javascript - 根据{喜欢的:true} ,距离然后通过avg_rating对数组对象进行多重排序

Question

我有一个如下所示的数组对象

arr = [ {distance:4, avg_rating: 5}, 
        {distance: 6, avg_rating: 4}, 
        {distance: 7, avg_rating: 2, liked: true}, 
        {distance: 6, avg_rating: 4 , liked: true}, 
        {distance: 6, avg_rating: 3 , liked: true},
        {distance: 4, avg_rating: 3}]

我希望首先获得 {liked: true}，然后按距离升序排序，如果距离相同，则按 avg_rating 排序。

我使用了 javascript 的排序方法，但它不起作用。

arr.sort(function(a, b){                                     
      if (b.liked)  return  1;
      if(!b.liked) return -1;
      if (a.distance < b.distance) return 1;
      if (a.distance > b.distance) return -1;
      if (a.avg_rating < b.avg_rating) return -1;
      if (a.avg_rating > b.avg_rating) return 1;
 })

我希望结果是

arr = [{distance: 6, avg_rating: 4 , liked: true},
       {distance: 6, avg_rating: 3 , liked: true}
       {distance: 7, avg_rating: 2, liked: true},
       {distance: 4, avg_rating: 5},
       {distance: 4, avg_rating: 3},
       {distance: 6, avg_rating: 4}]

因此，首先所有 { liked : true } 都在顶部并按距离升序排序(即所有附近的都在数组顶部)然后如果距离相同则按 avg_rating 降序排序(即所有相同距离值再次按 avg_rating 排序)

Answer 1

您可以链接所有条件并通过获取每个项目的增量来获得想要的订单。

对于第一对，取 logical NOT !获取用于减法的 bool 值。

其他的按照排序的顺序取，a - b (asc) and b - a (desc)。

var array = [{ distance: 4, avg_rating: 5 }, { distance: 6, avg_rating: 4 }, { distance: 7, avg_rating: 2, liked: true }, { distance: 6, avg_rating: 4, liked: true }, { distance: 6, avg_rating: 3, liked: true }, { distance: 4, avg_rating: 3 }];

array.sort((a, b) =>
    !a.liked - !b.liked ||
     a.distance - b.distance ||
     b.avg_rating - a.avg_rating
);

console.log(array);

.as-console-wrapper { max-height: 100% !important; top: 0; }