blocks|key|2944364|text|首先，您应该只在必要时才将它们作为字符串处理。大多数情况下，您应该在实际描述您正在处理的数据的数据类型中使用它们。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2944365|我建议您使用Joda+Time，它是一个比Date/Calendar好得多的API。在这种情况下，听起来应该使用LocalDate类型。然后，您可以使用：|offset|length|style|CODE|2944366|int+days+=+Days.daysBetween(date1,+date2).getDays();|code-block|syntax|javascript|2944367|entityMap|0|LINK|mutability|MUTABLE|url|http://www.joda.org/joda-time/|1|http://joda-time.sourceforge.net/api-release/org/joda/time/LocalDate.html^0|0|L|4|Q|8|1K|9|6|9|0|1K|9|1|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@$D|Y|E|Z|F|G]|$D|10|E|11|F|G]|$D|12|E|13|F|G]]|9|@$D|14|E|15|1|16]|$D|17|E|18|1|19]]|A|$]]|$1|H|3|I|5|J|7|1A|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]|U|$5|P|Q|R|A|$S|V]]]]

Well to start with, you should only deal with them as strings when you have to. Most of the time you should work with them in a data type which actually describes the data you're working with.

I would recommend that you use <a href="http://www.joda.org/joda-time/" rel="noreferrer">Joda Time</a>, which is a much better API than <code>Date</code>/<code>Calendar</code>. It sounds like you should use the <a href="http://joda-time.sourceforge.net/api-release/org/joda/time/LocalDate.html" rel="noreferrer"><code>LocalDate</code></a> type in this case. You can then use:

<pre><code>int days = Days.daysBetween(date1, date2).getDays();
</code></pre>

blocks|key|3158609|text|Java8及更高版本：ChronoUnit.between|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3158610|使用ChronoUnit实例以不同的单位(天、月、秒)计算时间量。|3158611|例如：|3158612|ChronoUnit.DAYS.between(startDate,endDate)|code-block|syntax|javascript|3158613|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/time/temporal/ChronoUnit.html^0|B|I|0|2|A|2|A|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]]]]

<h1>Java 8 and later: <code>ChronoUnit.between</code></h1>
Use instances of <a href="https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/time/temporal/ChronoUnit.html" rel="noreferrer"><code>ChronoUnit</code></a> to calculate amount of time in different units (days,months, seconds).
For Example:
<pre><code>ChronoUnit.DAYS.between(startDate,endDate)
</code></pre>

blocks|key|2944391|text|试试这段代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2944392|+++++Calendar+cal1+=+new+GregorianCalendar();
+++++Calendar+cal2+=+new+GregorianCalendar();

+++++SimpleDateFormat+sdf+=+new+SimpleDateFormat("ddMMyyyy");

+++++Date+date+=+sdf.parse("your+first+date");
+++++cal1.setTime(date)
+++++date+=+sdf.parse("your+second+date");
+++++cal2.setTime(date);

++++//cal1.set(2008,+8,+1);+
+++++//cal2.set(2008,+9,+31);
+++++System.out.println("Days=+"%2BdaysBetween(cal1.getTime(),cal2.getTime()));|code-block|syntax|javascript|2944393|此函数|2944394|+++++public+int+daysBetween(Date+d1,+Date+d2){
+++++++++++++return+(int)(+(d2.getTime()+-+d1.getTime())+/+(1000+*+60+*+60+*+24));
+++++}|2944395|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

try this code

<pre><code> Calendar cal1 = new GregorianCalendar();
 Calendar cal2 = new GregorianCalendar();

 SimpleDateFormat sdf = new SimpleDateFormat("ddMMyyyy");

 Date date = sdf.parse("your first date");
 cal1.setTime(date)
 date = sdf.parse("your second date");
 cal2.setTime(date);

 //cal1.set(2008, 8, 1); 
 //cal2.set(2008, 9, 31);
 System.out.println("Days= "+daysBetween(cal1.getTime(),cal2.getTime()));
</code></pre>

this function 

<pre><code> public int daysBetween(Date d1, Date d2){
 return (int)( (d2.getTime() - d1.getTime()) / (1000 * 60 * 60 * 24));
 }
</code></pre>

blocks|key|260730|text|我知道这个帖子已经有两年的历史了，但我还是看不到正确的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|260731|除非您想使用Joda或Java+8，并且如果您需要减少受夏令时影响的日期。|260732|所以我写了我自己的解决方案。重要的是，只有当你真的只关心日期时，它才能起作用，因为有必要丢弃时间信息，所以如果你想要像25.06.2014+-+01.01.2010+=+1636这样的东西，不管DST如何，它都应该起作用：|offset|length|style|CODE|260733|private+static+SimpleDateFormat+simpleDateFormat+=+new+SimpleDateFormat("dd.MM.yyyy");

public+static+long+getDayCount(String+start,+String+end)+{
++long+diff+=+-1;
++try+{
++++Date+dateStart+=+simpleDateFormat.parse(start);
++++Date+dateEnd+=+simpleDateFormat.parse(end);

