blocks|key|458574|text|java.util.Calendar的代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|458575|如果您必须使用java.util.Calendar，我怀疑您需要：|offset|length|style|CODE|458576|int+days+=+calendar.getActualMaximum(Calendar.DAY_OF_MONTH);|code-block|syntax|javascript|458577|Joda时间的代码|458578|然而，就我个人而言，我建议首先使用Joda+Time而不是java.util.{Calendar,+Date}，在这种情况下，您可以使用：|458579|int+days+=+chronology.dayOfMonth().getMaximumValue(date);|458580|请注意，与其单独解析字符串值，不如获取您正在使用的任何日期/时间API来解析它。在java.util.*中，您可以使用SimpleDateFormat；在Joda+Time中，您可以使用DateTimeFormatter。|458581|entityMap|0|LINK|mutability|MUTABLE|url|http://www.joda.org/joda-time/^0|0|7|I|0|0|0|T|Q|H|9|0|0|0|15|B|1N|G|2L|H|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@$D|14|E|15|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|16|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|18|8|@$D|19|E|1A|F|G]]|9|@$D|1B|E|1C|1|1D]]|A|$]]|$1|Q|3|R|5|J|7|1E|8|@]|9|@]|A|$K|L]]|$1|S|3|T|5|6|7|1F|8|@$D|1G|E|1H|F|G]|$D|1I|E|1J|F|G]|$D|1K|E|1L|F|G]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

<h2>Code for java.util.Calendar</h2>

If you have to use <code>java.util.Calendar</code>, I suspect you want:

<pre><code>int days = calendar.getActualMaximum(Calendar.DAY_OF_MONTH);
</code></pre>

<h2>Code for Joda Time</h2>

Personally, however, I'd suggest using <a href="http://www.joda.org/joda-time/" rel="noreferrer">Joda Time</a> instead of <code>java.util.{Calendar, Date}</code> to start with, in which case you could use:

<pre><code>int days = chronology.dayOfMonth().getMaximumValue(date);
</code></pre>

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 <code>java.util.*</code> you might use <code>SimpleDateFormat</code>; in Joda Time you'd use a <code>DateTimeFormatter</code>.

blocks|key|458606|text|您可以使用Calendar.getActualMaximum方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|458607|Calendar+calendar+=+Calendar.getInstance();
calendar.set(Calendar.YEAR,+year);
calendar.set(Calendar.MONTH,+month);
int+numDays+=+calendar.getActualMaximum(Calendar.DATE);|code-block|syntax|javascript|458608|entityMap^0|5|P|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can use <code>Calendar.getActualMaximum</code> method:

<pre><code>Calendar calendar = Calendar.getInstance();
calendar.set(Calendar.YEAR, year);
calendar.set(Calendar.MONTH, month);
int numDays = calendar.getActualMaximum(Calendar.DATE);
</code></pre>

blocks|key|458618|text|if+(month+==+4+%7C%7C+month+==+6+%7C%7C+month+==+9+%7C%7C+month+==+11)

daysInMonth+=+30;

else+

if+(month+==+2)+

daysInMonth+=+(leapYear)+?+29+:+28;

else+

daysInMonth+=+31;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|458619|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>if (month == 4 || month == 6 || month == 9 || month == 11)

daysInMonth = 30;

else 

if (month == 2) 

daysInMonth = (leapYear) ? 29 : 28;

else 

daysInMonth = 31;
</code></pre>

blocks|key|672833|text|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="+"%2B+m1%2B"-"%2By1;
++++++++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);


++++}

}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|672834|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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);


 }

}
</code></pre>

blocks|key|672846|text|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);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|672847|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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);
</code></pre>

blocks|key|672895|text|java.time.LocalDate|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|672896|从Java1.8开始，您可以在java.time.LocalDate上使用lengthOfMonth方法|672897|LocalDate+date+=+LocalDate.of(2010,+1,+19);
int+days+=+date.lengthOfMonth();|code-block|syntax|javascript|672898|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/time/LocalDate.html|1|https://docs.oracle.com/javase/8/docs/api/java/time/LocalDate.html#lengthOfMonth--^0|0|J|0|F|J|11|D|F|J|0|11|D|1|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@$9|14|A|15|1|16]|$9|17|A|18|1|19]]|E|$]]|$1|H|3|I|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

<h1><code>java.time.LocalDate</code></h1>

From Java 1.8, you can use the method <a href="https://docs.oracle.com/javase/8/docs/api/java/time/LocalDate.html#lengthOfMonth--" rel="noreferrer"><code>lengthOfMonth</code></a> on <a href="https://docs.oracle.com/javase/8/docs/api/java/time/LocalDate.html" rel="noreferrer"><code>java.time.LocalDate</code></a>:

