代码之家 › 专栏 › 技术社区 › Bruce the Hoon

使用JAVA从日期识别季节

date java

Bruce the Hoon · 技术社区 · 18 年前

从那以后我除了好运什么都没有了,为什么不再试一次呢?

我有一个应用程序需要根据一年中的季节(春天、夏天、冬天、秋天)显示不同的图像。这些季节我有非常具体的开始和结束日期。

我想从你们这些天才那里得到的是一个名为getSeason的方法,它将日期作为输入,并返回一个字符串值:春、夏、冬或秋。以下是日期范围及其相关季节:

春季:3/1-4/30
夏季:5/1-8/31
秋季:9 / 1-10 / 31
冬季:11/1-2/28

有人能提供一种工作方法来返回合适的季节吗? 谢谢大家!

9 回复 | 直到 7 年前

Vlad Holubiev 10 年前

似乎只要检查一下月份就可以了:

private static final String seasons[] = {
    "Winter", "Winter",
    "Spring", "Spring", "Spring",
    "Summer", "Summer", "Summer",
    "Fall", "Fall", "Fall",
    "Winter"
};
public String getSeason( Date date ) {
   return seasons[ date.getMonth() ];
}

// As stated above, getMonth() is deprecated, but if you start with a Date, 
// you'd have to convert to Calendar before continuing with new Java, 
// and that's not fast.

Basil Bourque 10 年前

这里有一些很好的答案,但它们已经过时了。java.time类使这项工作更加容易。

Java.时间

与最早版本的Java捆绑在一起的麻烦的旧类已经被 java.time 构建到Java 8和以后的类。见 Oracle Tutorial . 许多功能已经被移植到Java 6和7中。 ThreeTen-Backport 并进一步适应安卓 ThreeTenABP .

`Month`

考虑到这里用整个月来定义季节,我们可以利用 Month enum . 这样的枚举值比仅仅的整数值(1-12)要好,因为它们是类型安全的,并且可以保证有效值。

`EnumSet`

安 EnumSet 是跟踪枚举值子集的一种快速、紧凑的内存方法。

EnumSet<Month> spring = EnumSet.of( Month.MARCH , Month.APRIL );
EnumSet<Month> summer = EnumSet.of( Month.MAY , Month.JUNE , Month.JULY , Month.AUGUST );
EnumSet<Month> fall = EnumSet.of( Month.SEPTEMBER , Month.OCTOBER );
EnumSet<Month> winter = EnumSet.of( Month.NOVEMBER , Month.DECEMBER , Month.JANUARY , Month.FEBRUARY );

例如,我们得到特定时区的当前时刻。

ZoneId zoneId = ZoneId.of( "America/Montreal" );
ZonedDateTime zdt = ZonedDateTime.now( zoneId );

询问日期时间值 月份 .

Month month = Month.from( zdt );

寻找哪个季节 枚举集 通过调用 contains .

if ( spring.contains( month ) ) {
    â¦
} else if ( summer.contains( month ) ) {
    â¦
} else if ( fall.contains( month ) ) {
    â¦
} else if ( winter.contains( month ) ) {
    â¦
} else {
    // FIXME: Handle reaching impossible point as error condition.
}

定义您自己的SeasonEnum

如果您在代码库中使用这个季节概念,我建议您定义自己的枚举,即__season_。

基本枚举很简单: public enum Season { SPRING, SUMMER, FALL, WINTER; } . 但是我们也添加了一个静态方法 of 查找哪个月对应哪个季节。

package com.example.javatimestuff;

import java.time.Month;

public enum Season {
    SPRING, SUMMER, FALL, WINTER;

    static public Season of ( Month month ) {
        switch ( month ) {

            // Spring.
            case MARCH:  // Java quirk: An enum switch case label must be the unqualified name of an enum. So cannot use `Month.MARCH` here, only `MARCH`.
                return Season.SPRING;

            case APRIL:
                return Season.SPRING;

            // Summer.
            case MAY:
                return Season.SUMMER;

            case JUNE:
                return Season.SUMMER;

            case JULY:
                return Season.SUMMER;

            case AUGUST:
                return Season.SUMMER;

            // Fall.
            case SEPTEMBER:
                return Season.FALL;

            case OCTOBER:
                return Season.FALL;

            // Winter.
            case NOVEMBER:
                return Season.WINTER;

            case DECEMBER:
                return Season.WINTER;

            case JANUARY:
                return Season.WINTER;

            case FEBRUARY:
                return Season.WINTER;

            default:
                System.out.println ( "ERROR." );  // FIXME: Handle reaching impossible point as error condition.
                return null;
        }
    }

