, 2, 3, 1, 1 ]

负数的添加使它更加复杂，因此Math.abs()。

试试这个:

$("#element").text().length;

它在使用中的例子

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});

你应该使用最简单的字符串(stringLength)，可读性总是胜过速度。但如果你关心速度，下面有一些。

三种不同的方法，速度各不相同。

// 34ms
let weissteinLength = function(n) { 
    return (Math.log(Math.abs(n)+1) * 0.43429448190325176 | 0) + 1;
}

// 350ms
let stringLength = function(n) {
    return n.toString().length;
}

// 58ms
let mathLength = function(n) {
    return Math.ceil(Math.log(n + 1) / Math.LN10);
}

// Simple tests below if you care about performance.

let iterations = 1000000;
let maxSize = 10000;

// ------ Weisstein length.

console.log("Starting weissteinLength length.");
let startTime = Date.now();

for (let index = 0; index < iterations; index++) {
    weissteinLength(Math.random() * maxSize);
}

console.log("Ended weissteinLength length. Took : " + (Date.now() - startTime ) + "ms");


// ------- String length slowest.

console.log("Starting string length.");
startTime = Date.now();

for (let index = 0; index < iterations; index++) {
    stringLength(Math.random() * maxSize);
}

console.log("Ended string length. Took : " + (Date.now() - startTime ) + "ms");


// ------- Math length.

console.log("Starting math length.");
startTime = Date.now();

for (let index = 0; index < iterations; index++) {
    mathLength(Math.random() * maxSize);
}