Warning: file_get_contents(/data/phpspider/zhask/data//catemap/9/javascript/410.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
如何在javascript中删除dateT~~?_Javascript - Fatal编程技术网

如何在javascript中删除dateT~~?

如何在javascript中删除dateT~~?,javascript,Javascript,我为即将到来的星期六写了代码。所以我这样写,结果是2020-03-07T04:00:48.306Z。我找到了日期。我只需要日期。我可以用分割法,但我不想用这个。还有别的办法吗 function getSaturday(d) { d = new Date(d); var day = d.getDay(), diff = d.getDate()+5 - day + (day == 5 ? 7:1); console.log(diff); return

我为即将到来的星期六写了代码。所以我这样写,结果是2020-03-07T04:00:48.306Z。我找到了日期。我只需要日期。我可以用分割法,但我不想用这个。还有别的办法吗

function getSaturday(d) {
    d = new Date(d);
    var day = d.getDay(),
        diff = d.getDate()+5 - day + (day == 5 ? 7:1); 
    console.log(diff);
    return new Date(d.setDate(diff));
  }
getSaturday函数返回一个新的日期对象,即星期六

最简单的方法是使用split返回日期字符串

const d = getSaturday(...)
d.toISOString().split("T")[0] 
但既然你不想使用split。您可能想使用矩:一个非常流行的轻量级JavaScript日期库,用于解析、验证、操作和格式化日期

然后你可以像这样重写你的函数:

函数getSaturdayd{ 返回时间d.daystadix.formatYYYY-MM-DD } console.loggetSaturdaynew日期 尝试使用substr

function getSaturday(d) {
d = new Date(d);
var day = d.getDay(),
    diff = d.getDate()+5 - day + (day == 5 ? 7:1); 
console.log(diff);
return new Date(d.setDate(diff)).toISOString.substr(0,10);
}

函数getSaturdayd{ d=新日期; var day=d.getDay, diff=d.getDate+5天+day==5-7:1; console.logdiff; 返回新的Dated.setDatediff; } 函数dateSaturdayd{ 变量ds=[ d、 整整一年, d、 getMonth+1<10?`0${d.getMonth+1}`:d.getMonth+1.toString, d、 getDate<10?`0${d.getDate}`:d.getDate.toString ]; 返回ds.join- }
console.logdateSaturdaygetSaturday2020-03-04;实际上,这个函数返回一个日期,如果您想用另一种字符串格式,只需格式化它即可