假设我有以下复选框:

<input type="checkbox" value="1-25" />

为了得到定义我正在寻找的范围边界的两个数字,我使用下面的jQuery:

var value = $(this).val();
var lowEnd = Number(value.split('-')[0]);
var highEnd = Number(value.split('-')[1]);

然后,我如何创建一个包含lowEnd和highEnd之间的所有整数的数组,包括lowEnd和highEnd本身?对于这个特定的例子,显然,结果数组将是:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]

当前回答

_Array = (length) => Object.keys(Array.from({length}))

//_Array = [0, 1, 2, 3, 4]

其他回答

这里有3个函数,它们应该涵盖了我能想到的一切(包括对其他一些答案中的问题的修复):rangeInt(), range()和between()。在所有情况下都考虑升序和降序。

例子

范围英特()

包括端点并且只处理整数

rangeInt(1, 4)  // [1, 2, 3, 4] Ascending order
rangeInt(5, 2)  // [5, 4, 3, 2] Descending order
rangeInt(4, 4)  // [4]          Singleton set (i.e. not [4, 4])
rangeInt(-1, 1) // [-1, 0, 1]   Mixing positive and negative

range ()

与rangeInt()相同,除了

不限于整数 允许在第三个参数中指定数量的点

range(0, 10, 2)  // [0, 3.333, 6.666, 10] Gets endpoints and 2 points between
range(0, 1.5, 1) // [0, 0.75, 1.5]        Accepts fractions

间()

与range()相同,除了

不包括端点 没有单例集(将返回一个空数组)

between(0, 10, 2) // [3.333, 6.666]
between(-1, -1.5) // [-1.25]
between(4, 4, 99) // []

/**
 * Gets a set of integers that are evenly distributed along a closed interval
 * @param {int} begin - Beginning endpoint (inclusive)
 * @param {int} end   - Ending endpoint (inclusive)
 * @return {Array} Range of integers
 */
function rangeInt( begin, end )  {
    if ( !Number.isInteger(begin) || !Number.isInteger(end) ) {
        throw new Error('All arguments must be integers')
    }
    return range(begin, end, Math.abs(end - begin) - 1)
}

/**
 * Gets a set of numbers that are evenly distributed along a closed interval
 * @param {Number} begin  - Beginning endpoint (inclusive)
 * @param {Number} end    - Ending endpoint (inclusive)
 * @param {int}    points - How many numbers to retrieve from the open interval
 * @return {Array} Range of numbers
 */
function range( begin, end, points ) {
    if ( begin !== end ) {
        return [ begin, ...between(begin, end, points), end ]
    }
    else if ( Number.isFinite(begin) ) {
        return [ begin ] // singleton set
    }
    else throw new Error('Endpoints must be finite')
}

/**
 * Gets a subset of numbers that are evenly distributed along an open interval
 * @param {Number} begin  - Beginning endpoint (exclusive)
 * @param {Number} end    - Ending endpoint (exclusive)
 * @param {int}    points - How many numbers to retrieve from the interval
 * @return {Array} Retrieved numbers
 */
function between( begin, end, points = 1 ) {
    if ( !Number.isFinite(begin) || !Number.isFinite(end) || !Number.isFinite(points) ) {
        throw new Error('All arguments must be finite')
    }
    const set = []
    // Skip if an open interval does not exist
    if ( begin !== end ) {
        const step = (end - begin) / (points + 1)
        for ( let i = 0; i < points; i++ ) {
            set[i] = begin + (i + 1) * step
        }
    }
    return set
}
function createNumberArray(lowEnd, highEnd) {
    var start = lowEnd;
    var array = [start];
    while (start < highEnd) {
        array.push(start);
        start++;
    }
} 

我的循环版本;)

var lowEnd = 1;
var highEnd = 25;
var arr = [];
while(lowEnd <= highEnd){
   arr.push(lowEnd++);
}

最快的方式

而——在大多数浏览器上更快 直接设置变量比推设置快

功能:

var x=function(a,b,c,d){d=[];c=b-a+1;while(c--){d[c]=b--}return d},

theArray=x(lowEnd,highEnd);

or

var arr=[],c=highEnd-lowEnd+1;
while(c--){arr[c]=highEnd--}

编辑

可读版本

var arr = [],
c = highEnd - lowEnd + 1;
while ( c-- ) {
 arr[c] = highEnd--
}

Demo

http://jsfiddle.net/W3CUn/

对于持悲观态度的选民

性能

http://jsperf.com/for-push-while-set/2

ie更快,firefox快3倍

只有在aipad air上,for循环速度略快。

在win8, osx10.8, ubuntu14.04, ipad, ipad air, ipod上测试;

chrome,ff,即,safari,移动safari。

我希望看到旧ie浏览器的性能,其中for循环没有优化!

函数getRange (a, b) { ar = new Array(); Var y = a - b > 0 ?A - b: b - A; (我= 1,< y;我+ +) { ar.push (i + b); } 返回基于“增大化现实”技术; }