Number of days in particular month of particular year?

Question

How to know how many days has particular month of particular year?

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

FYI, the terribly troublesome old date-time classes such as java.util.Date, java.util.Calendar, and java.text.SimpleDateFormat are now legacy, supplanted by the java.time classes built into Java 8 and later. See Tutorial by Oracle. — Basil Bourque, Commented Aug 3, 2018 at 19:18

Basil Bourque · Accepted Answer · 2018-08-03 19:17:36Z

455

Java 8 and later

@Warren M. Nocos. If you are trying to use Java 8's new Date and Time API, you can use java.time.YearMonth class. See Oracle Tutorial.

// Get the number of days in that month
YearMonth yearMonthObject = YearMonth.of(1999, 2);
int daysInMonth = yearMonthObject.lengthOfMonth(); //28

Test: try a month in a leap year:

yearMonthObject = YearMonth.of(2000, 2);
daysInMonth = yearMonthObject.lengthOfMonth(); //29

Java 7 and earlier

Create a calendar, set year and month and use 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

Test: try a month in a leap year:

mycal = new GregorianCalendar(2000, Calendar.FEBRUARY, 1);
daysInMonth= mycal.getActualMaximum(Calendar.DAY_OF_MONTH);      // 29

edited Aug 3, 2018 at 19:17

Basil Bourque

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answered Jan 20, 2012 at 11:08

Hemant Metalia

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2

How to do this on Java 8's new Date and Time API?
– Warren Nocos
Commented Nov 9, 2015 at 11:36
2

@WarrenM.Nocos sorry for late response, but i wasnt active these months. please check the edit for the solution for java 8.
– Hemant Metalia
Commented Jan 20, 2016 at 6:55
1

As for prior to Java 8… Much of the java.time functionality is back-ported to Java 6 & Java 7 in the ThreeTen-Backport project. Further adapted for earlier Android in the ThreeTenABP project. See How to use ThreeTenABP….
– Basil Bourque
Commented Aug 3, 2018 at 19:20

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aug · Accepted Answer · 2015-11-11 19:19:56Z

Code for java.util.Calendar

If you have to use java.util.Calendar, I suspect you want:

int days = calendar.getActualMaximum(Calendar.DAY_OF_MONTH);

Code for Joda Time

Personally, however, I'd suggest using Joda Time instead of java.util.{Calendar, Date} to start with, in which case you could use:

int days = chronology.dayOfMonth().getMaximumValue(date);

Note that rather than parsing the string values individually, it would be better to get whichever date/time API you're using to parse it. In java.util.* you might use SimpleDateFormat; in Joda Time you'd use a DateTimeFormatter.

Aleks G · Accepted Answer · 2012-01-20 11:09:35Z

33

You can use Calendar.getActualMaximum method:

Calendar calendar = Calendar.getInstance();
calendar.set(Calendar.YEAR, year);
calendar.set(Calendar.MONTH, month);
int numDays = calendar.getActualMaximum(Calendar.DATE);

answered Jan 20, 2012 at 11:09

Aleks G

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Basil Bourque · Accepted Answer · 2017-03-02 05:03:58Z

30

`java.time.LocalDate`

From Java 1.8, you can use the method lengthOfMonth on java.time.LocalDate:

LocalDate date = LocalDate.of(2010, 1, 19);
int days = date.lengthOfMonth();

edited Mar 2, 2017 at 5:03

Basil Bourque

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answered Mar 7, 2016 at 17:54

Alexandre DuBreuil

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Shl · Accepted Answer · 2022-11-07 09:14:35Z

14

This is the mathematical way:

For year (e.g. 2012), month (1 to 12):

int daysInMonth = month !== 2 ? 
  31 - (((month - 1) % 7) % 2) : 
  28 + (year % 4 == 0 ? 1 : 0) - (year % 100 == 0 ? 1 : 0) + (year % 400 == 0 ? 1 : 0)

edited Nov 7, 2022 at 9:14

answered Oct 6, 2017 at 11:45

Shl

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OneCricketeer · Accepted Answer · 2022-01-05 23:48:20Z

