在java中根据时间显示早上、下午、晚上、晚上的消息

Showing Morning, afternoon, evening, night message based on Time in java

提问by Devrath

What i am trying to do::

我正在尝试做什么::

Show message based on

显示消息基于

• Good morning (12am-12pm)
• Good after noon (12pm -4pm)
• Good evening (4pm to 9pm)
• Good night ( 9pm to 6am)

• 早上好（12am-12pm）
• 中午之后好（12pm -4pm）
• 晚上好（下午 4 点至晚上 9 点）
• 晚安（晚上 9 点至早上 6 点）

---

---

CODE::

代码：：

I used 24-hr format to get this logic

我使用 24 小时格式来获取此逻辑

private void getTimeFromAndroid() {
        Date dt = new Date();
        int hours = dt.getHours();
        int min = dt.getMinutes();

        if(hours>=1 || hours<=12){
            Toast.makeText(this, "Good Morning", Toast.LENGTH_SHORT).show();
        }else if(hours>=12 || hours<=16){
            Toast.makeText(this, "Good Afternoon", Toast.LENGTH_SHORT).show();
        }else if(hours>=16 || hours<=21){
            Toast.makeText(this, "Good Evening", Toast.LENGTH_SHORT).show();
        }else if(hours>=21 || hours<=24){
            Toast.makeText(this, "Good Night", Toast.LENGTH_SHORT).show();
        }
    }

---

---

Question:

题：

• Is this this best way of doing it, If no which is the best way

• 这是最好的方法吗，如果不是，这是最好的方法

采纳答案by SMA

You should be doing something like:

你应该做这样的事情：

Calendar c = Calendar.getInstance();
int timeOfDay = c.get(Calendar.HOUR_OF_DAY);

if(timeOfDay >= 0 && timeOfDay < 12){
    Toast.makeText(this, "Good Morning", Toast.LENGTH_SHORT).show();        
}else if(timeOfDay >= 12 && timeOfDay < 16){
    Toast.makeText(this, "Good Afternoon", Toast.LENGTH_SHORT).show();
}else if(timeOfDay >= 16 && timeOfDay < 21){
    Toast.makeText(this, "Good Evening", Toast.LENGTH_SHORT).show();
}else if(timeOfDay >= 21 && timeOfDay < 24){
    Toast.makeText(this, "Good Night", Toast.LENGTH_SHORT).show();
}

回答by Konstantin Yovkov

I would shorten your if/elseifstatement to:

我会将您的if/elseif陈述缩短为：

String greeting = null;
if(hours>=1 && hours<=12){
    greeting = "Good Morning";
} else if(hours>=12 && hours<=16){
    greeting = "Good Afternoon";
} else if(hours>=16 && hours<=21){
    greeting = "Good Evening";
} else if(hours>=21 && hours<=24){
    greeting = "Good Night";
}
Toast.makeText(this, greeting, Toast.LENGTH_SHORT).show();

回答by user987339

You determine if it is in the first interval, and then all other intervals depends on the upper limit. So you can make it even shorter:

您确定它是否在第一个区间，然后所有其他区间取决于上限。所以你可以让它更短：

String greeting = null;
if(hours>=1 && hours<=11){
    greeting = "Good Morning";
} else if(hours<=15){
    greeting = "Good Afternoon";
} else if(hours<=20){
    greeting = "Good Evening";
} else if(hours<=24){
    greeting = "Good Night";
}
Toast.makeText(this, greeting, Toast.LENGTH_SHORT).show();

回答by m2k_72

try this code(get hours and get minute methods in Date class are deprecated.)

试试这个代码（不推荐使用 Date 类中的 get hours 和 get minute 方法。）

 private void getTimeFromAndroid() {
    Date dt = new Date();
    Calendar c = Calendar.getInstance();
    c.setTime(dt);
    int hours = c.get(Calendar.HOUR_OF_DAY);
    int min = c.get(Calendar.MINUTE);

    if(hours>=1 && hours<=12){
        Toast.makeText(this, "Good Morning", Toast.LENGTH_SHORT).show();
    }else if(hours>=12 && hours<=16){
        Toast.makeText(this, "Good Afternoon", Toast.LENGTH_SHORT).show();
    }else if(hours>=16 && hours<=21){
        Toast.makeText(this, "Good Evening", Toast.LENGTH_SHORT).show();
    }else if(hours>=21 && hours<=24){
        Toast.makeText(this, "Good Night", Toast.LENGTH_SHORT).show();
    }
}

回答by SREE

When I write

当我写

Calendar c = Calendar.getInstance();
int timeOfDay = c.get(Calendar.HOUR_OF_DAY);

I didn't get output and it doesn't show any error. Just the timeOfDaywon't get assigned any value in the code. I felt it was because of some threading while Calendar.getInstance()is executed. But when I collapsed the lines it worked for me. See the following code:

我没有得到输出，也没有显示任何错误。只是timeOfDay不会在代码中分配任何值。我觉得这是因为在Calendar.getInstance()执行时进行了一些线程处理。但是当我折叠线路时，它对我有用。请参阅以下代码：

int timeOfDay = Calendar.getInstance().get(Calendar.HOUR_OF_DAY);

if(timeOfDay >= 0 && timeOfDay < 12){
        greeting.setText("Good Morning");
}else if(timeOfDay >= 12 && timeOfDay < 16){
        greeting.setText("Good Afternoon");
}else if(timeOfDay >= 16 && timeOfDay < 21){
        greeting.setText("Good Evening");
}else if(timeOfDay >= 21 && timeOfDay < 24){
        greeting.setText("Good Morning");
}

回答by Yoshua Nahar

java.time

时间

I would advise to use Java 8 LocalTime.

