有一个时间数组，时次不连续，怎样补全比较好？

Question

一个时间数组，比如：

var times = [    '2018-5-15 00:00:00','2018-5-15 01:00:00','2018-5-15 02:00:00','2018-5-15 08:00:00','2018-5-16 03:00:00'
]

像这样，补全之间缺失的时次，怎样实现较好？

Answer 1

这看起来，不是一个补全的题，是按照间隔1小时生成下标0和下标length-1的时间间隔数组的问题。let times = [&nbsp; &nbsp; '2018-5-15 00:00:00', '2018-5-15 01:00:00', '2018-5-15 02:00:00', '2018-5-15 08:00:00', '2018-5-16 03:00:00']let oneHour = 60 * 60 * 1000;let now = new Date(times[0]);let max = new Date(times[times.length - 1]);let result = [];while (!(max < now)) {&nbsp; &nbsp; result.push(now)&nbsp; &nbsp; now = new Date(now.getTime() + oneHour);}console.log(result)还是用moment.js比较方便。

Answer 2

你的需求是补全两个时间段之间缺失的小时时间对吧，看到你的数组是排好序的，很简单呀，你可以取出数组首跟尾，然后转成时间戳，接着用个for循环，只要小于尾部时间戳的，就一直加一个小时，存入一个新数组，最后将新数组装回字符串，完成。

Answer 3

&nbsp;var times = [&nbsp; &nbsp; &nbsp; '2018-5-15 00:00:00','2018-5-15 01:00:00','2018-5-15 02:00:00','2018-5-15 08:00:00','2018-5-16 03:00:00'];&nbsp; var oneHour = 60*60*1000;&nbsp; var index = 0;&nbsp; while(index<times.length-1){&nbsp; &nbsp; &nbsp; var next = new Date(new Date(times[index]).getTime()+oneHour);&nbsp; &nbsp; &nbsp; if(next<new Date(times[index+1])){&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; times.splice(index+1,0,next);&nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; index++;&nbsp; }&nbsp; alert(times);