Question

有人可以帮我算法吗？

我有一个事件列表，我希望按当前日期获取下一个事件和之前的事件。

示例

我从sql数据库中获取所有事件，如下所示：

events = [
    {eventId:1, eventDate: "Wen Apr 01 2015 18:41:00 GMT+0300", eventPlace":'Dortmund, DE'},
    {eventId:2, eventDate: "Sun Apr 05 2015 23:41:00 GMT+0300", eventPlace":'Budapest, HU'},
    {eventId:3, eventDate: "Fri Apr 03 2015 13:41:00 GMT+0300" eventPlace":'Madrid, ES'},
    {eventId:4, eventDate: "Mon Jun 01 2015 22:00:00 GMT+0300" eventPlace":'London, EN'},
    .......
    {eventId:100, eventDate: "Mon Aug 31 2015 22:00:00 GMT+0300"}
    ]

如果当前日期是2015年4月5日星期日15:00:00我想要

    resultNextEv= {eventId:2, eventDate: "Sun Apr 05 2015 23:41:00 GMT+0300", eventPlace":'Budapest, HU'}
    resultLastEv= {eventId:3, eventDate: "Fri Apr 03 2015 13:41:00 GMT+0300" eventPlace":'Madrid, ES'}

获得此结果的最佳方法是什么？（我使用javascript / angular）有什么想法吗？

由于

Answer 1

首先，按日期对事件进行排序：

var sorted= events.sort(function(a,b) {
  return new Date(a.eventDate) - new Date(b.eventDate);
});

然后，遍历已排序的事件。如果日期小于目标日期，请将其设置为上一个事件。如果它大于目标日期，请将其设置为下一个事件，并停止迭代：

function getDates() {
  var sorted= events.sort(function(a,b) {
    return new Date(a.eventDate) - new Date(b.eventDate);
  });
  var date= new Date(document.getElementById('date').value),
      i,sdate;

  for(i = 0 ; i < sorted.length ; i++) {
    sdate= new Date(sorted[i].eventDate);
    if(sdate > date) break;
  }

  //do something with sorted[i-1] and sorted[i]
} //getDates

要获取最后的 N 事件，请使用sorted[i-1]，sorted[i-2]，... sorted[i-N]。

要获取下一个 N 事件，请使用sorted[i]，sorted[i+1]，... sorted[i+N-1]。

Fiddle

Answer 2

var events = [
    {eventId:1, eventDate: "Wen Apr 01 2015 18:41:00 GMT+0300", eventPlace:"Dortmund, DE"},
    {eventId:2, eventDate: "Sun Apr 05 2015 23:41:00 GMT+0300", eventPlace:"Budapest, HU"},
    {eventId:3, eventDate: "Fri Apr 03 2015 13:41:00 GMT+0300", eventPlace:"Madrid, ES"},
    {eventId:4, eventDate: "Mon Jun 01 2015 22:00:00 GMT+0300", eventPlace:"London, EN"}
    ];

 // sort the events 
 var sortedEvents = events.sort(function(a, b) {
  return new Date(a.eventDate) - new Date(b.eventDate);
  });

var now = new Date("Sun Apr 05 2015 15:00:00");
var futureEvents = [];
var previousEvents = [];

// finding out the next event. Go from the earliest to the newest. When you find the first later than "now", get it
for(var i = 0; i < sortedEvents.length; i++){
  if(new Date(sortedEvents[i].eventDate) > now) {
    futureEvents.push(sortedEvents[i]);
    }
}

// finding out the previous event. Go from the newest to the earliest and when you find the first earlier than "now", get it
for(var i = sortedEvents.length - 1; i >= 0; i--){
  if(new Date(sortedEvents[i].eventDate) < now) {
    previousEvents.push(sortedEvents[i]);
    }
}

console.log("now: " + now);
console.log("next event: " + futureEvents.length ? futureEvents[0].eventDate : undefined);
console.log("previous event: " + previousEvents.length ? previousEvents[0].eventDate : undefined);

Codepen：http://codepen.io/andrerpena/pen/WbBwbG

如果您希望获得多个未来事件或多个previos事件，只需遍历futureEvents或previousEvents即可。例如，接下来的两个事件是futureEvents[0]和futureEvents[1]

Answer 3

以下是使用开源项目jinqJs

进行操作的方法

＆＃13;

var events = [
    {eventId:1, eventDate: "Wen Apr 01 2015 18:41:00 GMT+0300", eventPlace:'Dortmund, DE'},
    {eventId:2, eventDate: "Sun Apr 05 2015 23:41:00 GMT+0300", eventPlace:'Budapest, HU'},
    {eventId:3, eventDate: "Fri Apr 03 2015 13:41:00 GMT+0300", eventPlace:'Madrid, ES'},
    {eventId:4, eventDate: "Mon Jun 01 2015 22:00:00 GMT+0300", eventPlace:'London, EN'},
    {eventId:100, eventDate: "Mon Aug 31 2015 22:00:00 GMT+0300", eventPlace:'London, EN'}
    ]
;


var result1 = jinqJs().from(events).orderBy('eventDate').where(function(row){return Date.parse(row.eventDate) < Date.parse('Sun Apr 05 2015 15:00:00');}).top(1).select();

var result2 = jinqJs().from(events).orderBy('eventDate').where(function(row){return Date.parse(row.eventDate) > Date.parse('Sun Apr 05 2015 15:00:00');}).bottom(1).select();

var result = jinqJs().from(result1,result2).select();

document.body.innerHTML += '<pre>' + JSON.stringify(result, null, 2) + '</pre><br><br>';

＆＃13;

<script src="https://rawgit.com/fordth/jinqJs/master/jinqjs.js"></script>

＆＃13;

algorithm：从数组中获取当前日期的最后和下一个事件（javascript / angular）

3 个答案: