Question

我有一个对象数组，我想从最近的对象获取最近的前一个ID。我可以获取最近的下一个ID，它可以正常工作，但对于前一个不起作用。它直接获取对象的第一个ID。是下面的代码。有人可以帮我吗？

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const array = [{id:4}, {id:10}, {id:15}];

const findClosesPrevtId = (x) => ( array.find( ({id}) => x <= id ) || {} ).id;
const findClosestNextId = (x) => ( array.find( ({id}) => x >= id ) || {} ).id;

console.log(findClosesPrevtId(5));
console.log(findClosestNextId(11));

Answer 1

如果您使用x <= i从左到右搜索，则第一个元素将充满find。使用findLast从右到左搜索。

不幸的是，我刚刚发现实际上还没有findLast（有reduceRight，lastIndexOf ...：/），所以您必须自己编写： / p>

  Object.defineProperty(Array.prototype, "findLast", {
    value(cb, context) {
       for(let i = this.length - 1; i >= 0; i--)
         if(cb.call(context, this[i], i, this))
            return this[i];
   }
 });

const findClosesPrevtId = (x) => ( array.findLast( ({id}) => x <= id ) || {} ).id;

Answer 2

我对您的代码进行了一些更改，现在应该可以使用了。看看。

    const array = [{id:3}, {id:4}, {id:10}, {id:15}];
    // you should order the list by id before you try to search, this incase you have not orginized list.
    // filter the list first and get the prev id to 5 
    // you should get 3 and 4 then 
    // slice(-1) to get the last element of the array which should be 4
    const findClosesPrevtId = (x) =>
    (array.filter(({id}) => id <= x ).slice(-1)[0] || {}).id;
    const findClosestNextId = (x) => 
    (array.filter(({id}) => id >= x )[0] || {}).id;

    console.log("Prev to 5:"+ findClosesPrevtId(5));
    console.log("Next to 11:" +findClosestNextId(11));

Answer 3

如果id大于所需ID，则可以存储贵重/ next元素并停止迭代。

const
    closestPrevious = id => {
        var last = {};
        array.some(o => o.id > id || (last = o, false))
        return last.id;
    },
    closestNext = id => {
        var next = array[0];
        array.some((o, i, { [i + 1]: n = {} }) => o.id > id || (next = n, false))
        return next.id;
    },
    array = [{ id: 4 }, { id: 10 }, { id: 15 }];

console.log(closestNext(5));      // 10
console.log(closestNext(11));     // 15

console.log(closestPrevious(5));  //  4
console.log(closestPrevious(11)); // 10

Answer 4

我发现更容易反转数组并将比较从>=切换到<=：

const findClosestNextId  = (x, arr) => 
  (arr.find ( ({id}) => id >= x) || {} ) .id

const findClosestPrevId  = (x, arr) => 
  (arr .slice(0) .reverse() .find ( ({id}) => id <= x) || {}) .id

const array = [{ id: 4 }, { id: 10 }, { id: 15 }];

console .log (
  findClosestNextId (5,  array), //=> 10
  findClosestNextId (11, array), //=> 15
  findClosestNextId (42, array), //=> undefined

  findClosestPrevId (5,  array), //=> 4
  findClosestPrevId (11, array), //=> 10
  findClosestPrevId (2,  array), //=> undefined
)

那里有slice调用，以防止它修改原始数组。如果没有找到元素，它将返回undefined。

如何从JavaScript中的对象数组中获取最接近的先前ID

JAVASCRIPT

4 个答案: