java - Parsing timestamp as LocalDateTime

Question

I want to parse a timestamp in the form of yyyy-MM-dd'T'HH:mm:ss as LocalDateTime. When doing so, it strips the seconds if they are 00.

As described here, I need to use a custom formatter

LocalDateTime date = LocalDateTime.parse("2008-10-02T12:30:00");
DateTimeFormatter f = DateTimeFormatter.ofPattern("yyyy-MM-dd'T'HH:mm:ss");

String dateString = date.toString();
String dateFormatted = date.format(f);

System.out.println(dateString); // 2008-10-02T12:30

This one works, but it returns a String:

System.out.println(dateFormatted); // 2008-10-02T12:30:00

When I parse the string to LocalDateTime it strips the 00 again:

LocalDateTime dateLDT = LocalDateTime.parse(dateFormatted, f);
System.out.println(dateLDT); // 2008-10-02T12:30

So how can I parse a date as LocalDateTime, instead of String, and keep the 00 at the end?

Answer 1

You should expect a difference in output between

LocalDateTime dateLDT = LocalDateTime.parse(dateFormatted, f);
System.out.println(dateLDT);

And

System.out.println(dateLDT.format(f)) //or f.format(dateLDT)

System.out.println(dateLDT); prints the value of dateLDT.toString(), which is not expected to produce the same output as your pattern.

When you look at LocalDateTime.toString(), you'll see that it delegates the time part to LocalTime.toString(), which prints seconds conditionally:

public String toString() {
    ...
    if (secondValue > 0 || nanoValue > 0) {
        buf.append(secondValue < 10 ? ":0" : ":").append(secondValue);
        ...
        }
    }
    return buf.toString();
}

It simply omits the seconds field if its value is 0.

What you need to do in this case is always use a DateTimeFormatter to format your date if you have to be certain about the output/input format.