在ES6中，您可以执行以下操作：

数组（N）.fill（）.map（（e，i）=>i+1）；

http://jsbin.com/molabiluwa/edit?js安慰

编辑：更新问题后，将数组（45）更改为数组（N）。

控制台日志(数组（45）.填充（0）.映射（（e，i）=>i+1）);

function range(start, end) {
    var foo = [];
    for (var i = start; i <= end; i++) {
        foo.push(i);
    }
    return foo;
}

然后由调用

var foo = range(1, 5);

在Javascript中没有内置的方法来实现这一点，但如果您需要多次执行，则可以创建一个非常有效的实用程序函数。

编辑：在我看来，以下是一个更好的范围函数。也许只是因为我对LINQ有偏见，但我认为它在更多情况下更有用。您的里程数可能有所不同。

function range(start, count) {
    if(arguments.length == 1) {
        count = start;
        start = 0;
    }

    var foo = [];
    for (var i = 0; i < count; i++) {
        foo.push(start + i);
    }
    return foo;
}

试试这个

[…Array.from（｛length:30｝）.keys（）]

Array.from({ length: (stop - start) / step + 1}, (_, i) => start + (i * step));

资料来源：https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/from

https://stackoverflow.com/a/49577331/8784402

使用Delta

对于javascript

smallest and one-liner

[...Array(N)].map((v, i) => from + i * step);

示例和其他备选方案

Array.from(Array(10).keys()).map(i => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]

[...Array(10).keys()].map(i => 4 + i * -2);
//=> [4, 2, 0, -2, -4, -6, -8, -10, -12, -14]

Array(10).fill(0).map((v, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]

Array(10).fill().map((v, i) => 4 + i * -2);
//=> [4, 2, 0, -2, -4, -6, -8, -10, -12, -14]

[...Array(10)].map((v, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]

Range Function

const range = (from, to, step) =>
  [...Array(Math.floor((to - from) / step) + 1)].map((_, i) => from + i * step);

range(0, 9, 2);
//=> [0, 2, 4, 6, 8]

// can also assign range function as static method in Array class (but not recommended )
Array.range = (from, to, step) =>
  [...Array(Math.floor((to - from) / step) + 1)].map((_, i) => from + i * step);

Array.range(2, 10, 2);
//=> [2, 4, 6, 8, 10]

Array.range(0, 10, 1);
//=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Array.range(2, 10, -1);
//=> []

Array.range(3, 0, -1);
//=> [3, 2, 1, 0]

As Iterators

class Range {
  constructor(total = 0, step = 1, from = 0) {
    this[Symbol.iterator] = function* () {
      for (let i = 0; i < total; yield from + i++ * step) {}
    };
  }
}

[...new Range(5)]; // Five Elements
//=> [0, 1, 2, 3, 4]
[...new Range(5, 2)]; // Five Elements With Step 2
//=> [0, 2, 4, 6, 8]
[...new Range(5, -2, 10)]; // Five Elements With Step -2 From 10
//=>[10, 8, 6, 4, 2]
[...new Range(5, -2, -10)]; // Five Elements With Step -2 From -10
//=> [-10, -12, -14, -16, -18]

// Also works with for..of loop
for (i of new Range(5, -2, 10)) console.log(i);
// 10 8 6 4 2

As Generators Only

const Range = function* (total = 0, step = 1, from = 0) {
  for (let i = 0; i < total; yield from + i++ * step) {}
};

Array.from(Range(5, -2, -10));
//=> [-10, -12, -14, -16, -18]

[...Range(5, -2, -10)]; // Five Elements With Step -2 From -10
//=> [-10, -12, -14, -16, -18]

// Also works with for..of loop
for (i of Range(5, -2, 10)) console.log(i);
// 10 8 6 4 2

// Lazy loaded way
const number0toInf = Range(Infinity);
number0toInf.next().value;
//=> 0
number0toInf.next().value;
//=> 1
// ...

带步长/增量的从到

using iterators

class Range2 {
  constructor(to = 0, step = 1, from = 0) {
    this[Symbol.iterator] = function* () {
      let i = 0,
        length = Math.floor((to - from) / step) + 1;
      while (i < length) yield from + i++ * step;
    };
  }
}
[...new Range2(5)]; // First 5 Whole Numbers
//=> [0, 1, 2, 3, 4, 5]

[...new Range2(5, 2)]; // From 0 to 5 with step 2
//=> [0, 2, 4]

[...new Range2(5, -2, 10)]; // From 10 to 5 with step -2
//=> [10, 8, 6]

using Generators

const Range2 = function* (to = 0, step = 1, from = 0) {
  let i = 0,
    length = Math.floor((to - from) / step) + 1;
  while (i < length) yield from + i++ * step;
};

[...Range2(5, -2, 10)]; // From 10 to 5 with step -2
//=> [10, 8, 6]

let even4to10 = Range2(10, 2, 4);
even4to10.next().value;
//=> 4
even4to10.next().value;
//=> 6
even4to10.next().value;
//=> 8
even4to10.next().value;
//=> 10
even4to10.next().value;
//=> undefined

对于字体

class _Array<T> extends Array<T> {
  static range(from: number, to: number, step: number): number[] {
    return Array.from(Array(Math.floor((to - from) / step) + 1)).map(
      (v, k) => from + k * step
    );
  }
}
_Array.range(0, 9, 1);

我没有看到任何基于递归函数的解决方案（我自己也从未编写过递归函数），所以这里是我的尝试。

注意array.push（something）返回数组的新长度：

(a=[]).push(a.push(a.push(0))) //  a = [0, 1, 2]

使用递归函数：

var a = (function f(s,e,a,n){return ((n?n:n=s)>e)?a:f(s,e,a?a:a=[],a.push(n)+s)})(start,end) // e.g., start = 1, end = 5

编辑：其他两种解决方案

var a = Object.keys(new Int8Array(6)).map(Number).slice(1)

and

var a = []
var i=setInterval(function(){a.length===5?clearInterval(i):a.push(a.length+1)})