<pre><code>LocalDate date = LocalDate.of(2010, 1, 19);
int days = date.lengthOfMonth();
</code></pre>

blocks|key|458707|text|下面的方法将为您提供特定月份的天数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|458708|public+static+int+getNoOfDaysInAMonth(String+date)+{
++++Calendar+cal+=+Calendar.getInstance();
++++cal.setTime(date);
++++return+(cal.getActualMaximum(Calendar.DATE));
}|code-block|syntax|javascript|458709|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

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

blocks|key|673003|text|您可以使用Calendar.getActualMaximum方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|673004|Calendar+calendar+=+Calendar.getInstance();
calendar.set(Calendar.YEAR,+year);
calendar.set(Calendar.MONTH,+month-1);
int+numDays+=+calendar.getActualMaximum(Calendar.DATE);|code-block|syntax|javascript|673005|month-1是因为month采用其原始的月数，而in方法采用Calendar.class中的参数，如下所示|673006|public+int+getActualMaximum(int+field)+{
+++throw+new+RuntimeException("Stub!");
}|673007|(int字段)如下所示。|673008|public+static+final+int+JANUARY+=+0;
public+static+final+int+NOVEMBER+=+10;
public+static+final+int+DECEMBER+=+11;|673009|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

You can use Calendar.getActualMaximum method:

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

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

<pre><code>public int getActualMaximum(int field) {
 throw new RuntimeException("Stub!");
}
</code></pre>

And the (int field) is like as below.

<pre><code>public static final int JANUARY = 0;
public static final int NOVEMBER = 10;
public static final int DECEMBER = 11;
</code></pre>

blocks|key|673030|text|这对我来说很好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|673031|This+is+a+Sample+Output|offset|length|673032|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+%25+4+==+0+&&+year+%25+100+!=+0)+%7C%7C+(year+%25+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+"+%2B+days+%2B+"+Days+in+Month+"+%2B+month+%2B+"+of+Year+"+%2B+year+%2B+".\n";+
++++++++System.out.println(response);+//+new+line+to+show+the+result+to+the+screen.+
++++}+
}+//abhinavsthakur00@gmail.com|code-block|syntax|javascript|673033|entityMap|0|LINK|mutability|MUTABLE|url|https://i.stack.imgur.com/592gh.png^0|0|0|N|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@$D|U|E|V|1|W]]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

This worked fine for me.

<a href="https://i.stack.imgur.com/592gh.png" rel="nofollow noreferrer">This is a Sample Output</a>

<pre><code>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 &amp;&amp; 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
</code></pre>

blocks|key|673088|text|在Java8中，您可以使用从日期字段获取ValueRange。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|673089|LocalDateTime+dateTime+=+LocalDateTime.now();

ChronoField+chronoField+=+ChronoField.MONTH_OF_YEAR;
long+max+=+dateTime.range(chronoField).getMaximum();|code-block|syntax|javascript|673090|这允许您在字段上进行参数化。|673091|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

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

<pre><code>LocalDateTime dateTime = LocalDateTime.now();

ChronoField chronoField = ChronoField.MONTH_OF_YEAR;
long max = dateTime.range(chronoField).getMaximum();
</code></pre>

This allows you to parameterize on the field.

blocks|key|458907|text|这是数学方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|458908|对于年份(例如2012)，月份(0到11)：|458909|int+daysInMonth+=+month+!==+2+?+
++31+-+(((month+-+1)+%25+7)+%25+2)+:+
++28+%2B+(year+%25+4+==+0+?+1+:+0)+-+(year+%25+100+==+0+?+1+:+0)+%2B+(year+%25+400+==+0+?+1+:+0)|code-block|syntax|javascript|458910|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is the mathematical way:
For year (e.g. 2012), month (0 to 11):
<pre><code>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)
</code></pre>

blocks|key|458940|text|public+class+Main+{

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

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

++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|458941|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public class Main {

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

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

 }
}
</code></pre>

blocks|key|458971|text|应避免使用过时的Calendar应用编程接口。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|458972|在Java8或更高版本中，可以使用YearMonth完成此操作。|458973|示例代码：|458974|int+year+=+2011;
int+month+=+2;
YearMonth+yearMonth+=+YearMonth.of(year,+month);
int+lengthOfMonth+=+yearMonth.lengthOfMonth();
System.out.println(lengthOfMonth);|code-block|syntax|javascript|458975|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/time/YearMonth.html^0|8|8|0|H|9|H|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]]|E|$]]|$1|H|3|I|5|6|7|15|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|16|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

The use of outdated <code>Calendar</code> API should be avoided.

In Java8 or higher version, this can be done with <a href="https://docs.oracle.com/javase/8/docs/api/java/time/YearMonth.html" rel="nofollow noreferrer"><code>YearMonth</code></a>.