++++//time+is+always+00:00:00,+so+rounding+should+help+to+ignore+the+missing+hour+when+going+from+winter+to+summer+time,+as+well+as+the+extra+hour+in+the+other+direction
++++diff+=+Math.round((dateEnd.getTime()+-+dateStart.getTime())+/+(double)+86400000);
++}+catch+(Exception+e)+{
++++//handle+the+exception+according+to+your+own+situation
++}
++return+diff;
}|code-block|syntax|javascript|260734|由于时间始终为00:00:00，因此在从冬季时间转换为夏季时间时使用double，然后使用Math.round()，将有助于忽略丢失的3600000毫秒(1小时)，以及如果从夏季时间转换为冬季时间，则忽略额外的1小时。|260735|这是一个小的JUnit4测试，我用来证明这一点：|260736|@Test
public+void+testGetDayCount()+{
++String+startDateStr+=+"01.01.2010";
++GregorianCalendar+gc+=+new+GregorianCalendar(locale);
++try+{
++++gc.setTime(simpleDateFormat.parse(startDateStr));
++}+catch+(Exception+e)+{
++}

++for+(long+i+=+0;+i+<+10000;+i%2B%2B)+{
++++String+dateStr+=+simpleDateFormat.format(gc.getTime());
++++long+dayCount+=+getDayCount(startDateStr,+dateStr);
++++assertEquals("dayCount+must+be+equal+to+the+loop+index+i:+",+i,+dayCount);
++++gc.add(GregorianCalendar.DAY_OF_YEAR,+1);
++}
}|260737|..。或者，如果您想了解它在“生活”中的作用，请将断言替换为：|260738|System.out.println("i:+"+%2B+i+%2B+"+%7C+"+%2B+dayCount+%2B+"+-+getDayCount("+%2B+startDateStr+%2B+",+"+%2B+dateStr+%2B+")");|260739|..。下面是输出应该是什么样子：|260740|++i:+0+%7C+0++-+getDayCount(01.01.2010,+01.01.2010)
++i:+1+%7C+1++-+getDayCount(01.01.2010,+02.01.2010)
++i:+2+%7C+2++-+getDayCount(01.01.2010,+03.01.2010)
++i:+3+%7C+3++-+getDayCount(01.01.2010,+04.01.2010)
++...
++i:+1636+%7C+1636++-+getDayCount(01.01.2010,+25.06.2014)
++...
++i:+9997+%7C+9997++-+getDayCount(01.01.2010,+16.05.2037)
++i:+9998+%7C+9998++-+getDayCount(01.01.2010,+17.05.2037)
++i:+9999+%7C+9999++-+getDayCount(01.01.2010,+18.05.2037)|260741|entityMap^0|0|0|1N|U|0|0|7|8|19|C|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|16|8|@$F|17|G|18|H|I]]|9|@]|A|$]]|$1|J|3|K|5|L|7|19|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|1A|8|@$F|1B|G|1C|H|I]|$F|1D|G|1E|H|I]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1F|8|@]|9|@]|A|$]]|$1|S|3|T|5|L|7|1G|8|@]|9|@]|A|$M|N]]|$1|U|3|V|5|6|7|1H|8|@]|9|@]|A|$]]|$1|W|3|X|5|L|7|1I|8|@]|9|@]|A|$M|N]]|$1|Y|3|Z|5|6|7|1J|8|@]|9|@]|A|$]]|$1|10|3|11|5|L|7|1K|8|@]|9|@]|A|$M|N]]|$1|12|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|13|$]]

I know this thread is two years old now, I still don't see a correct answer here.

Unless you want to use Joda or have Java 8 and if you need to subract dates influenced by daylight saving.

So I have written my own solution. The important aspect is that it only works if you really only care about dates because it's necessary to discard the time information, so if you want something like <code>25.06.2014 - 01.01.2010 = 1636</code>, this should work regardless of the DST:

<pre><code>private static SimpleDateFormat simpleDateFormat = new SimpleDateFormat("dd.MM.yyyy");

public static long getDayCount(String start, String end) {
 long diff = -1;
 try {
 Date dateStart = simpleDateFormat.parse(start);
 Date dateEnd = simpleDateFormat.parse(end);

 //time is always 00:00:00, so rounding should help to ignore the missing hour when going from winter to summer time, as well as the extra hour in the other direction
 diff = Math.round((dateEnd.getTime() - dateStart.getTime()) / (double) 86400000);
 } catch (Exception e) {
 //handle the exception according to your own situation
 }
 return diff;
}
</code></pre>

As the time is always <code>00:00:00</code>, using double and then <code>Math.round()</code> should help to ignore the missing 3600000 ms (1 hour) when going from winter to summer time, as well as the extra hour if going from summer to winter.

