DateTimeFormatter could not be parsed using "HH:mm E d MMM YYYY" pattern

Question

I am retrieving a date/time from an external data source, this is returned in the following format "14:30 Sat 05 May" with no year.

I've been trying to parse this to a LocalDateTime unsuccessfully. The data returned does not return a year as it is an assumption that we are always operating in the current year.

//date to parse
String time = "14:30 Sat 05 May";

//specify date format matching above string
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("HH:mm E d MMM YYYY") ;

//we do not have a year returned but i can make the assumption we use the current year
LocalDateTime formatDateTime = LocalDateTime.parse(time, formatter).withYear(2018);

The above code throws the following exception

Exception in thread "main" java.time.format.DateTimeParseException: Text '14:30 Sat 05 May' could not be parsed at index 16

Any help appreciated.

Answer 1

Default year

Specify a default year in your DateTimeFormatter, using the DateTimeFormatterBuilder class by calling parseDefaulting and specifying the year-field with ChronoField.YEAR.

DateTimeFormatter formatter = new DateTimeFormatterBuilder()
    .appendPattern("HH:mm E d MMM")
    .parseDefaulting(ChronoField.YEAR, 2018)  // <------
    .toFormatter(Locale.ENGLISH);

With this formatter instead of yours:

LocalDateTime.parse( "14:30 Sat 05 May" , formatter ) 

…I get:

2018-05-05T14:30

See that code run live at IdeOne.com.

Points to note:

• Your format pattern string needs to match the parsed string end-to-end. So when your date-time string doesn’t have a year in it, don’t include YYYY in your format pattern.
• In any case don’t use uppercase YYYY here. It’s for week-based year and only useful with week numbers. If your string had had a year in it, you should have used uuuu or lowercase yyyy.
• Make it a habit to give explicit locale to your formatter so you know it also works on other computers, and on yours when one day you play with its settings.