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

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));

我的代码高尔夫同事想出了这个(ES6),包容的:

(s,f)=>[...Array(f-s+1)].map((e,i)=>i+s)

非包容性:

(s,f)=>[...Array(f-s)].map((e,i)=>i+s)

我的实施

export function stringRange(a: string, b: string) {
    let arr = [a + ''];

    const startPrefix = a.match(/([\D])+/g);
    const endPrefix = b.match(/([\D])+/g);

    if ((startPrefix || endPrefix) && (Array.isArray(startPrefix) && startPrefix[0]) !== (Array.isArray(endPrefix) && endPrefix[0])) {
        throw new Error('Series number does not match');
    }

    const startNum = a.match(/([\d])+/g);
    const endNum = b.match(/([\d])+/g);

    if (!startNum || !endNum) {
        throw new Error('Range is not valid');
    }

    let start = parseInt(startNum[0], 10);
    let end = parseInt(endNum[0], 10);

    if (start > end) {
        throw new Error('Ending value should be lessesr that starting value');
    }

    while (start !== end) {
        start++;
        arr.push(startPrefix ? startPrefix[0] + (start + '').padStart(startNum[0].length, '0') : start + '');

    }

    return arr;
}

样本结果

// console.log(range('0', '10'));
// console.log(range('10', '10')); 
// console.log(range('10', '20'));
// console.log(range('10', '20000'));
// console.log(range('ABC10', 'ABC23'));
// console.log(range('ABC10', 'ABC2300'));
// console.log(range('ABC10', 'ABC09')); --> Failure case
// console.log(range('10', 'ABC23')); --> Failure case
// console.log(range('ABC10', '23')); --> Failure case

至于为给定范围生成数字数组,我使用以下方法:

function range(start, stop)
{
    var array = [];

    var length = stop - start; 

    for (var i = 0; i <= length; i++) { 
        array[i] = start;
        start++;
    }

    return array;
}

console.log(range(1, 7));  // [1,2,3,4,5,6,7]
console.log(range(5, 10)); // [5,6,7,8,9,10]
console.log(range(-2, 3)); // [-2,-1,0,1,2,3]

显然,它不适用于字母数组。

…更大范围,使用生成器功能。

function range(s, e, str){
  // create generator that handles numbers & strings.
  function *gen(s, e, str){
    while(s <= e){
      yield (!str) ? s : str[s]
      s++
    }
  }
  if (typeof s === 'string' && !str)
    str = 'abcdefghijklmnopqrstuvwxyz'
  const from = (!str) ? s : str.indexOf(s)
  const to = (!str) ? e : str.indexOf(e)
  // use the generator and return.
  return [...gen(from, to, str)]
}

// usage ...
console.log(range('l', 'w'))
//=> [ 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w' ]

console.log(range(7, 12))
//=> [ 7, 8, 9, 10, 11, 12 ]

// first 'o' to first 't' of passed in string.
console.log(range('o', 't', "ssshhhooooouuut!!!!"))
// => [ 'o', 'o', 'o', 'o', 'o', 'u', 'u', 'u', 't' ]

// only lowercase args allowed here, but ...
console.log(range('m', 'v').map(v=>v.toUpperCase()))
//=> [ 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V' ]

// => and decreasing range ...
console.log(range('m', 'v').map(v=>v.toUpperCase()).reverse())

// => ... and with a step
console.log(range('m', 'v')
          .map(v=>v.toUpperCase())
          .reverse()
          .reduce((acc, c, i) => (i % 2) ? acc.concat(c) : acc, []))

// ... etc, etc.

希望这有用。