7

if (month == 4 || month == 6 || month == 9 || month == 11) {
   daysInMonth = 30;
} else if (month == 2) {
    daysInMonth = (leapYear) ? 29 : 28;
else { 
    daysInMonth = 31;
}

edited Jan 5, 2022 at 23:48

OneCricketeer

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answered Mar 2, 2013 at 8:25

zoidbot

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Calendar c = Calendar.getInstance();c.set(Calendar.DAY_OF_MONTH, c.getActualMaximum(Calendar.DAY_OF_MONTH)); // you can set the year before you get the actual maximum coz they might not be the same. ie feb of 2011 and 2012 is not same length (leap year)
– Rose
Commented Jul 10, 2013 at 22:02

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adiga · Accepted Answer · 2019-06-26 10:23:48Z

4

I would go for a solution like this:

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();
}

If the month isn't February (92% of the cases), it depends on the month only and it is more efficient not to involve the year. This way, you don't have to call logic to know whether it is a leap year and you don't need to get the year in 92% of the cases. And it is still clean and very readable code.

edited Jun 26, 2019 at 10:23

adiga

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answered Jun 26, 2019 at 10:11

Stefan Mondelaers

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2

I’d prefer to leave the whole logic to a proven library method — I think you’re optimizing very prematurely, and the library methods aren’t that inefficient. Still upvoted for the use of the modern java.time.
– Anonymous
Commented Jun 26, 2019 at 10:52
@OleV.V. True, in many cases leaving the optimization to a proven library can be better. But, in this case, existing libraries will require to pass a month and year. So that means I need to do whatever it takes to get the year anyway even if the method won't use the value in 92% of the cases. So, that's a part it can't optimize for me. My reasoning is similar to the reason why you shouldn't do a method call to get a value to be passed to a logger which might be disabled. No way that the logger can optimize that.
– Stefan Mondelaers
Commented Jun 26, 2019 at 11:27

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Ajay Chauhan · Accepted Answer · 2018-08-14 09:23:32Z

3

Simple as that,no need to import anything

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;
}

answered Aug 14, 2018 at 9:23

Ajay Chauhan

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This is OK, if you don't need historical dates or dates in far future. For February of years which are multiples of 100 but not multiples of 400, it will be wrong. But I agree, in most applications this will do and it is efficient.
– Stefan Mondelaers
Commented Jun 26, 2019 at 11:35

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Jamie Lister · Accepted Answer · 2017-08-29 04:37:25Z

2

In Java8 you can use get ValueRange from a field of a date.

LocalDateTime dateTime = LocalDateTime.now();

ChronoField chronoField = ChronoField.MONTH_OF_YEAR;
long max = dateTime.range(chronoField).getMaximum();

This allows you to parameterize on the field.

answered Aug 29, 2017 at 4:37

Jamie Lister

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Pochmurnik · Accepted Answer · 2019-03-04 08:27:55Z

Lets make it as simple if you don't want to hardcode the value of year and month and you want to take the value from current date and time:

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 );

Ommadawn · Accepted Answer · 2019-05-20 10:45:43Z

2

// 1 means Sunday ,2 means Monday .... 7 means Saturday
//month starts with 0 (January)

MonthDisplayHelper monthDisplayHelper = new MonthDisplayHelper(2019,4);
int numbeOfDaysInMonth = monthDisplayHelper.getNumberOfDaysInMonth();

edited May 20, 2019 at 10:45

Ommadawn

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answered May 20, 2019 at 9:52

user11527435

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1

It's worth to note that it's for Android (android.util.MonthDisplayHelper)
– barbsan
Commented May 20, 2019 at 10:22

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adiga · Accepted Answer · 2019-06-26 10:24:31Z

1

Following method will provide you the no of days in a particular month

public static int getNoOfDaysInAMonth(String date) {
    Calendar cal = Calendar.getInstance();
    cal.setTime(date);
    return (cal.getActualMaximum(Calendar.DATE));
}

edited Jun 26, 2019 at 10:24

adiga

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answered Jun 29, 2016 at 13:15

Tanmay kumar shaw

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whats the format of the string date should be passed ?
– Mahmoud Ayman
Commented Mar 19, 2021 at 8:01

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madhi vanan · Accepted Answer · 2014-09-10 12:05:40Z

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);


    }

}

What value does this answer provide over the existing and accepted answers? Also, please add some explanation or narrative along with your code. StackOverflow is more than a snippet library. — Basil Bourque, Commented Sep 10, 2014 at 16:29

jHilscher · Accepted Answer · 2017-11-14 07:59:12Z

0

public class Main {

    private static LocalDate local=LocalDate.now();
    public static void main(String[] args) {

            int month=local.lengthOfMonth();
            System.out.println(month);

