Hii,
我在磨难方面有投入时间。 在我的申请中,我想包括一个数字站。 时间将像每秒一个数字锁一样发生动态变化。
Hii,
我在磨难方面有投入时间。 在我的申请中,我想包括一个数字站。 时间将像每秒一个数字锁一样发生动态变化。
你们可以把摩擦隔1 000个,然后用60个时间接听,再用60个时间取时。 或者,甚至更好,使用模块:
hours = parseInt(milliseconds / 3600) % 24;
minutes = parseInt(milliseconds / 60) % 60;
seconds = (milliseconds / 1000) % 60;
然而,如果你想像02:00 PM 这样的时间,你必须知道你开始计算微秒(即当小秒是“0”时,时间是什么时候)。
我不肯定你的问题是什么,就像你需要帮助的具体部分一样。 即便是你不熟悉瓦查特语,通过按每个增加的因素进行区分,并且把其余因素考虑在内,这样做也非常简单:
var input = ...; // your input time
var millis = input % 1000;
input /= 1000;
var seconds = input % 60;
input /= 60;
var minutes = input % 60;
input /= 60;
var hours = input % 24; // I presume this will be less than 24 anyway)
var entireTime = hours + : + minutes + : + seconds;
这样做的另一种办法是创建<代码>Date,反对将投入时间移至施工人员;这将代表过去时代的摩擦数量,从而按其价值进行印刷。 视你有什么样的日期格式框架,这可能是一种更为简单的方法,而且肯定会允许在操纵价值方面有更多的灵活性。
只是一种想法——确保你充分了解实际投入。 投入(time, in milliseconds)相对不寻常;我希望这种投入实际上将成为duration。 诚然,这可能是过去午夜的摩擦次数,但确实不能确保它不会占用过去某些其他任意起点的摩擦。
固定支流为10^-3。 因此,每秒一千秒,每分钟60 000次,每小时3600 000次。 因此,1 000毫秒为1秒,6万毫秒为1分钟,3 600 000毫秒为1小时。
这意味着,将摩擦数量从1 000个减少到1 000个,你获得第二秒数,60 000个,你获得几分钟,3 600 000个。
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