js将"2018-08-15T08:04:52.000+0000"这种格式的时间转化为正常格式2019-08-15 15:02:07

Posted 2022-08-15 psxiao

第一种方法：

var time="2018-05-19T08:04:52.000+0000";

 var d = new Date(time);

var times=d.getFullYear() + ‘-‘ + (d.getMonth() + 1) + ‘-‘ + d.getDate() + ‘ ‘ + d.getHours() + ‘:‘ + d.getMinutes() + ‘:‘ + d.getSeconds();

输出 2018-05-19 15:59:10

发现问题：

   如果时间是09:00:00 ,你这样转之后显示的是9:0:0

         2018-05-19T00:00:00.000+0000

转换后："2018-5-19 8:0:0"

解决办法（网友提供）：加个判断就可以了 例如月(d.getMonth() + 1 < 10 ? "0" + (d.getMonth() + 1) : d.getMonth() + 1)

eg：

dateTime(rowData)
  // console.log(rowData)
  var d = new Date(rowData)
  // var a= d.getFullYear() + ‘-‘ + (d.getMonth() + 1) + ‘-‘ + d.getDate() + ‘ ‘ + d.getHours() + ‘:‘ + d.getMinutes() + ‘:‘ + d.getSeconds();              /*d.getMonth() + 1 < 10 ? "0" + (d.getMonth() + 1) : d.getMonth() + 1*/
  var a= d.getFullYear() + ‘-‘ + (d.getMonth() + 1 < 10 ? "0" + (d.getMonth() + 1) : d.getMonth() + 1) + ‘-‘ + (d.getDate()<10 ? "0" +d.getDate():d.getDate())+ ‘ ‘ + (d.getHours()<10 ?"0"+d.getDate():d.getDate()) + ‘:‘ + (d.getMinutes()<10 ? "0" + d.getMinutes(): d.getMinutes()) + ‘:‘ + (d.getSeconds()<10 ? "0" + d.getSeconds() : d.getSeconds() );
  // console.log(a)
  return a
,

另一种办法：

       myFunction();
       function myFunction()
           
           var dateee = new Date("2018-05-19T00:00:00.000+0000").toJSON();

//        var dateee = new Date("2017-07-09T09:46:49.667").toJSON();
        
        var date = new Date(+new Date(dateee)+8*3600*1000).toISOString().replace(/T/g,‘ ‘).replace(/\.[\d]3Z/,‘‘)  
        
           alert(date);
           console.log("时间2==="+date);