我发现了一种使用ES6默认参数实现这一点的新方法。它非常漂亮，因为它允许一个参数或两个参数。这里是：

function random(n, b = 0) {
    return Math.random() * (b-n) + n;
}

根据@Francisc的答案中的int版本添加具有固定精度版本的float：

function randomFloatFromInterval (min, max, fractionDigits) {
  const fractionMultiplier = Math.pow(10, fractionDigits)
  return Math.round(
    (Math.random() * (max - min) + min) * fractionMultiplier,
  ) / fractionMultiplier
}

so:

randomFloatFromInterval(1,3,4) // => 2.2679, 1.509, 1.8863, 2.9741, ...

对于int答案

randomFloatFromInterval(1,3,0) // => 1, 2, 3

函数randomIntFromInterval（min，max）｛//包括min和maxreturn Math.floor（Math.random（）*（max-min+1）+min）}常量rndInt=随机IntFromInterval（1，6）控制台日志（rndInt）

它的“额外”之处在于它允许不以1开头的随机间隔。例如，你可以得到一个从10到15的随机数。灵活性

尽管有许多答案，但结果几乎相同。我想补充一下我的答案并解释一下它的作用。因为理解它的工作方式比复制粘贴单行代码更重要。生成随机数只是简单的数学。

代码：

function getR(lower, upper) {

  var percent = (Math.random() * 100);
  // this will return number between 0-99 because Math.random returns decimal number from 0-0.9929292 something like that
  //now you have a percentage, use it find out the number between your INTERVAL :upper-lower 
  var num = ((percent * (upper - lower) / 100));
  //num will now have a number that falls in your INTERVAL simple maths
  num += lower;
  //add lower to make it fall in your INTERVAL
  //but num is still in decimal
  //use Math.floor>downward to its nearest integer you won't get upper value ever
  //use Math.ceil>upward to its nearest integer upper value is possible
  //Math.round>to its nearest integer 2.4>2 2.5>3   both lower and upper value possible
  console.log(Math.floor(num), Math.ceil(num), Math.round(num));
}

其他解决方案：

（数学随机（）*6|0）+1~~（数学随机（）*6）+1

联机尝试

这个简单的功能很方便，在任何情况下都可以使用（经过充分测试）。此外，结果的分布已经过充分测试，100%正确。

function randomInteger(pMin = 1, pMax = 1_000_000_000)
//Author: Axel Gauffre. 
//Here: https://stackoverflow.com/a/74636954/5171000
//Inspired by: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/random#getting_a_random_number_between_two_values
//
//This function RETURNS A RANDOM INTEGER between pMin (INCLUDED) and pMax (INCLUDED).
//  - pMin and pMax should be integers.
//  - HOWEVER, if pMin and/or pMax are FLOATS, they will be ROUNDED to the NEAREST integer.
//  - NEGATIVE values ARE supported.
//  - The ORDER of the 2 arguments has NO consequence: If pMin > pMax, then pMin and pMax will simply be SWAPPED.
//  - If pMin is omitted, it will DEFAULT TO 1.
//  - If pMax is omitted, it will DEFAULT TO 1 BILLION.
//
//This function works in ANY cases (fully tested).
//Also, the distribution of the results has been fully tested and is 100% correct.
{
    pMin = Math.round(pMin);
    pMax = Math.round(pMax);
    if (pMax < pMin) { let t = pMin; pMin = pMax; pMax = t;}
    return Math.floor(Math.random() * (pMax+1 - pMin) + pMin);
}