How to parse date with optional characters in format

Question

I have the following two dates:

• 8 Oct. 2009
• 13 May 2010

I am using Jackson to convert the date from an rest api to joda Datetime.

Answer 1

Given OP's new comment and requirements, the solution is to use a custom deserializer:

You would do something like this:

@JsonDeserialize(using = MyDateDeserializer.class)
class MyClassThatHasDateField {...}

See tutorial here: http://www.baeldung.com/jackson-deserialization

See an example here: Custom JSON Deserialization with Jackson

OLD ANSWER:

You can use Java's SimpleDateFormat and either:

1. Use a regex to choose the proper pattern
2. Simply try them and catch (and ignore) the exception

Example:

String[] formats = { "dd MMM. yyyy", "dd MM yyyy" };

for (String format : formats)
{
    try
    {
        return new SimpleDateFormat( format ).parse( theDateString );
    }
    catch (ParseException e) {}
}

OR

String[] formats = { "dd MMM. yyyy", "dd MM yyyy" };
String[] patterns = { "\\d+ [a-zA-Z]+\. \d{4}", "\\d+ [a-zA-Z]+ \d{4}" };

for ( int i = 0; i < patterns.length; i++ )
{
  // Create a Pattern object
  Pattern r = Pattern.compile(patterns[ i ] );

  // Now create matcher object.
  Matcher m = r.matcher( theDateString );

  if (m.find( )) {
     return new SimpleDateFormat( formats[ i ] ).parse( theDateString );
  }
}

Answer 2

You could just define two formatters and use the one with a dot, if it is present in the string.

String dateString = "8 Oct. 2009";

DateTimeFormatter formatWithDot    = DateTimeFormat.forPattern("dd MMM. yyyy");
DateTimeFormatter formatWithoutDot = DateTimeFormat.forPattern("dd MMM yyyy");

DateTime date = dateString.indexOf('.') > -1
    ? formatWithDot.parseDateTime(dateString)
    : formatWithoutDot.parseDateTime(dateString);

Answer 3

There would also be a different way by fixing the string for parsing:

String date = "13 May. 2009";
DateTime result = DateTime.parse(date.replace(".",""),
DateTimeFormat.forPattern("dd MMM yyyy"));