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

在纠正JSON之后,这应该为你工作了:

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


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

其他回答

您的数据需要更正:

var array = [{id: 1, date: "Mar 12 2012 10:00:00 AM"},{id: 2, date: "Mar 28 2012 08:00:00 AM"}];

在修正数据之后,你可以使用这段代码:

function sortFunction(a,b){  
    var dateA = new Date(a.date).getTime();
    var dateB = new Date(b.date).getTime();
    return dateA > dateB ? 1 : -1;  
}; 

var array = [{id: 1, date: "Mar 12 2012 10:00:00 AM"},{id: 2, date: "Mar 28 2012 08:00:00 AM"}];
array.sort(sortFunction);​

我个人使用以下方法来排序日期。

let array = ["July 11, 1960", "February 1, 1974", "July 11, 1615", "October 18, 1851", "November 12, 1995"];

array.sort(function(date1, date2) {
   date1 = new Date(date1);
   date2 = new Date(date2);
   if (date1 > date2) return 1;
   if (date1 < date2) return -1;
})
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)) 
});

@Phrogz的回答都很棒,但这里有一个很棒的、更简洁的回答:

array.sort(function(a,b){return a.getTime() - b.getTime()});

用箭头函数的方式

array.sort((a,b)=>a.getTime()-b.getTime());

在Javascript中排序日期

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;
        });