How to convert the input format of date to a specific new format in android

Issue

How to convert the value in the below input to the format of 9:15 pm, 2nd May 2023

Input: 2023-11-20T17:09:08.000Z

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What I tried: I tried to use SimpleDateFormat, But I couldn't get the appropriate format

val curFormater = SimpleDateFormat("dd-MM-yyyy")

Any other attempts result in a parse error ... Is there a proper standard way to achieve this?

Solution

There is. As rzwitserloot already said, immediately ditch the obsolete SimplateDateFormat. Use classes from the java.time package instead.

Then you need to do the following:

1. Parse the string

to a proper date and time model. I picked OffsetDateTime here, as your input string contains a date component, a time component and a time zone thingy (the Z).

OffsetDateTime dateTime = OffsetDateTime.parse(input);

Since the input format is conform to the ISO 8601 standard, you don't have to specify a format; it will parse the string out-of-the-box.

2. Define the right output format

and you can use format to form the desired text. This looks like it:

String output = dateTime.format(DateTimeFormatter.ofPattern("h:mm a, d MMMM uuuu", Locale.UK));

Ordinal suffix

But then there is a small problem. You are using an English ordinal suffix (st, nd, rd, th), and there is no such formatting pattern to output such. Fortunately, you can use a DateTimeFormatterBuilder to have a more fine-grained control over the output. We can use appendText(Map<Long, String>) to look up a value mapped to a certain comnponent field.

final DateTimeFormatter formatter = new DateTimeFormatterBuilder()
    .appendPattern("h:mm a, d")
    .appendText(ChronoField.DAY_OF_MONTH, ordinalSuffixes)
    .appendPattern(" MMMM uuuu")
    .toFormatter(Locale.UK);

This will use the day of the month as a long to lookup that specific value in the given map. So we will need to make sure that we have a map which returns the correct values for the correct keys.

Map<Long, String> ordinalSuffixes = IntStream.rangeClosed(1, 31)
    .mapToObj(n -> Map.entry((long) n, switch (n) {
        case 1, 21, 31 -> "st";
        case 2, 22 -> "nd";
        case 3, 23 -> "rd";
        default -> "th";
    }))
    .collect(Collectors.toUnmodifiableMap(Map.Entry::getKey, Map.Entry::getValue));

Answered By - MC Emperor