如何通过特定索引获取未来项目和过去项目，就像在轮播中一样

Question

我正在尝试创建一个自定义轮播。我被算法困住了。我需要通过特定索引获取未来和过去的项目。旋转木马可以无限移动。示例： var exampleArray = [1,2,3,4,5,6,7,8,9];切片（示例数组，0）；-->> currentItem = [1], futureItems = [2,3,4,5], PastItems = [6,7,8,9]slice(exampleArray, 4); -->> currentItem = [5], futureItems = [6,7,8,9], PastItems = [1,2,3,4]slice(exampleArray, 7); -->> 当前项目 = [8]，未来项目 = [9,1,2,3]，过去项目 = [4,5,6,7]

Answer 1

旋转数组以将所需的数据放入currentItem中间；那么你可以将中间之前的切片作为pastItems，将中间元素作为currentItem，将中间之后的切片作为futureItems：

const slicer = (arr, idx) => {

  let len = arr.length;

  let mid = Math.floor(len/2);

  let rotated = arr.slice(0);

  rotated = rotated.concat(rotated.splice(0, (mid + idx + 1) % len));

  return [rotated.slice(0, mid), [rotated[mid]], rotated.slice(mid+1)]

}

const exampleArray = [1,2,3,4,5,6,7,8,9];

[pastItems, currentItem, futureItems] = slicer(exampleArray, 0);

console.log(pastItems);

console.log(currentItem);

console.log(futureItems);

[pastItems, currentItem, futureItems] = slicer(exampleArray, 4);

console.log(pastItems);

console.log(currentItem);

console.log(futureItems);

[pastItems, currentItem, futureItems] = slicer(exampleArray, 7);

console.log(pastItems);

console.log(currentItem);

console.log(futureItems);