如何知道特定年份的特定月份有多少天？
String date = "2010-01-19";
String[] ymd = date.split("-");
int year = Integer.parseInt(ymd[0]);
int month = Integer.parseInt(ymd[1]);
int day = Integer.parseInt(ymd[2]);
Calendar calendar = Calendar.getInstance();
calendar.set(Calendar.YEAR,year);
calendar.set(Calendar.MONTH,month);
int daysQty = calendar.getDaysNumber(); // Something like this
答案 0 :(得分：308)
@Warren M. Nocos。
如果您尝试使用Java 8的新Date and Time API，则可以使用java.time.YearMonth类。请参阅Oracle Tutorial。
// Get the number of days in that month
YearMonth yearMonthObject = YearMonth.of(1999, 2);
int daysInMonth = yearMonthObject.lengthOfMonth(); //28
测试：在闰年尝试一个月：
yearMonthObject = YearMonth.of(2000, 2);
daysInMonth = yearMonthObject.lengthOfMonth(); //29
创建日历，设置年份和月份并使用getActualMaximum
int iYear = 1999;
int iMonth = Calendar.FEBRUARY; // 1 (months begin with 0)
int iDay = 1;
// Create a calendar object and set year and month
Calendar mycal = new GregorianCalendar(iYear, iMonth, iDay);
// Get the number of days in that month
int daysInMonth = mycal.getActualMaximum(Calendar.DAY_OF_MONTH); // 28
测试：在闰年尝试一个月：
mycal = new GregorianCalendar(2000, Calendar.FEBRUARY, 1);
daysInMonth= mycal.getActualMaximum(Calendar.DAY_OF_MONTH); // 29
答案 1 :(得分：37)
如果你必须使用java.util.Calendar，我怀疑你想要：
int days = calendar.getActualMaximum(Calendar.DAY_OF_MONTH);
但就个人而言，我建议先使用Joda Time代替java.util.{Calendar, Date}，在这种情况下您可以使用：
int days = chronology.dayOfMonth().getMaximumValue(date);
请注意，不是单独解析字符串值，而是最好使用您用来解析它的日期/时间API。在java.util.*中，您可以使用SimpleDateFormat;在Joda Time中你使用DateTimeFormatter。
答案 2 :(得分：21)
您可以使用Calendar.getActualMaximum方法：
Calendar calendar = Calendar.getInstance();
calendar.set(Calendar.YEAR, year);
calendar.set(Calendar.MONTH, month);
int numDays = calendar.getActualMaximum(Calendar.DATE);
答案 3 :(得分：17)
java.time.LocalDate 从Java 1.8开始，您可以使用lengthOfMonth上的方法java.time.LocalDate：
LocalDate date = LocalDate.of(2010, 1, 19);
int days = date.lengthOfMonth();
答案 4 :(得分：7)
if (month == 4 || month == 6 || month == 9 || month == 11)
daysInMonth = 30;
else
if (month == 2)
daysInMonth = (leapYear) ? 29 : 28;
else
daysInMonth = 31;
答案 5 :(得分：3)
这是数学方式：
年，月（1至12）：
int daysInMonth = month == 2 ?
