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?
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
可以使用带有数组、for循环和Math.random()变量的函数来解决这个问题。for循环将数字推入数组,数组将包含您范围内的所有数字。然后Math.random()根据数组的长度随机选择一个。
function randNumInRange(min, max) {
var range = []
for(var count = min; count <= max; count++) {
range.push(count);
}
var randNum = Math.floor(Math.random() * range.length);
alert(range[randNum]);
}
其他回答
函数范围(firstNum,lastNum){let rangeList=[];如果(firstNum>lastNum){return console.error(“第一个数字不能大于最后一个数字”);}让counter=firstNum;while(计数器<=lastNum){rangeList.push(计数器);计数器++;}返回范围列表;}
如果您只想使用范围来重复一个过程n次,您可以简单地使用此代码
[...Array(10)].map((item, index) => (
console.log("item:", index)
))
保持简单:
// Generator
function* iter(a, b, step = 1) {
for (let i = b ? a : 0; i < (b || a); i += step) {
yield i
}
}
const range = (a, b, step = 1) =>
typeof a === 'string'
? [...iter(a.charCodeAt(), b.charCodeAt() + 1)].map(n => String.fromCharCode(n))
: [...iter(a, b, step)]
range(4) // [0, 1, 2, 3]
range(1, 4) // [1, 2, 3]
range(2, 20, 3) // [2, 5, 8, 11, 14, 17]
range('A', 'C') // ['A', 'B', 'C']
/**
* @param {!number|[!number,!number]} sizeOrRange Can be the `size` of the range (1st signature) or a
* `[from, to]`-shape array (2nd signature) that represents a pair of the *starting point (inclusive)* and the
* *ending point (exclusive)* of the range (*mathematically, a left-closed/right-open interval: `[from, to)`*).
* @param {!number} [fromOrStep] 1st signature: `[from=0]`. 2nd signature: `[step=1]`
* @param {!number} [stepOrNothing] 1st signature: `[step=1]`. 2nd signature: NOT-BEING-USED
* @example
* range(5) ==> [0, 1, 2, 3, 4] // size: 5
* range(4, 5) ==> [5, 6, 7, 8] // size: 4, starting from: 5
* range(4, 5, 2) ==> [5, 7, 9, 11] // size: 4, starting from: 5, step: 2
* range([2, 5]) ==> [2, 3, 4] // [2, 5) // from: 2 (inclusive), to: 5 (exclusive)
* range([1, 6], 2) ==> [1, 3, 5] // from: 1, to: 6, step: 2
* range([1, 7], 2) ==> [1, 3, 5] // from: 1, to: 7 (exclusive), step: 2
* @see {@link https://stackoverflow.com/a/72388871/5318303}
*/
export function range (sizeOrRange, fromOrStep, stepOrNothing) {
let from, to, step, size
if (sizeOrRange instanceof Array) { // 2nd signature: `range([from, to], step)`
[from, to] = sizeOrRange
step = fromOrStep ?? 1
size = Math.ceil((to - from) / step)
} else { // 1st signature: `range(size, from, step)`
size = sizeOrRange
from = fromOrStep ?? 0
step = stepOrNothing ?? 1
}
return Array.from({length: size}, (_, i) => from + i * step)
}
示例:
控制台日志(范围(5),//[0,1,2,3,4]//size:5范围([2,5]),//[2,3,4]//[2、5)//从:2(含)到:5(不含)范围(4,2),//[2,3,4,5]//大小:4,从:2开始范围([1,6],2),//[1,3,5]//从:1到:6,步骤:2范围([1,7],2),//[1,3,5]//从:1到:7(不含),步骤:2)<脚本>函数范围(sizeOrRange、fromOrStep、stepOrNothing){让从、到、步长、大小if(sizeOrRange instanceof Array){//第二个签名:`range([from,to],step)`[from,to]=sizeOrRange步骤=来自或步骤??1.size=数学ceil((to-from)/步长)}else{//第一个签名:`range(大小,从,步)`size=sizeOrRangefrom=来自或步骤??0step=stepOrNothing??1.}return Array.from({length:size},(_,i)=>from+i*step)}</script>
Python风格的方式:
range = (start, end, step) => {
let arr = []
for(let n=start;n<end;n+=(step||1)) arr.push(n)
return arr;
}