This is a small JUnit4 test I use to prove it:

<pre><code>@Test
public void testGetDayCount() {
 String startDateStr = "01.01.2010";
 GregorianCalendar gc = new GregorianCalendar(locale);
 try {
 gc.setTime(simpleDateFormat.parse(startDateStr));
 } catch (Exception e) {
 }

 for (long i = 0; i &lt; 10000; i++) {
 String dateStr = simpleDateFormat.format(gc.getTime());
 long dayCount = getDayCount(startDateStr, dateStr);
 assertEquals("dayCount must be equal to the loop index i: ", i, dayCount);
 gc.add(GregorianCalendar.DAY_OF_YEAR, 1);
 }
}
</code></pre>

... or if you want to see what it does 'life', replace the assertion with just:

<pre><code>System.out.println("i: " + i + " | " + dayCount + " - getDayCount(" + startDateStr + ", " + dateStr + ")");
</code></pre>

... and this is what the output should look like:

<pre><code> i: 0 | 0 - getDayCount(01.01.2010, 01.01.2010)
 i: 1 | 1 - getDayCount(01.01.2010, 02.01.2010)
 i: 2 | 2 - getDayCount(01.01.2010, 03.01.2010)
 i: 3 | 3 - getDayCount(01.01.2010, 04.01.2010)
 ...
 i: 1636 | 1636 - getDayCount(01.01.2010, 25.06.2014)
 ...
 i: 9997 | 9997 - getDayCount(01.01.2010, 16.05.2037)
 i: 9998 | 9998 - getDayCount(01.01.2010, 17.05.2037)
 i: 9999 | 9999 - getDayCount(01.01.2010, 18.05.2037)
</code></pre>

blocks|key|2944387|text|下面是一个小程序，它可能会对你有所帮助：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2944388|import+java.util.*;

public+class+DateDifference+{
++++public+static+void+main(String+args[]){
++++++++DateDifference+difference+=+new+DateDifference();
++++}

++++DateDifference()+{
++++++++Calendar+cal1+=+new+GregorianCalendar();
++++++++Calendar+cal2+=+new+GregorianCalendar();

++++++++cal1.set(2010,+12,+1);+
++++++++cal2.set(2011,+9,+31);
++++++++System.out.println("Days=+"%2BdaysBetween(cal1.getTime(),cal2.getTime()));
++++}

++++public+int+daysBetween(Date+d1,+Date+d2)+{
++++++++return+(int)(+(d2.getTime()+-+d1.getTime())+/+(1000+*+60+*+60+*+24));
++++}
}|code-block|syntax|javascript|2944389|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|$]]

here's a small program which may help you:

<pre><code>import java.util.*;

public class DateDifference {
 public static void main(String args[]){
 DateDifference difference = new DateDifference();
 }

 DateDifference() {
 Calendar cal1 = new GregorianCalendar();
 Calendar cal2 = new GregorianCalendar();

 cal1.set(2010, 12, 1); 
 cal2.set(2011, 9, 31);
 System.out.println("Days= "+daysBetween(cal1.getTime(),cal2.getTime()));
 }

 public int daysBetween(Date d1, Date d2) {
 return (int)( (d2.getTime() - d1.getTime()) / (1000 * 60 * 60 * 24));
 }
}
</code></pre>

blocks|key|257520|text|//+http://en.wikipedia.org/wiki/Julian_day
public+static+int+julianDay(int+year,+int+month,+int+day)+{
++int+a+=+(14+-+month)+/+12;
++int+y+=+year+%2B+4800+-+a;
++int+m+=+month+%2B+12+*+a+-+3;
++int+jdn+=+day+%2B+(153+*+m+%2B+2)/5+%2B+365*y+%2B+y/4+-+y/100+%2B+y/400+-+32045;
++return+jdn;
}

public+static+int+diff(int+y1,+int+m1,+int+d1,+int+y2,+int+m2,+int+d2)+{
++return+julianDay(y1,+m1,+d1)+-+julianDay(y2,+m2,+d2);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|257521|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>// http://en.wikipedia.org/wiki/Julian_day
public static int julianDay(int year, int month, int day) {
 int a = (14 - month) / 12;
 int y = year + 4800 - a;
 int m = month + 12 * a - 3;
 int jdn = day + (153 * m + 2)/5 + 365*y + y/4 - y/100 + y/400 - 32045;
 return jdn;
}

public static int diff(int y1, int m1, int d1, int y2, int m2, int d2) {
 return julianDay(y1, m1, d1) - julianDay(y2, m2, d2);
}
</code></pre>

blocks|key|3158552|text|这个函数对我有好处：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3158553|public+static+int+getDaysCount(Date+begin,+Date+end)+{
++++Calendar+start+=+org.apache.commons.lang.time.DateUtils.toCalendar(begin);
++++start.set(Calendar.MILLISECOND,+0);
++++start.set(Calendar.SECOND,+0);
++++start.set(Calendar.MINUTE,+0);
++++start.set(Calendar.HOUR_OF_DAY,+0);

