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
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
其他回答
private String getDay(Date date){
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("EEEE");
//System.out.println("DAY "+simpleDateFormat.format(date).toUpperCase());
return simpleDateFormat.format(date).toUpperCase();
}
private String getDay(String dateStr){
//dateStr must be in DD-MM-YYYY Formate
Date date = null;
String day=null;
try {
date = new SimpleDateFormat("DD-MM-YYYY").parse(dateStr);
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("EEEE");
//System.out.println("DAY "+simpleDateFormat.format(date).toUpperCase());
day = simpleDateFormat.format(date).toUpperCase();
} catch (ParseException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return day;
}
是的。根据具体情况:
You can use java.util.Calendar: Calendar c = Calendar.getInstance(); c.setTime(yourDate); int dayOfWeek = c.get(Calendar.DAY_OF_WEEK); if you need the output to be Tue rather than 3 (Days of week are indexed starting at 1 for Sunday, see Calendar.SUNDAY), instead of going through a calendar, just reformat the string: new SimpleDateFormat("EE").format(date) (EE meaning "day of week, short version") if you have your input as string, rather than Date, you should use SimpleDateFormat to parse it: new SimpleDateFormat("dd/M/yyyy").parse(dateString) you can use joda-time's DateTime and call dateTime.dayOfWeek() and/or DateTimeFormat. edit: since Java 8 you can now use java.time package instead of joda-time
Calendar cal = Calendar.getInstance(desired date);
cal.setTimeInMillis(System.currentTimeMillis());
int dayOfWeek = cal.get(Calendar.DAY_OF_WEEK);
通过提供当前时间戳获取日值。
简单地使用SimpleDateFormat。
SimpleDateFormat sdf = new SimpleDateFormat("dd/MM/yyyy", java.util.Locale.ENGLISH);
Date myDate = sdf.parse("28/12/2013");
sdf.applyPattern("EEE, d MMM yyyy");
String sMyDate = sdf.format(myDate);
结果是:2013年12月28日星期六
默认构造函数接受“默认”区域设置,因此在需要特定模式时要小心使用它。
public SimpleDateFormat(String pattern) {
this(pattern, Locale.getDefault(Locale.Category.FORMAT));
}
import java.text.SimpleDateFormat;
import java.util.Scanner;
class DayFromDate {
public static void main(String args[]) {
System.out.println("Enter the date(dd/mm/yyyy):");
Scanner scan = new Scanner(System.in);
String Date = scan.nextLine();
try {
boolean dateValid = dateValidate(Date);
if(dateValid == true) {
SimpleDateFormat df = new SimpleDateFormat( "dd/MM/yy" );
java.util.Date date = df.parse( Date );
df.applyPattern( "EEE" );
String day= df.format( date );
if(day.compareTo("Sat") == 0 || day.compareTo("Sun") == 0) {
System.out.println(day + ": Weekend");
} else {
System.out.println(day + ": Weekday");
}
} else {
System.out.println("Invalid Date!!!");
}
} catch(Exception e) {
System.out.println("Invalid Date Formats!!!");
}
}
static public boolean dateValidate(String d) {
String dateArray[] = d.split("/");
int day = Integer.parseInt(dateArray[0]);
int month = Integer.parseInt(dateArray[1]);
int year = Integer.parseInt(dateArray[2]);
System.out.print(day + "\n" + month + "\n" + year + "\n");
boolean leapYear = false;
if((year % 4 == 0) && (year % 100 != 0) || (year % 400 == 0)) {
leapYear = true;
}
if(year > 2099 || year < 1900)
return false;
if(month < 13) {
if(month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12) {
if(day > 31)
return false;
} else if(month == 4 || month == 6 || month == 9 || month == 11) {
if(day > 30)
return false;
} else if(leapYear == true && month == 2) {
if(day > 29)
return false;
} else if(leapYear == false && month == 2) {
if(day > 28)
return false;
}
return true;
} else return false;
}
}