    }
}

edited Nov 14, 2017 at 7:59

jHilscher

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answered Nov 14, 2017 at 6:58

Nodirbek Hasanboev

1

6

Add some explanations as well.
– BlackBeard
Commented Nov 14, 2017 at 6:59
1

Welcome to Stack Overflow! While this code snippet may be the solution, including an explanation really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion.
– yivi
Commented Nov 14, 2017 at 9:40

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xingbin · Accepted Answer · 2018-04-29 12:34:24Z

0

The use of outdated Calendar API should be avoided.

In Java8 or higher version, this can be done with YearMonth.

Example code:

int year = 2011;
int month = 2;
YearMonth yearMonth = YearMonth.of(year, month);
int lengthOfMonth = yearMonth.lengthOfMonth();
System.out.println(lengthOfMonth);

edited Apr 29, 2018 at 12:34

answered Apr 29, 2018 at 11:41

xingbin

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Call requires API level 26 (current min is 21): java.time.YearMonth#lengthOfMonth
– Vlad
Commented May 4, 2018 at 14:58

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adiga · Accepted Answer · 2019-06-26 10:24:15Z

You can use Calendar.getActualMaximum method:

Calendar calendar = Calendar.getInstance();
calendar.set(Calendar.YEAR, year);
calendar.set(Calendar.MONTH, month-1);
int numDays = calendar.getActualMaximum(Calendar.DATE);

And month-1 is Because of month takes its original number of month while in method takes argument as below in Calendar.class

public int getActualMaximum(int field) {
   throw new RuntimeException("Stub!");
}

And the (int field) is like as below.

public static final int JANUARY = 0;
public static final int NOVEMBER = 10;
public static final int DECEMBER = 11;

magiccrafter · Accepted Answer · 2019-11-17 22:54:17Z

0

An optimal and performant variance:

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;
        }
    }
}

For more details on the leap algorithm check here

answered Nov 17, 2019 at 22:54

magiccrafter

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David Buck · Accepted Answer · 2021-08-29 10:41:27Z

An alternative solution is to use a Calendar object. Get the current date and set the day so it is the first of the month. Then add one month and take away one day to get the last day of the current month. Finally fetch the day to get the number of days in the month.

Calendar today = getInstance(TimeZone.getTimeZone("UTC"));

Calendar currMonthLastDay = getInstance(TimeZone.getTimeZone("UTC"));
currMonthLastDay.clear();
currMonthLastDay.set(YEAR, today.get(YEAR));
currMonthLastDay.set(MONTH, today.get(MONTH));
currMonthLastDay.set(DAY_OF_MONTH, 1); 
currMonthLastDay.add(MONTH, 1);
currMonthLastDay.add(DAY_OF_MONTH, -1);

Integer daysInMonth = currMonthLastDay.get(DAY_OF_MONTH);

pushkin · Accepted Answer · 2018-11-13 19:02:17Z

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;
    }
}

Pochmurnik · Accepted Answer · 2019-03-04 08:45:52Z

This worked fine for me.

This is a Sample Output

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. 
    } 
} //[email protected]

adiga · Accepted Answer · 2019-06-26 10:24:44Z

-1

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);

edited Jun 26, 2019 at 10:24

adiga

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answered Sep 24, 2014 at 7:07

Naga Budigam

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1

No need to answer to a question that is already accepted, until that answer invalid any how.
– Deepak
Commented Sep 24, 2014 at 7:29

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Tionio · Accepted Answer · 2020-11-26 14:11:39Z

-1

Number of days in particular year - Java 8+ solution

Year.now().length()

answered Nov 26, 2020 at 14:11

Tionio

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1

Does not address the Question. The Question is about length of month, not year. And the .length() method has been covered multiple times already in other Answers.
– Basil Bourque
Commented Nov 27, 2020 at 5:39

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Number of days in particular month of particular year?

22 Answers 22

Java 8 and later

Java 7 and earlier

Code for java.util.Calendar

Code for Joda Time

`java.time.LocalDate`

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Collectives™ on Stack Overflow

22 Answers 22

Java 8 and later

Java 7 and earlier

Code for java.util.Calendar

Code for Joda Time

java.time.LocalDate

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`java.time.LocalDate`

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