java SimpleDateFormat

Question

in Java, how to parse a date string that contains a letter that does not represent a pattern?

"2007-11-02T14:46:03+01:00"

String date ="2007-11-02T14:46:03+01:00";
String format = "yyyy-MM-ddTHH:mm:ssz";
new SimpleDateFormat(format).parse(date);

Exception in thread "main" java.lang.IllegalArgumentException: Illegal pattern character 'T'
    at java.text.SimpleDateFormat.compile(SimpleDateFormat.java:769)
    at java.text.SimpleDateFormat.initialize(SimpleDateFormat.java:576)
    at java.text.SimpleDateFormat.(SimpleDateFormat.java:501)
    at java.text.SimpleDateFormat.(SimpleDateFormat.java:476)

Answer 1

No formatter needed

It’s time to post the modern answer, the answer that uses java.time, the modern Java date and time API. Your format is ISO 8601, and the classes of java.time generally parse the most common ISO 8601 variants as their default, that is, without any explicit formatter.

    String date ="2007-11-02T14:46:03+01:00";
    OffsetDateTime dateTime = OffsetDateTime.parse(date);
    System.out.println(dateTime);

Output is:

2007-11-02T14:46:03+01:00

Yes, java.time also gives ISO 8601 format back from the toString methods, implicitly called when we print an object.

Enclose literal letters in single quotes

To answer the question as asked, you may enclose letters in single quotes to make DateTimeFormatter take them as literal letters rather than format specifiers. There would be no point whatsoever in doing the following in real code, but for the sake of demonstration:

    DateTimeFormatter isoFormatter
            = DateTimeFormatter.ofPattern("uuuu-MM-dd'T'HH:mm:ssXXX");
    String date ="2007-11-02T14:46:03+01:00";
    OffsetDateTime dateTime = OffsetDateTime.parse(date, isoFormatter);

The result is the same as before.

Link

Oracle tutorial: Date Time explaining how to use java.time.