假设我有一个包含几个对象的数组:

var array = [{id: 1, date: Mar 12 2012 10:00:00 AM}, {id: 2, date: Mar 8 2012 08:00:00 AM}];

如何按日期元素排序这个数组,从最接近当前日期和时间的日期下来?请记住,数组可能有许多对象,但为了简单起见,我使用2。

我会使用排序函数和自定义比较器吗?


当前回答

如果像我一样,你有一个日期格式为YYYY[-MM[-DD]]的数组,你想在不太特定的日期之前排序更具体的日期,我想出了这个方便的函数:

function sortByDateSpecificity(a, b) {
  const aLength = a.date.length
  const bLength = b.date.length
  const aDate = a.date + (aLength < 10 ? '-12-31'.slice(-10 + aLength) : '')
  const bDate = b.date + (bLength < 10 ? '-12-31'.slice(-10 + bLength) : '')
  return new Date(aDate) - new Date(bDate)
}

其他回答

你可以在下划线js中使用sortBy。

http://underscorejs.org/#sortBy

示例:

var log = [{date: '2016-01-16T05:23:38+00:00', other: 'sample'}, 
           {date: '2016-01-13T05:23:38+00:00',other: 'sample'}, 
           {date: '2016-01-15T11:23:38+00:00', other: 'sample'}];

console.log(_.sortBy(log, 'date'));

如果像我一样,你有一个日期格式为YYYY[-MM[-DD]]的数组,你想在不太特定的日期之前排序更具体的日期,我想出了这个方便的函数:

function sortByDateSpecificity(a, b) {
  const aLength = a.date.length
  const bLength = b.date.length
  const aDate = a.date + (aLength < 10 ? '-12-31'.slice(-10 + aLength) : '')
  const bDate = b.date + (bLength < 10 ? '-12-31'.slice(-10 + bLength) : '')
  return new Date(aDate) - new Date(bDate)
}
Adding absolute will give better results

var datesArray =[
      {"some":"data1","date": "2018-06-30T13:40:31.493Z"},
      {"some":"data2","date": "2018-07-04T13:40:31.493Z"},
      {"some":"data3","date": "2018-06-27T13:40:54.394Z"}
   ]

var sortedJsObjects = datesArray.sort(function(a,b){ 
    return Math.abs(new Date(a.date) - new Date(b.date)) 
});

简单的一行解决方案为我排序日期:

sort((a, b) => (a < b ? 1 : -1))

谢谢Ganesh Sanap。按日期字段从旧到新对项目进行排序。使用它

 myArray = [{transport: "Air",
             load: "Vatican Vaticano",
             created: "01/31/2020"},
            {transport: "Air",
             load: "Paris",
             created: "01/30/2020"}] 

        myAarray.sort(function(a, b) {
            var c = new Date(a.created);
            var d = new Date(b.created);
            return c-d;
        });