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

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

带增量/步长

smallest and one-liner
[...Array(N)].map((_, i) => from + i * step);

示例和其他备选方案

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

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

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

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

https://stackoverflow.com/a/64599169/8784402

用一行代码生成字符列表

constcharList=(a,z,d=1)=>(a=a.charCodeAt(),z=z.charCodeAt(),[…数组(Math.floor((z-a)/d)+1)].map((_,i)=>String.fromCharCode(a+i*d)));console.log(“从A到G”,charList('A','G'));console.log(“从A到Z,步长/增量为2”,charList('A','Z',2));console.log(“从Z到P的反向顺序”,charList('Z','P',-1));console.log(“从0到5”,charList(“0”,“5”,1));console.log(“从9到5”,charList('9','5',-1));console.log(“从0到8,步骤2”,charList('0','8',2));console.log(“从α到ω”,charList(“α”,“ω”));console.log(“印地语字符来自क 到ह“,charList('क', 'ह'));console.log(“从А到Е的俄语字符”,charList(“А”,“Е”));

For TypeScript
const charList = (p: string, q: string, d = 1) => {
  const a = p.charCodeAt(0),
    z = q.charCodeAt(0);
  return [...Array(Math.floor((z - a) / d) + 1)].map((_, i) =>
    String.fromCharCode(a + i * d)
  );
};

其他回答

我刚刚通过Object.defineProperty在Array上创建了这个polyfill,以创建整数或字符串的范围。Object.defineProperty是创建polyfills的一种更安全的方法。

更安全的polyfill

if (!Array.range) {
  Object.defineProperty(Array, 'range', {
    value: function (from, to, step) {
      if (typeof from !== 'number' && typeof from !== 'string') {
        throw new TypeError('The first parameter should be a number or a character')
      }

      if (typeof to !== 'number' && typeof to !== 'string') {
        throw new TypeError('The second parameter should be a number or a character')
      }

      var A = []
      if (typeof from === 'number') {
        A[0] = from
        step = step || 1
        while (from + step <= to) {
          A[A.length] = from += step
        }
      } else {
        var s = 'abcdefghijklmnopqrstuvwxyz'
        if (from === from.toUpperCase()) {
          to = to.toUpperCase()
          s = s.toUpperCase()
        }
        s = s.substring(s.indexOf(from), s.indexOf(to) + 1)
        A = s.split('')
      }
      return A
    }
  })
} else {
  var errorMessage = 'DANGER ALERT! Array.range has already been defined on this browser. '
  errorMessage += 'This may lead to unwanted results when Array.range() is executed.'
  console.log(errorMessage)
}

示例

Array.range(1, 3)

// Return: [1, 2, 3]
Array.range(1, 3, 0.5)

// Return: [1, 1.5, 2, 2.5, 3]
Array.range('a', 'c')

// Return: ['a', 'b', 'c']
Array.range('A', 'C')

// Return: ['A', 'B', 'C']
Array.range(null)
Array.range(undefined)
Array.range(NaN)
Array.range(true)
Array.range([])
Array.range({})
Array.range(1, null)

// Return: Uncaught TypeError: The X parameter should be a number or a character

它适用于字符和数字,通过可选步骤向前或向后移动。

var range = function(start, end, step) {
    var range = [];
    var typeofStart = typeof start;
    var typeofEnd = typeof end;

    if (step === 0) {
        throw TypeError("Step cannot be zero.");
    }

    if (typeofStart == "undefined" || typeofEnd == "undefined") {
        throw TypeError("Must pass start and end arguments.");
    } else if (typeofStart != typeofEnd) {
        throw TypeError("Start and end arguments must be of same type.");
    }

    typeof step == "undefined" && (step = 1);

    if (end < start) {
        step = -step;
    }

    if (typeofStart == "number") {

        while (step > 0 ? end >= start : end <= start) {
            range.push(start);
            start += step;
        }

    } else if (typeofStart == "string") {

        if (start.length != 1 || end.length != 1) {
            throw TypeError("Only strings with one character are supported.");
        }

        start = start.charCodeAt(0);
        end = end.charCodeAt(0);

        while (step > 0 ? end >= start : end <= start) {
            range.push(String.fromCharCode(start));
            start += step;
        }

    } else {
        throw TypeError("Only string and number types are supported");
    }

    return range;

}

jsFiddle。

如果扩充本机类型是您的事情,那么将其分配给Array.range。

var范围=函数(开始、结束、步骤){var范围=[];var typeofStart=启动类型;var typeofEnd=结束类型;如果(步骤==0){throw TypeError(“步长不能为零。”);}if(类型开始==“undefined”| |类型结束==“未定义”){throw TypeError(“必须传递开始和结束参数。”);}否则如果(typeofStart!=typeofEnd){throw TypeError(“开始和结束参数必须是相同的类型。”);}步骤类型==“未定义”&&(步骤=1);if(结束<开始){step=-步骤;}if(开始类型==“number”){while(步骤>0?结束>=开始:结束<=开始){范围.推(启动);开始+=步骤;}}否则if(typeofStart==“string”){如果(start.length!=1 | | end.length;=1){throw TypeError(“仅支持带有一个字符的字符串。”);}start=start.charCodeAt(0);end=end.charCodeAt(0);while(步骤>0?结束>=开始:结束<=开始){range.push(String.fromCharCode(开始));开始+=步骤;}}其他{throw TypeError(“仅支持字符串和数字类型”);}返回范围;}console.log(范围(“A”,“Z”,1));console.log(范围(“Z”,“A”,1));console.log(范围(“A”,“Z”,3));console.log(范围(0,25,1));console.log(范围(0,25,5));console.log(范围(20,5,5));

标准的Javascript没有生成范围的内置函数。有几个javascript框架添加了对这些功能的支持,或者正如其他人所指出的那样,您可以一直使用自己的功能。

如果您想再次检查,确定的资源是ECMA-262标准。

我发现了一个与PHP中的函数相当的JS范围函数,在这里工作得非常棒。向前和向后工作,可以处理整数、浮点数和字母!

function range(low, high, step) {
  //  discuss at: http://phpjs.org/functions/range/
  // original by: Waldo Malqui Silva
  //   example 1: range ( 0, 12 );
  //   returns 1: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
  //   example 2: range( 0, 100, 10 );
  //   returns 2: [0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
  //   example 3: range( 'a', 'i' );
  //   returns 3: ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']
  //   example 4: range( 'c', 'a' );
  //   returns 4: ['c', 'b', 'a']

  var matrix = [];
  var inival, endval, plus;
  var walker = step || 1;
  var chars = false;

  if (!isNaN(low) && !isNaN(high)) {
    inival = low;
    endval = high;
  } else if (isNaN(low) && isNaN(high)) {
    chars = true;
    inival = low.charCodeAt(0);
    endval = high.charCodeAt(0);
  } else {
    inival = (isNaN(low) ? 0 : low);
    endval = (isNaN(high) ? 0 : high);
  }

  plus = ((inival > endval) ? false : true);
  if (plus) {
    while (inival <= endval) {
      matrix.push(((chars) ? String.fromCharCode(inival) : inival));
      inival += walker;
    }
  } else {
    while (inival >= endval) {
      matrix.push(((chars) ? String.fromCharCode(inival) : inival));
      inival -= walker;
    }
  }

  return matrix;
}

这是缩小版:

function range(h,c,b){var i=[];var d,f,e;var a=b||1;var g=false;if(!isNaN(h)&&!isNaN(c)){d=h;f=c}else{if(isNaN(h)&&isNaN(c)){g=true;d=h.charCodeAt(0);f=c.charCodeAt(0)}else{d=(isNaN(h)?0:h);f=(isNaN(c)?0:c)}}e=((d>f)?false:true);if(e){while(d<=f){i.push(((g)?String.fromCharCode(d):d));d+=a}}else{while(d>=f){i.push(((g)?String.fromCharCode(d):d));d-=a}}return i};

一个有趣的挑战是编写最短的函数来实现这一点。救援递归!

function r(a,b){return a>b?[]:[a].concat(r(++a,b))}

在大范围内往往速度较慢,但幸运的是量子计算机即将问世。

另一个额外的好处是,它令人困惑。因为我们都知道隐藏代码以防被窥探是多么重要。

要真正彻底混淆功能,请执行以下操作:

function r(a,b){return (a<b?[a,b].concat(r(++a,--b)):a>b?[]:[a]).sort(function(a,b){return a-b})}