Java SimpleDateFormat: Pattern - ParseException [duplicate]

问题

This question already has an answer here:

Datetime parsing error (1 answer)

Closed last year.

I am struggling with a date string, I need to parse into the java ‘Date’ object. Here is what I have got so far:

try {

String value =   "2017‎-‎11‎-‎23T14:00:49.184000000Z";
String pattern = "yyyy-MM-dd'T'HH:mm:ss.SSSSSSSSS'Z'";

SimpleDateFormat parser = new SimpleDateFormat(pattern);

Date date = parser.parse(value);


} catch (ParseException e) {e.printStackTrace();}

It currently throws a ParseException “Unparseable date” and I can’t get it to work.

Any help is highly appreciated!

Thanks

回答1:

Use Instant from java.time package (java 8) instead, it should look like below

String value = "2017-11-23T14:00:49.184000000Z";
Instant instant = Instant.parse(value);
Date date = Date.from(instant);
System.out.println(date);

回答2:

you can use timeZone as well like this as another solution.

TimeZone tz = TimeZone.getTimeZone("Asia/Calcutta");
            Calendar cal = Calendar.getInstance(tz);
            SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSSSSSSSS'Z'");
            sdf.setCalendar(cal);
            cal.setTime(sdf.parse("2017-11-23T14:58:00.184000000Z"));
            Date date = cal.getTime();
            System.out.println(date);

来源：https://stackoverflow.com/questions/47926442/java-simpledateformat-pattern-parseexception