比较java.time.ZonedDateTime的实例,忽略Java 8中比较的秒和毫秒时刻

Question

我正在寻找Java 8中Joda Time的等效方法,比较org.joda.time.DateTime(指定时区)的实例,忽略比较中的秒和毫秒,如下所示.

DateTimeFormatter formatter = DateTimeFormat.forPattern("dd-MMM-yyyy hh:mm:ss:SSS a Z").withZone(DateTimeZone.forID("Asia/Kolkata"));
DateTime first = formatter.parseDateTime("16-Feb-2012 12:03:45:999 AM +05:30");
DateTime second = formatter.parseDateTime("16-Feb-2012 12:03:55:999 AM +05:30");

DateTimeComparator comparator = DateTimeComparator.getInstance(DateTimeFieldType.minuteOfHour());
int compare = comparator.compare(first, second);
System.out.println("compare : " + compare);

比较返回0意味着在忽略比较的秒和毫秒时刻之后,两个对象都被认为是相等的.

DateTimeFieldType此处忽略幅度小于指定的下限的字段.

在Java 8中使用Java Time API执行相同操作的等效方法是什么？

_{说实话,我尝试用Java 8实现同样的功能.}

Answer 1

由于Java-8引入了lambdas和方法引用,Comparator因此几乎没有必要使用专用类,因此它们不存在java.time.你可以写:

Comparator<ZonedDateTime> comparator = Comparator.comparing(
      zdt -> zdt.truncatedTo(ChronoUnit.MINUTES));

完整的例子:

DateTimeFormatter formatter = DateTimeFormatter.ofPattern("dd-MMM-yyyy hh:mm:ss:SSS a X")
        .withLocale(Locale.ENGLISH).withZone(ZoneId.of("Asia/Kolkata"));
ZonedDateTime first = ZonedDateTime.parse("16-Feb-2012 12:03:45:999 AM +0530", formatter);
ZonedDateTime second = ZonedDateTime.parse("16-Feb-2012 12:03:55:999 AM +0530", formatter);
Comparator<ZonedDateTime> comparator = Comparator.comparing(
        zdt -> zdt.truncatedTo(ChronoUnit.MINUTES));
System.out.println(comparator.compare(first, second));

Answer 2

试试ChronoUnit课程.

long minutes = ChronoUnit.MINUTES.between(first, second);