Question

我有一个像

这样的对象数组

[ {"id":0,"start":0,"duration":117,"slide":4,"view":0},
  {"id":0,"start":0,"duration":12,"slide":1,"view":0},
  {"id":0,"start":0,"duration":8,"slide":3,"view":0},
  {"id":0,"start":0,"duration":7,"slide":2,"view":0}
];

我希望得到3个具有前3个持续时间值的对象。

Answer 1

您可以使用Array#sort对它们进行排序，降序并取前三个元素。

var array = [{ id: 0, start: 0, duration: 7, slide: 2, view: 0 }, { id: 0, start: 0, duration: 12, slide: 1, view: 0 }, { id: 0, start: 0, duration: 117, slide: 4, view: 0 }, { id: 0, start: 0, duration: 8, slide: 3, view: 0 }],
    top3duration = array.sort(function (a, b) { 
        return b.duration - a.duration;
    }).slice(0, 3);

console.log(top3duration);

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

Answer 2

 yourArray.sort(function(a,b){ 
      return b.duration - a.duration;
 }).splice(0,3);

 console.log(yourArray);   // Will give top 3 objects of high duration.

但是你的mainArray会被修改。如果你想要那些作为新数组然后去切片（0,3）并分配给一个新的数组作为[Fallenreaper的]答案

Answer 3

var myList = [ {"id":0,"start":0,"duration":117,"slide":4,"view":0},
      {"id":0,"start":0,"duration":12,"slide":1,"view":0},
      {"id":0,"start":0,"duration":8,"slide":3,"view":0},
      {"id":0,"start":0,"duration":7,"slide":2,"view":0}
      ];

outList = myList.sort(function(a,b){ return b.duration - a.duration; }).slice(0,3);

return outList;

Answer 4

我将通过几个简单的步骤来完成你的工作。

var myList = [ {"id":0,"start":0,"duration":117,"slide":4,"view":0},
  {"id":0,"start":0,"duration":12,"slide":1,"view":0},
  {"id":0,"start":0,"duration":8,"slide":3,"view":0},
  {"id":0,"start":0,"duration":7,"slide":2,"view":0}
  ];
outList = myList.sort(function(a,b){ return b.duration - a.duration; }).slice(0,3);
return outList;

来自对象数组的前2个最大值

4 个答案: