How to Parse Date in java

Question

Unparseable date: "Tue Jul 03 16:59:51 IST 2018" Exception

String date="Tue Jul 03 16:59:51 IST 2018";

i want to parse it. My code is

SimpleDateFormat newformat=SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssZ")
Date d=newformat.parse(date);

hum, you do you expect your pattern to parse your string ? it's just impossible, string is *Day Month year*, and pattern is *year-month-day*, you should learn to build pattern — azro, Jul 03 '18 at 09:35
`SimpleDateFormat` looks at the first element of the format pattern string, `yyyy`, this means 4-digit year. Then it looks at the first element of your date string, `Tue`. `Tue` cannot be a four digit year. Therefore it throws the exception (it probably only needs to look at the `T` to decide). — Ole V.V., Jul 03 '18 at 13:51
I recommend you avoid the `SimpleDateFormat` class. It is not only long outdated, it is also notoriously troublesome. Today we have so much better in [`java.time`, the modern Java date and time API](https://docs.oracle.com/javase/tutorial/datetime/) and its `DateTimeFormatter`. — Ole V.V., Jul 04 '18 at 08:33

score 2 · Answer 1 · answered Jul 03 '18 at 09:45

2

You are using old and quite problematic classes, especially Date.

For your example, perhaps consider using LocalDateTime

String date = "Tue Jul 03 16:59:51 IST 2018";
DateTimeFormatter format = DateTimeFormatter
        .ofPattern("EEE MMM dd HH:mm:ss z yyyy");
LocalDateTime dateTime = LocalDateTime.parse(date, format);

answered Jul 03 '18 at 09:45

achAmháin

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I can’t decide whether I want to recommend `ZonedDateTime` over `LocalDateTime` here. The string includes a time zone abbreviation, so it could be considered a shame to throw it away from the outset. On the other hand, three and four letter time zone abbreviations are generally ambiguous, so if trying to parse it, we risk a result that conflicts with what was intended. `LocalDateTime` evades that problem. – Ole V.V. Jul 04 '18 at 10:47
1

@OleV.V. Funnily enough I wrote the answer initially with `ZonedDateTime` but edited it to `LocalDateTime`. – achAmháin Jul 04 '18 at 10:55

score 1 · Answer 2 · answered Jul 03 '18 at 09:41

Since your pattern has to match the string you want to parse you need to adjust the pattern as following:

EEE MMM dd HH:mm:ss z yyyy

Infos gathered from the docs.

You should also use the new java.time API introduced with Java 8.

String s = "Tue Jul 03 16:59:51 IST 2018";
//Java 7 way
SimpleDateFormat newformat = new SimpleDateFormat("EEE MMM dd HH:mm:ss z yyyy");
Date d = newformat.parse(s);
//Java 8 way
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("EEE MMM dd HH:mm:ss z yyyy");
LocalDateTime dateTime = LocalDateTime.parse(s, formatter);

How to Parse Date in java

2 Answers2