2

I've seen answers that let me convert when the format is like

"14/02/1952 14:52:22"

by using DateTimeFormatter

DateTimeFormatter formatter = DateTimeFormat.forPattern("dd/MM/yyyy HH:mm:ss");
DateTime dt = formatter.parseDateTime(string);

How do I apply the same idea to strings of this format:

"Sunday, September 29, 2013 7:59:58 AM PDT"

I am trying 

DateTimeFormatter formatter = DateTimeFormat.forPattern"EEEE, MMMM dd, yyyy h:mm:ss a zzz";
DateTime dt = formatter.parseDateTime(string);

but it's complaining about this.

Popcorn
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1 Answers1

6

2018/Java 10

String value = "Sunday, September 29, 2013 7:59:58 AM PDT";
String format = "EEEE, MMMM dd, yyyy h:mm:ss a zzz";

DateTimeFormatter formatter = new DateTimeFormatterBuilder()
        .parseCaseInsensitive()
        .appendPattern(format)
        .toFormatter(Locale.US);
ZonedDateTime zdt = ZonedDateTime.parse(value, formatter);
System.out.println(zdt);
System.out.println(zdt.format(formatter));

Which prints

2013-09-29T07:59:58-07:00[America/Los_Angeles]
Sunday, September 29, 2013 7:59:58 AM PDT

Original answer

You need to change the expected format to meet the requirements as specified by the SimpleDateFormat

String value = "Sunday, September 29, 2013 7:59:58 AM PDT";
String format = "EEEE, MMMM dd, yyyy h:mm:ss a zzz";
SimpleDateFormat sdf = new SimpleDateFormat(format);
try {
    Date date = sdf.parse(value);
    System.out.println(date);
    System.out.println(sdf.format(date));
} catch (ParseException ex) {
    ex.printStackTrace();
}

Which outputs...

Mon Sep 30 00:59:58 EST 2013
Sunday, September 29, 2013 7:59:58 AM PDT
MadProgrammer
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  • thanks, apparently this doesn't work for DateTimeFormatter: http://stackoverflow.com/questions/6393765/joda-datetime-parser-error – Popcorn Sep 30 '13 at 01:23
  • @Popcorn So stick with `SimpleDateFormat` for the time being. Get the value into a `Date` object and then start taking advantage of the JodaTime API's from there – MadProgrammer Sep 30 '13 at 01:30