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);

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()。

我一直在node.js中使用这个功能，这是我目前为止最快的实现:

var nLength = function(n) { 
    return (Math.log(Math.abs(n)+1) * 0.43429448190325176 | 0) + 1; 
}

它应该处理正整数和负整数(也是指数形式)，并应该以浮点数形式返回整数部分的长度。

下面的参考文献应该提供一些关于该方法的见解: Eric Weisstein;“数字长度。”来自MathWorld—Wolfram Web资源。

我相信一些位操作可以取代数学。但是jsperf显示Math. abs。Abs在大多数js引擎中工作得很好。

更新:正如评论中提到的，这个解决方案有一些问题:(

Update2(解决方案):我相信在某些时候精度问题开始出现，Math.log(…)*0.434…只是表现出人意料。但是，如果Internet Explorer或移动设备不是你的菜，你可以用Math代替这个操作。log10函数。在Node.js中，我写了一个快速的基本测试函数nLength = (n) => 1 + Math.log10(Math.abs(n) + 1) | 0;还有数学。Log10它像预期的那样工作。请注意数学。Log10不是普遍支持的。

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;
}

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});