我构建了一个kml文件,并且有一行时间戳:
<gx:TimeStamp>
<when>2002-07-09T19:00:00-08:00</when>
</gx:TimeStamp>
我需要转换时间,如:“1430477311”到“2002-07-09T19:00:00-08:00”格式
怎么样? (java code) tnx很多答案 0 :(得分:2)
您想要从时间格式转换为XML日期格式(ISO-8601) -
long timeStamp = 1430477311L;
java.util.Date yourDate = new java.util.Date(timeStamp*1000); //ms
SimpleDateFormat yyyyMMddTHHmmssSDF = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssX");
String date = yyyyMMddTHHmmssSDF.format(yourDate);
答案 1 :(得分:1)
只需将时间戳传递给Date
类型:
Timestamp stamp = new Timestamp(inputTimestamp);
Date date = new Date(stamp.getTime());
//change the to the format that you need
DateFormat df = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss");
String timeStr = df.format(date);