在PHP中,您可以。。。

range(1, 3); // Array(1, 2, 3)
range("A", "C"); // Array("A", "B", "C")

也就是说,有一个函数可以通过传递上下限来获得一系列数字或字符。

JavaScript本机是否有内置的功能?如果没有,我将如何实施?


当前回答

使用TypeScript设置(应用程序范围):

declare global {
  interface Function {
    range(count: number, start_with: number): number[];
  }
}

Function.prototype.range = function (
  count: number,
  start_with: number = 0
): number[] {
  return [...Array(count).keys()].map((key) => key + start_with);
};

使用JS设置:

Function.prototype.range = function(count, start_with=0){
    return [...Array(count).keys()].map((key) => key + start_with);
}

使用示例:

Function.range(2,0) //Will return [0,1]
Function.range(2,1) //Will return [1,2]
Function.range(2,-1) //Will return [-1,0]

其他回答

我想补充一点,我认为这是一个非常可调的版本,速度非常快。

const range = (start, end) => {
    let all = [];
    if (typeof start === "string" && typeof end === "string") {
        // Return the range of characters using utf-8 least to greatest
        const s = start.charCodeAt(0);
        const e = end.charCodeAt(0);
        for (let i = s; i <= e; i++) {
            all.push(String.fromCharCode(i));
        }
    } else if (typeof start === "number" && typeof end === "number") {
        // Return the range of numbers from least to greatest
        for(let i = end; i >= start; i--) {
            all.push(i);
        }
    } else {
        throw new Error("Did not supply matching types number or string.");
    }
    return all;
}
// usage
const aTod = range("a", "d");

如果您愿意,也可以使用打字机

const range = (start: string | number, end: string | number): string[] | number[] => {
    const all: string[] | number[] = [];
    if (typeof start === "string" && typeof end === "string") {
        const s: number = start.charCodeAt(0);
        const e: number = end.charCodeAt(0);
        for (let i = s; i <= e; i++) {
            all.push(String.fromCharCode(i));
        }
    } else if (typeof start === "number" && typeof end === "number") {
        for (let i = end; i >= start; i--) {
            all.push(i);
        }
    } else {
        throw new Error("Did not supply matching types number or string.");
    }
    return all;
}
// Usage
const negTenToten: number[] = range(-10, 10) as number[];

受到其他答案的影响。用户已离开。

已经给出了非常好的答案,但我没有看到ES6迭代器的完整使用来实现范围的完整实现,所以这里是:

/**
 * inspired by Python's built-in range utility function
 * implemented using ES6 Iterable, Iterator  protolols (interfaces)
 */
class Range {
  constructor(...args) {
    this.start = args.length <= 1 ? 0 : args[0];
    this.end = args.length <= 1 ? args[0] || 0 : args[1];
    this.step = args.length <= 2 ? 1 : args[2];
  }

  [Symbol.iterator]() {
    return this;
  }

  next() {
    if (this.end > this.start) {
      const result = { done: false, value: this.start };
      this.start = this.start + this.step;
      return result;
    } else return { done: true, value: undefined };
  }
}

/**
 * Symbol.iterator is part of the couple of inbuilt symbols
 * available in JavaScript. It allows for hooking into the
 * inbuilt for of iteration mechanism. This is why the
 * Symbol.iterator comes into play when talking about
 * iterables and iterators in JavaScript.
 */

function range(...args) {
  return new Range(...args);
}

console.log([...range(4)]);        // [0, 1, 2, 3]
console.log([...range(2, 5)]);     // [2, 3, 4]
console.log([...range(1, 10, 3)]); // [1, 4, 7]

一个可以在任一方向工作的衬垫:

const range = (a,b)=>Array(Math.abs(a-b)+1).fill(a).map((v,i)=>v+i*(a>b?-1:1));

请参阅实际操作:

常量范围=(a,b)=>数组(数学.abs(a-b)+1).fill(a).map((v,i)=>v+i*(a>b?-1:1));console.log(范围(1,4));console.log(范围(4,1));

我正在分享我的实现,以防它对某人有所帮助。

function Range(start_or_num, end = null, increment = 1) {
    const end_check = end === null
    const start = end_check  ? 0 : start_or_num
    const count = end_check ? start_or_num : Math.round((end - start) / increment) + 1
     const filterFunc = end_check  ?  x => x >= start : x => x < end && x >= start

    return [...Array.from(
        Array(count).keys(), x => increment * (x - 1) + start
    )
    ].filter(filterFunc)
}
// usage
// console.log(Range(4, 10, 2)) // [4, 6, 8]
// console.log(Range(5, 10 )) //[5, 6, 7, 8, 9]
// console.log(Range(10 ))// [0, 1, 2, 3, 4, 5, 6, 7, 8]

解决方案:

//best performance
var range = function(start, stop, step) {
    var a = [start];
    while (start < stop) {
        start += step || 1;
        a.push(start);
    }
    return a;
};

//or
var range = function(start, end) {
    return Array(++end-start).join(0).split(0).map(function(n, i) {
        return i+start
    });
}