例如,我有日期:“23/2/2010”(2010年2月23日)。我想把它传递给一个返回星期几的函数。我该怎么做呢?
在这个例子中,函数应该返回String "Tue"。
此外,如果只需要日期顺序,如何检索?
例如,我有日期:“23/2/2010”(2010年2月23日)。我想把它传递给一个返回星期几的函数。我该怎么做呢?
在这个例子中,函数应该返回String "Tue"。
此外,如果只需要日期顺序,如何检索?
当前回答
添加了另一种完全符合OP要求的方法,而不使用最新的内置方法:
public static String getDay(String inputDate) {
String dayOfWeek = null;
String[] days = new String[]{"Mon", "Tue", "Wed", "Thu", "Fri", "Sat", "Sun"};
try {
SimpleDateFormat format1 = new SimpleDateFormat("dd/MM/yyyy");
Date dt1 = format1.parse(inputDate);
dayOfWeek = days[dt1.getDay() - 1];
} catch(Exception e) {
System.out.println(e);
}
return dayOfWeek;
}
其他回答
//to get day of any date
import java.util.Scanner;
import java.util.Calendar;
import java.util.Date;
public class Show {
public static String getDay(String day,String month, String year){
String input_date = month+"/"+day+"/"+year;
Date now = new Date(input_date);
Calendar calendar = Calendar.getInstance();
calendar.setTime(now);
int final_day = (calendar.get(Calendar.DAY_OF_WEEK));
String finalDay[]={"SUNDAY","MONDAY","TUESDAY","WEDNESDAY","THURSDAY","FRIDAY","SATURDAY"};
System.out.println(finalDay[final_day-1]);
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String month = in.next();
String day = in.next();
String year = in.next();
getDay(day, month, year);
}
}
另一个“有趣”的方法是使用末日算法。这是一个更长的方法,但如果您不需要用给定的日期创建Calendar对象,那么它也更快。
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.ArrayList;
import java.util.Calendar;
import java.util.HashMap;
import java.util.Map;
import java.util.Random;
/**
*
* @author alain.janinmanificat
*/
public class Doomsday {
public static HashMap<Integer, ArrayList<Integer>> anchorDaysMap = new HashMap<>();
public static HashMap<Integer, Integer> doomsdayDate = new HashMap<>();
public static String weekdays[] = new DateFormatSymbols(Locale.FRENCH).getWeekdays();
/**
* @param args the command line arguments
*/
public static void main(String[] args) throws ParseException, ParseException {
// Map is fed manually but we can use this to calculate it : http://en.wikipedia.org/wiki/Doomsday_rule#Finding_a_century.27s_anchor_day
anchorDaysMap.put(Integer.valueOf(0), new ArrayList<Integer>() {
{
add(Integer.valueOf(1700));
add(Integer.valueOf(2100));
add(Integer.valueOf(2500));
}
});
anchorDaysMap.put(Integer.valueOf(2), new ArrayList<Integer>() {
{
add(Integer.valueOf(1600));
add(Integer.valueOf(2000));
add(Integer.valueOf(2400));
}
});
anchorDaysMap.put(Integer.valueOf(3), new ArrayList<Integer>() {
{
add(Integer.valueOf(1500));
add(Integer.valueOf(1900));
add(Integer.valueOf(2300));
}
});
anchorDaysMap.put(Integer.valueOf(5), new ArrayList<Integer>() {
{
add(Integer.valueOf(1800));
add(Integer.valueOf(2200));
add(Integer.valueOf(2600));
}
});
//Some reference date that always land on Doomsday
doomsdayDate.put(Integer.valueOf(1), Integer.valueOf(3));
doomsdayDate.put(Integer.valueOf(2), Integer.valueOf(14));
doomsdayDate.put(Integer.valueOf(3), Integer.valueOf(14));
doomsdayDate.put(Integer.valueOf(4), Integer.valueOf(4));
doomsdayDate.put(Integer.valueOf(5), Integer.valueOf(9));
doomsdayDate.put(Integer.valueOf(6), Integer.valueOf(6));
doomsdayDate.put(Integer.valueOf(7), Integer.valueOf(4));
doomsdayDate.put(Integer.valueOf(8), Integer.valueOf(8));
doomsdayDate.put(Integer.valueOf(9), Integer.valueOf(5));
doomsdayDate.put(Integer.valueOf(10), Integer.valueOf(10));
doomsdayDate.put(Integer.valueOf(11), Integer.valueOf(7));
doomsdayDate.put(Integer.valueOf(12), Integer.valueOf(12));
long time = System.currentTimeMillis();
for (int i = 0; i < 100000; i++) {
//Get a random date
int year = 1583 + new Random().nextInt(500);
int month = 1 + new Random().nextInt(12);
int day = 1 + new Random().nextInt(7);
//Get anchor day and DoomsDay for current date
int twoDigitsYear = (year % 100);
int century = year - twoDigitsYear;
int adForCentury = getADCentury(century);
int dd = ((int) twoDigitsYear / 12) + twoDigitsYear % 12 + (int) ((twoDigitsYear % 12) / 4);
//Get the gap between current date and a reference DoomsDay date
int referenceDay = doomsdayDate.get(month);
int gap = (day - referenceDay) % 7;
int result = (gap + adForCentury + dd) % 7;
if(result<0){
result*=-1;
}
String dayDate= weekdays[(result + 1) % 8];
//System.out.println("day:" + dayDate);
}
System.out.println("time (ms) : " + (System.currentTimeMillis() - time)); //time (ms) : 80
time = System.currentTimeMillis();
for (int i = 0; i < 100000; i++) {
Calendar c = Calendar.getInstance();
//I should have used random date here too, but it's already slower this way
c.setTime(new SimpleDateFormat("dd/MM/yyyy").parse("12/04/1861"));
// System.out.println(new SimpleDateFormat("EE").format(c.getTime()));
int result2 = c.get(Calendar.DAY_OF_WEEK);
// System.out.println("day idx :"+ result2);
}
System.out.println("time (ms) : " + (System.currentTimeMillis() - time)); //time (ms) : 884
}
public static int getADCentury(int century) {
for (Map.Entry<Integer, ArrayList<Integer>> entry : anchorDaysMap.entrySet()) {
if (entry.getValue().contains(Integer.valueOf(century))) {
return entry.getKey();
}
}
return 0;
}
}
您可以尝试以下代码:
import java.time.*;
public class Test{
public static void main(String[] args) {
DayOfWeek dow = LocalDate.of(2010,Month.FEBRUARY,23).getDayOfWeek();
String s = String.valueOf(dow);
System.out.println(String.format("%.3s",s));
}
}
方法下面检索七天,并返回短名称的天在列表数组在Kotlin,你可以重新格式化然后在Java格式,只是提出想法日历可以返回短名称
private fun getDayDisplayName():List<String>{
val calendar = Calendar.getInstance()
val dates= mutableListOf<String>()
dates.clear()
val s= calendar.getDisplayName(DAY_OF_WEEK, SHORT, Locale.US)
dates.add(s)
for(i in 0..5){
calendar.roll( Calendar.DATE, -1)
dates.add(calendar.getDisplayName(DAY_OF_WEEK, SHORT, Locale.US))
}
return dates.toList()
}
结果是这样的
I/System.out: Wed
Tue
Mon
Sun
I/System.out: Sat
Fri
Thu
我用这个
String[] weekdays = "Sunday Monday Tuesday Wednesday Thursday Friday Saturday".split(" ");
然后
weekdays[calendar.get(Calendar.DAY_OF_WEEK) - 1]
获取特定的工作日