我将时间作为Unix时间戳存储在MySQL数据库中,并将其发送给一些JavaScript代码。我怎样才能抽出时间?

例如,HH/MM/SS格式。


当前回答

您可以使用以下格式(源代码):

const date = new Date(yourTimestamp).toLocaleDateString('de-DE', {
    weekday: 'long',
    day: '2-digit',
    month: 'long',
    year: 'numeric'
})

结果:

Sonntag, 01. Januar 2023

其他回答

您可以使用以下格式(源代码):

const date = new Date(yourTimestamp).toLocaleDateString('de-DE', {
    weekday: 'long',
    day: '2-digit',
    month: 'long',
    year: 'numeric'
})

结果:

Sonntag, 01. Januar 2023

试试看:

      new Date(1638525320* 1e3).toISOString()  //2021-12-03T09:55:20.000Z

最短的

(new Date(ts*1000)+'').slice(16,24)

设ts=1549312452;让时间=(新日期(ts*1000)+“”).切片(16,24);console.log(时间);

函数timeConverter(UNIX_timestamp){var a=新日期(UNIX_timestamp*1000);var months=['Jan','Feb','Mar','Pr','May','Jun','Jul','Aug','Step','Oct','Nov','Dec'];var year=a.getFullYear();var month=月[a.getMonth()];var date=a.getDate();var hour=a.getHours();var最小值=a.获取最小值();var sec=a.getSeconds();var time=日期+“”+月份+“”+year+“”“+小时+”:“+分钟+”:”+秒;返回时间;}console.log(timeConverter(0));

function getTIMESTAMP() {
  var date = new Date();
  var year = date.getFullYear();
  var month = ("0" + (date.getMonth() + 1)).substr(-2);
  var day = ("0" + date.getDate()).substr(-2);
  var hour = ("0" + date.getHours()).substr(-2);
  var minutes = ("0" + date.getMinutes()).substr(-2);
  var seconds = ("0" + date.getSeconds()).substr(-2);

  return year + "-" + month + "-" + day + " " + hour + ":" + minutes + ":" + seconds;
}

//2016-01-14 02:40:01