++++Calendar+finish+=+org.apache.commons.lang.time.DateUtils.toCalendar(end);
++++finish.set(Calendar.MILLISECOND,+999);
++++finish.set(Calendar.SECOND,+59);
++++finish.set(Calendar.MINUTE,+59);
++++finish.set(Calendar.HOUR_OF_DAY,+23);

++++long+delta+=+finish.getTimeInMillis()+-+start.getTimeInMillis();
++++return+(int)+Math.ceil(delta+/+(1000.0+*+60+*+60+*+24));
}|code-block|syntax|javascript|3158554|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|$]]

This function is good for me:

<pre><code>public static int getDaysCount(Date begin, Date end) {
 Calendar start = org.apache.commons.lang.time.DateUtils.toCalendar(begin);
 start.set(Calendar.MILLISECOND, 0);
 start.set(Calendar.SECOND, 0);
 start.set(Calendar.MINUTE, 0);
 start.set(Calendar.HOUR_OF_DAY, 0);

 Calendar finish = org.apache.commons.lang.time.DateUtils.toCalendar(end);
 finish.set(Calendar.MILLISECOND, 999);
 finish.set(Calendar.SECOND, 59);
 finish.set(Calendar.MINUTE, 59);
 finish.set(Calendar.HOUR_OF_DAY, 23);

 long delta = finish.getTimeInMillis() - start.getTimeInMillis();
 return (int) Math.ceil(delta / (1000.0 * 60 * 60 * 24));
}
</code></pre>

blocks|key|255485|text|我(到目前为止)计算天数差异的最佳解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|255486|//++This+assumes+that+you+already+have+two+Date+objects:+startDate,+endDate
//++Also,+that+you+want+to+ignore+any+time+portions

Calendar+startCale=new+GregorianCalendar();
Calendar+endCal=new+GregorianCalendar();

startCal.setTime(startDate);
endCal.setTime(endDate);

endCal.add(Calendar.YEAR,-startCal.get(Calendar.YEAR));
endCal.add(Calendar.MONTH,-startCal.get(Calendar.MONTH));
endCal.add(Calendar.DATE,-startCal.get(Calendar.DATE));

int+daysDifference=endCal.get(Calendar.DAY_OF_YEAR);|code-block|syntax|javascript|255487|但是，请注意，这假设相差不到一年！|255488|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|$]]

My best solution (so far) for calculating the number of days difference: 

<pre><code>// This assumes that you already have two Date objects: startDate, endDate
// Also, that you want to ignore any time portions

Calendar startCale=new GregorianCalendar();
Calendar endCal=new GregorianCalendar();

startCal.setTime(startDate);
endCal.setTime(endDate);

endCal.add(Calendar.YEAR,-startCal.get(Calendar.YEAR));
endCal.add(Calendar.MONTH,-startCal.get(Calendar.MONTH));
endCal.add(Calendar.DATE,-startCal.get(Calendar.DATE));

int daysDifference=endCal.get(Calendar.DAY_OF_YEAR);
</code></pre>

Note, however, that this assumes less than a year's difference!

blocks|key|2944446|text|如果你对java感到厌烦，你可以把它作为查询的一部分发送给db2：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2944447|select+date1,+date2,+days(date1)+-+days(date2)+from+table|code-block|syntax|javascript|2944448|将返回date1、date2以及两者之间的天数差。|2944449|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|$]]

If you're sick of messing with java you can just send it to db2 as part of your query:

<pre><code>select date1, date2, days(date1) - days(date2) from table
</code></pre>

will return date1, date2 and the difference in days between the two.

<blockquote>
 Possible Duplicate: 
 <a href="https://stackoverflow.com/questions/1555262/calculating-the-difference-between-two-java-date-instances">Calculating the Difference Between Two Java Date Instances</a> 
</blockquote>



In Java, I want to calculate the number of days between two dates.

In my database they are stored as a <code>DATE</code> datatype, but in my code they are strings.

I want to calculate the no of days between those two strings.

Java, Calculate the number of days between two dates

数据库

Java

 可能重复：在Java中，我想计算两个日期之间的天数。在我的数据库中，它们被存储为DATE数据类型，但在我的代码中，它们是字符串。我想计算这两个字符串之间的天数。

问Java，计算两个日期之间的天数
EN

回答 9

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问Java，计算两个日期之间的天数EN

回答 9

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问Java，计算两个日期之间的天数
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