28 + (year % 4 == 0 ? 1:0) - (year % 100 == 0 ? (year % 400 == 0 ? 0 : 1) : 0) :
31 - (month-1) % 7 % 2;
答案 6 :(得分：1)
在Java8中，您可以使用日期字段中的get ValueRange。
LocalDateTime dateTime = LocalDateTime.now();
ChronoField chronoField = ChronoField.MONTH_OF_YEAR;
long max = dateTime.range(chronoField).getMaximum();
这允许您在该字段上进行参数化。
答案 7 :(得分：1)
我会寻求这样的解决方案：
int monthNr = getMonth();
final Month monthEnum = Month.of(monthNr);
int daysInMonth;
if (monthNr == 2) {
int year = getYear();
final boolean leapYear = IsoChronology.INSTANCE.isLeapYear(year);
daysInMonth = monthEnum.length(leapYear);
} else {
daysInMonth = monthEnum.maxLength();
}
如果月份不是2月（占案例的92％），则仅取决于月份，并且不涉及年份会更有效。这样，您无需致电逻辑即可知道这是否是a年，并且无需在92％的情况下获得年份。 而且它仍然是干净且可读性强的代码。
答案 8 :(得分：1)
// 1 means Sunday ,2 means Monday .... 7 means Saturday
//month starts with 0 (January)
MonthDisplayHelper monthDisplayHelper = new MonthDisplayHelper(2019,4);
int numbeOfDaysInMonth = monthDisplayHelper.getNumberOfDaysInMonth();
答案 9 :(得分：1)
就这么简单，不需要导入任何东西
public static int getMonthDays(int month, int year) {
int daysInMonth ;
if (month == 4 || month == 6 || month == 9 || month == 11) {
daysInMonth = 30;
}
else {
if (month == 2) {
daysInMonth = (year % 4 == 0) ? 29 : 28;
} else {
daysInMonth = 31;
}
}
return daysInMonth;
}
答案 10 :(得分：0)
这对我来说很好。
import java.util.*;
public class DaysInMonth {
public static void main(String args []) {
Scanner input = new Scanner(System.in);
System.out.print("Enter a year:");
int year = input.nextInt(); //Moved here to get input after the question is asked
System.out.print("Enter a month:");
int month = input.nextInt(); //Moved here to get input after the question is asked
int days = 0; //changed so that it just initializes the variable to zero
boolean isLeapYear = (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
switch (month) {
case 1:
days = 31;
break;
case 2:
if (isLeapYear)
days = 29;
else
days = 28;
break;
case 3:
days = 31;
break;
case 4:
days = 30;
break;
case 5:
days = 31;
break;
case 6:
days = 30;
break;
case 7:
days = 31;
break;
case 8:
days = 31;
break;
case 9:
days = 30;
break;
case 10:
days = 31;
break;
case 11:
days = 30;
break;
case 12:
days = 31;
break;
default:
String response = "Have a Look at what you've done and try again";
System.out.println(response);
System.exit(0);
}
String response = "There are " + days + " Days in Month " + month + " of Year " + year + ".\n";
System.out.println(response); // new line to show the result to the screen.
}
} //abhinavsthakur00@gmail.com
答案 11 :(得分：0)
您可以使用Calendar.getActualMaximum方法：
Calendar calendar = Calendar.getInstance();
calendar.set(Calendar.YEAR, year);
calendar.set(Calendar.MONTH, month-1);
int numDays = calendar.getActualMaximum(Calendar.DATE);
而且月份1是因为月份采用其原始月份而在方法中采用如下参数在Calendar.class中
public int getActualMaximum(int field) {
throw new RuntimeException("Stub!");
}
（int字段）如下所示。
public static final int JANUARY = 0;
public static final int NOVEMBER = 10;
public static final int DECEMBER = 11;
答案 12 :(得分：0)
public class Main {
private static LocalDate local=LocalDate.now();
public static void main(String[] args) {
int month=local.lengthOfMonth();
System.out.println(month);
}
}
答案 13 :(得分：0)
应避免使用过时的Calendar API。
在Java8或更高版本中，可以使用YearMonth完成此操作。
示例代码：
int year = 2011;
int month = 2;
YearMonth yearMonth = YearMonth.of(year, month);
int lengthOfMonth = yearMonth.lengthOfMonth();
System.out.println(lengthOfMonth);
答案 14 :(得分：0)
以下方法将为您提供特定月份的天数