}

下面是如何使用该枚举。

ZoneId zoneId = ZoneId.of ( "America/Montreal" );
ZonedDateTime zdt = ZonedDateTime.now ( zoneId );
Month month = Month.from ( zdt );
Season season = Season.of ( month );

转储到控制台。

System.out.println ( "zdt: " + zdt + " |  month: " + month + " | season: " + season );

ZDT:2016-06-25T18:23:14.695-04:00[美国/蒙特利尔]月:6月季:夏季

helloandre 18 年前

我觉得受人恩惠,但受宠若惊。所以我来做。

这不仅检查月份,而且检查每月的某一天。

import java.util.*

public String getSeason(Date today, int year){

    // the months are one less because GC is 0-based for the months, but not days.
    // i.e. 0 = January.
    String returnMe = "";

    GregorianCalender dateToday = new GregorianCalender(year, today.get(Calender.MONTH_OF_YEAR), today.get(Calender.DAY_OF_MONTH);
    GregorianCalender springstart = new GregorianCalender(year, 2, 1);
    GregorianCalender springend = new GregorianCalender(year, 3, 30);
    GregorianCalender summerstart = new GregorianCalender(year, 4, 1);
    GregorianCalender summerend = new GregorianCalender(year, 7, 31);
    GregorianCalender fallstart = new GregorianCalender(year, 8, 1);
    GregorianCalender fallend = new GregorianCalender(year, 9, 31);
    GregorianCalender winterstart = new GregorianCalender(year, 10, 1);
    GregorianCalender winterend = new GregorianCalender(year, 1, 28);

    if ((dateToday.after(springstart) && dateToday.before(springend)) || dateToday.equals(springstart) || dateToday.equals(springend)){
        returnMe = "Spring";

    else if ((dateToday.after(summerstart) && dateToday.before(summerend)) || dateToday.equals(summerstart) || dateToday.equals(summerend)){
        returnMe = "Summer";

    else if ((dateToday.after(fallstart) && dateToday.before(fallend)) || dateToday.equals(fallstart) || dateToday.equals(fallend)){
        returnMe = "Fall";

    else if ((dateToday.after(winterstart) && dateToday.before(winterend)) || dateToday.equals(winterstart) || dateToday.equals(winterend)){
        returnMe = "Winter";

    else {
        returnMe = "Invalid";
    }
    return returnMe;
}

我相信这很可怕,可以改进。在评论中告诉我。

nsayer 18 年前

嗯,它可以像

String getSeason(int month) {
    switch(month) {
          case 11:
          case 12:
          case 1:
          case 2:
                return "winter";
          case 3:
          case 4:
                return "spring";
          case 5:
          case 6:
          case 7:
          case 8:
                return "summer";
          default:
                return "autumn";
      }
}

在评论中,我被指责为一个更好的解决方案:枚举:

public static Enum Season {
    WINTER(Arrays.asList(11,12,1,2)),
    SPRING(Arrays.asList(3,4)),
    SUMMER(Arrays.asList(5,6,7,8)),
    AUTUMN(Arrays.asList(9,10));

    Season(List<Integer> months) {
        this.monthlist = months;
    }
    private List<Integer> monthlist;
    public boolean inSeason(int month) {
        return this.monthlist.contains(month);  // if months are 0 based, then insert +1 before the )
    }

    public static Season seasonForMonth(int month) {
        for(Season s: Season.values()) {
            if (s.inSeason(month))
                 return s;
        }
        throw new IllegalArgumentException("Unknown month");
    }
}

Jérôme Verstrynge 15 年前

import java.util.Scanner;

public class lab6project1
{
 public static void main(String[] args)
 {
  Scanner keyboard = new Scanner(System.in);