我建议使用 Java 8 LocalTime。

Maybe create a class like this to handle your time of day problem.

也许创建一个这样的类来处理您一天中的时间问题。

public class GreetingMaker { // think of a better name then this.

  private static final LocalTime MORNING = LocalTime.of(0, 0, 0);
  private static final LocalTime AFTER_NOON = LocalTime.of(12, 0, 0);
  private static final LocalTime EVENING = LocalTime.of(16, 0, 0);
  private static final LocalTime NIGHT = LocalTime.of(21, 0, 0);

  private LocalTime now;

  public GreetingMaker(LocalTime now) {
    this.now = now;
  }

  public void printTimeOfDay() { // or return String in your case
    if (between(MORNING, AFTER_NOON)) {
      System.out.println("Good Morning");
    } else if (between(AFTER_NOON, EVENING)) {
      System.out.println("Good Afternoon");
    } else if (between(EVENING, NIGHT)) {
      System.out.println("Good Evening");
    } else {
      System.out.println("Good Night");
    }
  }

  private boolean between(LocalTime start, LocalTime end) {
    return (!now.isBefore(start)) && now.isBefore(end);
  }

}

回答by Meno Hochschild

Using Time4J(or Time4A on Android) enables following solutions which do not need any if-else-statements:

使用Time4J（或 Android 上的 Time4A）可以启用以下不需要任何 if-else 语句的解决方案：

ChronoFormatter<PlainTime> parser =
    ChronoFormatter.ofTimePattern("hh:mm a", PatternType.CLDR, Locale.ENGLISH);
PlainTime time = parser.parse("10:05 AM");

Map<PlainTime, String> table = new HashMap<>();
table.put(PlainTime.of(1), "Good Morning");
table.put(PlainTime.of(12), "Good Afternoon");
table.put(PlainTime.of(16), "Good Evening");
table.put(PlainTime.of(21), "Good Night");
ChronoFormatter<PlainTime> customPrinter=
    ChronoFormatter
      .setUp(PlainTime.axis(), Locale.ENGLISH)
      .addDayPeriod(table)
      .build();
System.out.println(customPrinter.format(time)); // Good Morning

There is also another pattern-based way to let the locale decide in a standard way based on CLDR-data how to format the clock time:

还有另一种基于模式的方法可以让语言环境根据 CLDR 数据以标准方式决定如何格式化时钟时间：

ChronoFormatter<PlainTime> parser =
    ChronoFormatter.ofTimePattern("hh:mm a", PatternType.CLDR, Locale.ENGLISH);
PlainTime time = parser.parse("10:05 AM");

ChronoFormatter<PlainTime> printer1 =
    ChronoFormatter.ofTimePattern("hh:mm B", PatternType.CLDR, Locale.ENGLISH);
System.out.println(printer1.format(time)); // 10:05 in the morning

ChronoFormatter<PlainTime> printer2 =
    ChronoFormatter.ofTimePattern("B", PatternType.CLDR, Locale.ENGLISH)
        .with(Attributes.OUTPUT_CONTEXT, OutputContext.STANDALONE);
System.out.println(printer2.format(time)); // morning

The only other library known to me which can also do this (but in an awkward way) is ICU4J.

我所知道的唯一可以做到这一点的其他图书馆（但以一种尴尬的方式）是 ICU4J。

回答by Sanjay Hadiya

 private String getStringFromMilli(long millis) {

    Calendar c = Calendar.getInstance();
    c.setTimeInMillis(millis);
    int hours = c.get(Calendar.HOUR_OF_DAY);

    if(hours >= 1 && hours <= 12){
        return "MORNING";
    }else if(hours >= 12 && hours <= 16){
        return "AFTERNOON";
    }else if(hours >= 16 && hours <= 21){
        return "EVENING";
    }else if(hours >= 21 && hours <= 24){
        return "NIGHT";
    }
    return null;
}

回答by Vishal Sonawane

For anyone who is looking for the latest Kotlinsyntax for @SMA's answer, here is the helper function :

对于正在为@SMA的答案寻找最新Kotlin语法的任何人，这里是辅助函数：

fun getGreetingMessage():String{
    val c = Calendar.getInstance()
    val timeOfDay = c.get(Calendar.HOUR_OF_DAY)

    return when (timeOfDay) {
           in 0..11 -> "Good Morning"
           in 12..15 -> "Good Afternoon"
           in 16..20 -> "Good Evening"
           in 21..23 -> "Good Night"
           else -> "Hello"
      }
    }

回答by awaik

If somebody looking the same for Dart and Flutter: the code without if statements - easy to read and edit.

如果有人对 Dart 和 Flutter 有相同的看法：没有 if 语句的代码 - 易于阅读和编辑。

main() {
  int hourValue = DateTime.now().hour;
  print(checkDayPeriod(hourValue));
}

String checkDayPeriod(int hour) {
  int _res = 21;
  Map<int, String> dayPeriods = {
    0: 'Good night',
    12: 'Good morning',
    16: 'Good afternoon',
    21: 'Good evening',
  };

  dayPeriods.forEach(
    (key, value) {
      if (hour < key && key <= _res) _res = key;
    },
  );

  return dayPeriods[_res];
}