SimpleDateFormat with pattern yyyyMMddhhmmss unable to parse date "20160327020727"

Question

I am getting an exception when parsing date 20160327020727 with format yyyyMMddhhmmss. Note that the lenient is set to false.

    SimpleDateFormat df = new SimpleDateFormat("yyyyMMddhhmmss");
    df.setLenient(false);
    try {
        Date dt = df.parse("20160327020727");
    } catch (ParseException e) {
        e.printStackTrace();
    }

It is parsing other dates with the same format and it is working as expected. Why is this happening?

java.text.ParseException: Unparseable date: "20160327020727" — Amer Hukic, Aug 19 '16 at 15:24
Possibly something to do with you using `hh` which is the 12 hour format. Do you want `HH` for the 24 hour format. — greg-449, Aug 19 '16 at 15:30
One error is that it probably should not be `hh` (**12** hour format) but `HH` (**24** hour format). — Joop Eggen, Aug 19 '16 at 15:30

score 7 · Accepted Answer · answered Aug 19 '16 at 15:32

7

CET changes to summer time the last Sunday of march, so there is no 2AM this day.

You go from 1:59 to 3:00

answered Aug 19 '16 at 15:32

Akah

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DevNico · Answer 2 · 2016-08-19T15:35:01.223

2

You are getting an error because that time does not exist in your default time zone.

Try setting the timezone to UTC by doing df.setTimeZone(TimeZone.getTimeZone("UTC"));

In CET on the last Sunday of march it changes to summertime -> No 2AM on that day.

edited Aug 19 '16 at 15:35

answered Aug 19 '16 at 15:25

DevNico

138
1
12

score -1 · Answer 3 · edited Nov 27 '17 at 09:46

-1

Change it to "yyyyMMdd HHmmss", so you can parse it easily.

edited Nov 27 '17 at 09:46

Bugs

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answered Nov 27 '17 at 09:24

RayChang

1

SimpleDateFormat with pattern yyyyMMddhhmmss unable to parse date "20160327020727"

3 Answers3