I have the following code:
String ModifiedDate = "1993-06-08T18:27:02.000Z" ;
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSSZ");
Date ModDate = sdf.parse(ModifiedDate);
I am getting the following exception even though my date format is fine...
java.text.ParseException: Unparseable date: "1993-06-08T18:27:02.000Z"
at java.text.DateFormat.parse(DateFormat.java:337)
The Z pattern latter indicates an RFC 822 time zone. Your string
String ModifiedDate = "1993-06-08T18:27:02.000Z" ;
does not contain such a time zone. It contains a Z literally.
You'll want a date pattern, that similarly to the literal T, has a literal Z.
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSS'Z'");
If you meant for Z to indicate Zulu time, add that as a timezone when constructing the SimpleDateFormat
sdf.setTimeZone(TimeZone.getTimeZone("Zulu"));;
The modern Date-Time API, released with Java-8 in March 2014, supplanted the legacy date-time API (the java.util Date-Time API and their formatting API, SimpleDateFormat). Since then, it is strongly recommended to stop using the error-prone legacy API and switch to java.time API.
Solution using java.time API: java.time API is based on ISO 8601 and therefore you do not need a DateTimeFormatter to parse a date-time string which is already in ISO 8601 format (e.g. your date-time string, 1993-06-08T18:27:02.000Z).
import java.time.Instant;
import java.time.OffsetDateTime;
import java.time.ZonedDateTime;
class Main {
public static void main(String[] args) {
String strModifiedDate = "1993-06-08T18:27:02.000Z";
Instant instant = Instant.parse(strModifiedDate);
System.out.println(instant);
// It can also be directly parsed into a ZonedDateTime
ZonedDateTime zdt = ZonedDateTime.parse(strModifiedDate);
System.out.println(zdt);
// or even into an OffsetDateTime
OffsetDateTime odt = OffsetDateTime.parse(strModifiedDate);
System.out.println(odt);
}
}
Output:
1993-06-08T18:27:02Z
1993-06-08T18:27:02Z
1993-06-08T18:27:02Z
If for any reason, you need java.util.Date instance, you can get it as follows:
Date date = Date.from(instant);
Learn more about the modern Date-Time API from Trail: Date Time.
Here is the solution using the legacy date-time API. First of all, note that 'Z' is not the same as Z. 'Z' is just a character literal whereas Z is the timezone designator for zero-timezone offset (or UTC). So, never use 'Z' in the date-time formatting/parsing pattern; otherwise, it will be interpreted merely as a literal character, and not as a timezone designator.
The correct letter to be used in the pattern for timezone offset is X.
Demo:
class Main {
public static void main(String[] args) throws ParseException {
String strModifiedDate = "1993-06-08T18:27:02.000Z";
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSSX");
Date modifiedDate = sdf.parse(strModifiedDate);
System.out.println(modifiedDate);
}
}
The answer by Sotirios Delimanolis is correct. The Z means Zulu time, a zero offset from UTC (+00:00). In other words, not adjusted to any time zone.
FYI, the Joda-Time library make this work much easier, as does the new java.time package in Java 8.
The format you are using is defined by the ISO 8601 standard. Joda-Time and java.time both parse & generate ISO 8601 strings by default.
A DateTime in Joda-Time knows its own assigned time zone. So as part of the process of parsing, specify a time zone to adjust.
DateTimeZone timeZone = DateTimeZone.forID( "Europe/Paris" );
DateTime dateTime = new DateTime( "1993-06-08T18:27:02.000Z", timeZone );
String output = dateTime.toString();
You can keep the DateTime object in Universal Time if desired.
DateTime dateTime = new DateTime( "1993-06-08T18:27:02.000Z", DateTimeZone.UTC );
When required by other classes, you can generate a java.util.Date object.
java.util.Date date = dateTime.toDate();