public static int getNoOfDaysInAMonth(String date) {
Calendar cal = Calendar.getInstance();
cal.setTime(date);
return (cal.getActualMaximum(Calendar.DATE));
}
答案 15 :(得分：0)
String MonthOfName = "";
int number_Of_DaysInMonth = 0;
//year,month
numberOfMonth(2018,11); // calling this method to assign values to the variables MonthOfName and number_Of_DaysInMonth
System.out.print("Number Of Days: "+number_Of_DaysInMonth+" name of the month: "+ MonthOfName );
public void numberOfMonth(int year, int month) {
switch (month) {
case 1:
MonthOfName = "January";
number_Of_DaysInMonth = 31;
break;
case 2:
MonthOfName = "February";
if ((year % 400 == 0) || ((year % 4 == 0) && (year % 100 != 0))) {
number_Of_DaysInMonth = 29;
} else {
number_Of_DaysInMonth = 28;
}
break;
case 3:
MonthOfName = "March";
number_Of_DaysInMonth = 31;
break;
case 4:
MonthOfName = "April";
number_Of_DaysInMonth = 30;
break;
case 5:
MonthOfName = "May";
number_Of_DaysInMonth = 31;
break;
case 6:
MonthOfName = "June";
number_Of_DaysInMonth = 30;
break;
case 7:
MonthOfName = "July";
number_Of_DaysInMonth = 31;
break;
case 8:
MonthOfName = "August";
number_Of_DaysInMonth = 31;
break;
case 9:
MonthOfName = "September";
number_Of_DaysInMonth = 30;
break;
case 10:
MonthOfName = "October";
number_Of_DaysInMonth = 31;
break;
case 11:
MonthOfName = "November";
number_Of_DaysInMonth = 30;
break;
case 12:
MonthOfName = "December";
number_Of_DaysInMonth = 31;
}
}
答案 16 :(得分：0)
如果您不想对年份和月份的值进行硬编码，而又想从当前日期和时间中获取该值，则可以轻松进行以下操作：
Date d = new Date();
String myDate = new SimpleDateFormat("dd/MM/yyyy").format(d);
int iDayFromDate = Integer.parseInt(myDate.substring(0, 2));
int iMonthFromDate = Integer.parseInt(myDate.substring(3, 5));
int iYearfromDate = Integer.parseInt(myDate.substring(6, 10));
YearMonth CurrentYear = YearMonth.of(iYearfromDate, iMonthFromDate);
int lengthOfCurrentMonth = CurrentYear.lengthOfMonth();
System.out.println("Total number of days in current month is " + lengthOfCurrentMonth );
答案 17 :(得分：0)
String date = "11-02-2000";
String[] input = date.split("-");
int day = Integer.valueOf(input[0]);
int month = Integer.valueOf(input[1]);
int year = Integer.valueOf(input[2]);
Calendar cal=Calendar.getInstance();
cal.set(Calendar.YEAR,year);
cal.set(Calendar.MONTH,month-1);
cal.set(Calendar.DATE, day);
//since month number starts from 0 (i.e jan 0, feb 1),
//we are subtracting original month by 1
int days = cal.getActualMaximum(Calendar.DAY_OF_MONTH);
System.out.println(days);
答案 18 :(得分：0)
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Date;
/*
* 44. Return the number of days in a month
* , where month and year are given as input.
*/
public class ex44 {
public static void dateReturn(int m,int y)
{
int m1=m;
int y1=y;
String str=" "+ m1+"-"+y1;
System.out.println(str);
SimpleDateFormat sd=new SimpleDateFormat("MM-yyyy");
try {
Date d=sd.parse(str);
System.out.println(d);
Calendar c=Calendar.getInstance();
c.setTime(d);
System.out.println(c.getActualMaximum(Calendar.DAY_OF_MONTH));
} catch (ParseException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
public static void main(String[] args) {
dateReturn(2,2012);
}
}
答案 19 :(得分：0)
最佳和高效的方差：
public static int daysInMonth(int month, int year) {
if (month != 2) {
return 31 - (month - 1) % 7 % 2;
}
else {
if ((year & 3) == 0 && ((year % 25) != 0 || (year & 15) == 0)) { // leap year
return 29;
} else {
return 28;
}
}
}
有关跳跃算法的更多详细信息，请检查here
答案 20 :(得分：0)
特定年份的天数-Java 8+解决方案
Year.now().length()