TA貢獻(xiàn)1802條經(jīng)驗 獲得超6個贊
這看起來，不是一個補(bǔ)全的題，是按照間隔1小時生成下標(biāo)0和下標(biāo)length-1的時間間隔數(shù)組的問題。
let 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'
]
let oneHour = 60 * 60 * 1000;
let now = new Date(times[0]);
let max = new Date(times[times.length - 1]);
let result = [];
while (!(max < now)) {
result.push(now)
now = new Date(now.getTime() + oneHour);
}
console.log(result)
還是用moment.js比較方便。
TA貢獻(xiàn)1856條經(jīng)驗 獲得超5個贊
TA貢獻(xiàn)1824條經(jīng)驗 獲得超6個贊
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'];
var oneHour = 60*60*1000;
var index = 0;
while(index<times.length-1){
var next = new Date(new Date(times[index]).getTime()+oneHour);
if(next<new Date(times[index+1])){
times.splice(index+1,0,next);
}
index++;
}
alert(times);
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