  System.out.println("This program reports the season for a given day and month");
  System.out.println("Please enter the month and day as integers with a space between the month and day");

  int month = keyboard.nextInt();
  int day = keyboard.nextInt();


  if ( (month == 1) || (month == 2))
   System.out.println("The season is Winter");

  else if ( (month == 4) || (month == 5))
   System.out.println("The season is Spring");

  else if ( (month == 7) || (month == 8))
   System.out.println("The season is Summer");

  else if ( (month == 10)|| (month == 11))
   System.out.println("The season is Fall");

  else if ( (month == 3) && (day <= 19 ))
   System.out.println("The season is Winter");

  else if ( (month == 3) && (day >= 20 ))
   System.out.println("The season is Spring");

  else if ( (month == 6) && (day <= 20 ))
   System.out.println("The season is Spring");

  else if ( (month == 6) && (day >= 21 ))
   System.out.println("The season is Summer");

  else if ( (month == 9) && (day <= 20 ))
   System.out.println("The season is Summer");

  else if ( (month == 9) && (day >= 21 ))
   System.out.println("The season is Autumn");

  else if ( (month == 12) && (day <= 21 ))
   System.out.println("The season is Autumn");

  else if ( (month == 12) && (day >= 22 ))
   System.out.println("The season is Winter");


 }

}

ECE 12 年前

尝试使用哈希表或枚举。您可以将日期转换为某个值(1月1日为1,…),然后为某个字段创建容器。或者可以对月份进行枚举。一月:冬天,二月:冬天,…七月:夏天,等等

Jorn 18 年前

因为在这个范围内,所有季节都是完整的月份,所以您可以切换到日期的月份:

switch (date.getMonth()) {
    case Calendar.JANUARY:
    case Calendar.FEBRUARY:
         return "winter";
    case Calendar.MARCH:
         return "spring";
    //etc
}

我建议使用所有12个日历常量来完成整个切换,而不是最后一个日历常量的默认值。然后可以确保输入正确,例如

default:
 throw new IllegalArgumentException();

最后。

根据您的用例,您可能还希望在季节中使用枚举,而不是简单的字符串。

注意Debug .GETMUNTH()方法,你应该使用java. UTI.Calase.GET(日历。月)来代替。(只需使用calendar.setdate(yourdate)将日期转换为日历)

Volodymyr Machekhin 10 年前

简单解

 Calendar calendar = Calendar.getInstance();
 calendar.setTimeInMillis(timeInMills);
 int month = calendar.get(Calendar.MONTH);
 CurrentSeason = month == 11 ? 0 : (month + 1) / 3;

Meno Hochschild 7 年前

您问题的标题非常笼统,因此大多数用户会首先想到天文季节。尽管您问题的详细内容仅限于自定义的日期范围,但这一限制可能是由于无法计算天文情况造成的,因此我敢对这个旧问题发表答案,也适用于天文情况。

这里的大多数答案都是基于完整的月份。我在这里举两个例子来说明天文季节和基于任意日期范围的季节。

a)任意日期范围到季节的映射

在这里 我们肯定需要额外的信息,具体的时区 或者抵消,否则我们不能翻译一个瞬间(就像老式的 java.util.Date -输入的实例)使用月和日的组合到本地表示。为了简单起见,我假设系统时区。

    // your input
    java.util.Date d = new java.util.Date();
    ZoneId tz = ZoneId.systemDefault();

    // extract the relevant month-day
    ZonedDateTime zdt = d.toInstant().atZone(tz);
    MonthDay md = MonthDay.of(zdt.getMonth(), zdt.getDayOfMonth());

    // a definition with day-of-month other than first is possible here
    MonthDay beginOfSpring = MonthDay.of(3, 1);
    MonthDay beginOfSummer = MonthDay.of(5, 1);
    MonthDay beginOfAutumn = MonthDay.of(9, 1);
    MonthDay beginOfWinter = MonthDay.of(11, 1);

    // determine the season
    Season result;

    if (md.isBefore(beginOfSpring)) {
        result = Season.WINTER;
    } else if (md.isBefore(beginOfSummer)) {
        result = Season.SPRING;
    } else if (md.isBefore(beginOfAutumn)) {
        result = Season.SUMMER;
    } else if (md.isBefore(beginOfWinter)) {
        result = Season.FALL;
    } else {
        result = Season.WINTER;
    }

    System.out.println(result);

我使用了一个简单的helper枚举 public enum Season { SPRING, SUMMER, FALL, WINTER; } .

b)天文季节

这里我们还需要一个额外的信息,即如果季节是在北半球或南半球。我的图书馆 Time4J 基于预定义的枚举提供以下解决方案 AstronomicalSeason 使用版本v5.2:

    // your input
    java.util.Date d = new java.util.Date();
    boolean isSouthern = false;

    Moment m = TemporalType.JAVA_UTIL_DATE.translate(d);
    AstronomicalSeason result = AstronomicalSeason.of(m);

    if (isSouthern) { // switch to southern equivalent if necessary
        result = result.onSouthernHemisphere();
    }

    System.out.println(result);