java – LocalDate减去Period得到错误的结果

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LocalDate减去一个Period(如“28年,1个月和27天”),得到错误的结果.

但减去一个时期(只有天数单位,如“10282”天)得到正确的结果.
有什么需要注意的吗?

public static void main(String[] args) {
    printAgeAndBirthday(1989,2,22);

    printBirthdayFromPeriod(28,1,27);
}
  private static void printBirthdayFromPeriod(int years,int months,int days) {
    final Period period = Period.of(years,months,days);
    final LocalDate now = LocalDate.now();
    final LocalDate birthday = now.minus(28,ChronoUnit.YEARS)
            .minus(1,ChronoUnit.MONTHS)
            .minus(27,ChronoUnit.DAYS);

    System.out.println("your birthday is : "+ birthday);//1989-02-19
    System.out.println("your birthday is : "+ now.minusYears(28).minusMonths(1).minusDays(27));//1989-02-19
    System.out.println("your birthday is : "+ now.minus(period));//1989-02-19
    System.out.println("your birthday is : "+period.subtractFrom(now));//1989-02-19
    System.out.println("your birthday is : "+ now.minus(Period.ofDays(10282)));//1989-02-22
}

private static void printAgeAndBirthday(int year,int month,int dayOfMonth) {
    LocalDate today = LocalDate.now();
    LocalDate birthday = LocalDate.of(year,month,dayOfMonth);

    Period p = Period.between(birthday,today);
    long p2 = ChronoUnit.DAYS.between(birthday,today);
    System.out.printf("You are %d years,%d months,and %d days old. (%d days total)%n",p.getYears(),p.getMonths(),p.getDays(),p2);

    LocalDate nextBDay = birthday.withYear(today.getYear());

    //If your birthday has occurred this year already,add 1 to the year.
    if (nextBDay.isBefore(today) || nextBDay.isEqual(today)) {
        nextBDay = nextBDay.plusYears(1);
    }

    Period p_1 = Period.between(today,nextBDay);
    long p_2 = ChronoUnit.DAYS.between(today,nextBDay);
    System.out.printf("There are %d months,and %d days until your next birthday. (%d total)%n",p_1.getMonths(),p_1.getDays(),p_2);
}

控制台日志:

You are 28 years,1 months,and 27 days old. (10282 days total)
There are 10 months,and 4 days until your next birthday. (310 total)
your birthday is : 1989-02-19
your birthday is : 1989-02-19
your birthday is : 1989-02-19
your birthday is : 1989-02-19
your birthday is : 1989-02-22

java版本:jdk1.8.0_45

解决方法

您的案例可以简化为
LocalDate date1 = LocalDate.of(2017,22),date2 = LocalDate.of(2017,4,18);
Period p = Period.between(date1,date2);
System.out.println("date1 + p: "+date1.plus(p));
System.out.println("date2 - p: "+date2.minus(p));

这将打印

date1 + p: 2017-04-18
date2 - p: 2017-02-19

换句话说,年数是无关紧要的(除非其中一年是闰年而另一年不是,但在这里,两者都不是).以下说明了问题:

February                       March                                                                                        April
19 20 21 22 23 24 25 26 27 28  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18
 ↑        │                                                                                   ↑                                                                                ↑
 │        └──────────────────────────── plus one Month ───────────────────────────────────────┴───────────────────────── plus 27 days ─────────────────────────────────────────┤
 │                                                                                ↑                                                                                            ↓
 └───────────────────────── minus 27 days ────────────────────────────────────────┴─────────────────── minus one month ────────────────────────────────────────────────────────┘

如果你改变方向,这将改变:

Period p2 = Period.between(date2,date1);
System.out.println("date1 - p2: "+date1.minus(p2));
System.out.println("date2 + p2: "+date2.plus(p2));

这将打印

date1 - p2: 2017-04-15
date2 + p2: 2017-02-22

因此,当您以年,月和日表示期间时,方向变得相关.相反,两个日期之间的平均天数是不变的:

LocalDate date1 = LocalDate.of(2017,18);
Period p = Period.ofDays((int)ChronoUnit.DAYS.between(date1,date2));
System.out.println("date1 + p: "+date1.plus(p));
System.out.println("date2 - p: "+date2.minus(p));
date1 + p: 2017-04-18
date2 - p: 2017-02-22
原文链接:https://www.f2er.com/java/125498.html

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