Example code:

<pre><code>int year = 2011;
int month = 2;
YearMonth yearMonth = YearMonth.of(year, month);
int lengthOfMonth = yearMonth.lengthOfMonth();
System.out.println(lengthOfMonth);
</code></pre>

blocks|key|673236|text|就这么简单，不需要导入任何东西|type|unstyled|depth|inlineStyleRanges|entityRanges|data|673237|public+static+int+getMonthDays(int+month,+int+year)+{
++++int+daysInMonth+;
++++if+(month+==+4+%7C%7C+month+==+6+%7C%7C+month+==+9+%7C%7C+month+==+11)+{
++++++++daysInMonth+=+30;
++++}
++++else+{
++++++++if+(month+==+2)+{
++++++++++++daysInMonth+=+(year+%25+4+==+0)+?+29+:+28;
++++++++}+else+{
++++++++++++daysInMonth+=+31;
++++++++}
++++}
++++return+daysInMonth;
}|code-block|syntax|javascript|673238|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Simple as that,no need to import anything

<pre><code>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;
}
</code></pre>

blocks|key|673242|text|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:+"%2Bnumber_Of_DaysInMonth%2B"+++name+of+the+month:+"%2B++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+%25+400+==+0)+%7C%7C+((year+%25+4+==+0)+&&+(year+%25+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;
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|673243|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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) &amp;&amp; (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;
 }
}
</code></pre>

blocks|key|459073|text|如果您不想硬编码年和月的值，而想从当前日期和时间获取值，那么让它变得简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|459074|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+"+%2B+lengthOfCurrentMonth+);|code-block|syntax|javascript|459075|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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:

<pre><code>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 );
</code></pre>

blocks|key|673304|text|//+1+means+Sunday+,2+means+Monday+....+7+means+Saturday
//month+starts+with+0+(January)

MonthDisplayHelper+monthDisplayHelper+=+new+MonthDisplayHelper(2019,4);
int+numbeOfDaysInMonth+=+monthDisplayHelper.getNumberOfDaysInMonth();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|673305|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

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

MonthDisplayHelper monthDisplayHelper = new MonthDisplayHelper(2019,4);
int numbeOfDaysInMonth = monthDisplayHelper.getNumberOfDaysInMonth();
</code></pre>

blocks|key|459130|text|我会选择这样的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|459131|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();
}|code-block|syntax|javascript|459132|如果月份不是2月份(92%25的案例)，则仅取决于月份，不涉及年份会更有效。这样，你不需要调用逻辑来知道它是否是闰年，并且在92%25的情况下你不需要得到年份。而且它仍然是干净的、可读性很好的代码。|459133|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I would go for a solution like this:

<pre><code>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();
}
</code></pre>

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.

blocks|key|459153|text|一种最优且性能良好的方差：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|459154|public+static+int+daysInMonth(int+month,+int+year)+{
++++if+(month+!=+2)+{
++++++++return+31+-+(month+-+1)+%25+7+%25+2;
++++}
++++else+{
++++++++if+((year+&+3)+==+0+&&+((year+%25+25)+!=+0+%7C%7C+(year+&+15)+==+0))+{+//+leap+year
++++++++++++return+29;
++++++++}+else+{
++++++++++++return+28;
++++++++}
++++}
}|code-block|syntax|javascript|459155|有关leap算法的更多详细信息，请查看here|offset|length|459156|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/3220163/how-to-find-leap-year-programatically-in-c/11595914#11595914^0|0|0|J|4|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

An optimal and performant variance:

<pre><code>public static int daysInMonth(int month, int year) {
 if (month != 2) {
 return 31 - (month - 1) % 7 % 2;
 }
 else {
 if ((year &amp; 3) == 0 &amp;&amp; ((year % 25) != 0 || (year &amp; 15) == 0)) { // leap year
 return 29;
 } else {
 return 28;
 }
 }
}
</code></pre>

For more details on the leap algorithm check <a href="https://stackoverflow.com/questions/3220163/how-to-find-leap-year-programatically-in-c/11595914#11595914">here</a>

blocks|key|459177|text|另一种解决方案是使用Calendar对象。获取当前日期并将日期设置为该月的第一天。然后加上一个月，再去掉一天，就得到了当前月份的最后一天。最后，获取日期以获取该月的天数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|459178|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);|code-block|syntax|javascript|459179|entityMap^0|A|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

An alternative solution is to use a <code>Calendar</code> 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.
<pre><code>Calendar today = getInstance(TimeZone.getTimeZone(&quot;UTC&quot;));

Calendar currMonthLastDay = getInstance(TimeZone.getTimeZone(&quot;UTC&quot;));
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);
</code></pre>

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

<pre><code>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
</code></pre>

Number of days in particular month of particular year?

如何知道某一年的某月有多少天？String date = "2010-01-19";String[] ymd = date.split("-");int year = Integer.parseInt(ymd[0]);int month = Integer.parseInt(ymd[1]);int day = Inte...

问特定年份中特定月份的天数？
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