java里面怎么求两个日期之间的天数？
我这里有个实现的原代码，但不会使用。。请求大家翻译下。。。说说使用时。。。时间的格式应该是什么？
package corejava;
import java.util.*;
import java.io.*;
/**
Stores dates and perform date arithmetic.
This is another date class, but more convenient that
<tt> java.util.Date </tt> or <tt> java.util.Calendar </tt>
@version 1.20 5 Oct 1998
@author Cay Horstmann
*/
public class Day implements Cloneable, Serializable
{ /**
Constructs today 's date
*/
public Day()
{ GregorianCalendar todaysDate
= new GregorianCalendar();
year = todaysDate.get(Calendar.YEAR);
month = todaysDate.get(Calendar.MONTH) + 1;
day = todaysDate.get(Calendar.DAY_OF_MONTH);
}
/**
Constructs a specific date
@param yyyy year (full year, e.g., 1996,
<i> not </i> starting from 1900)
@param m month
@param d day
@exception IllegalArgumentException if yyyy m d not a
valid date
*/
public Day(int yyyy, int m, int d)
{ year = yyyy;
month = m;
day = d;
if (!isValid())
throw new IllegalArgumentException();
}
/**
Advances this day by n days. For example.
d.advance(30) adds thirdy days to d
@param n the number of days by which to change this
day (can be < 0)
*/
public void advance(int n)
{ fromJulian(toJulian() + n);
}
/**
Gets the day of the month
@return the day of the month (1...31)
*/
public int getDay()
{ return day;
}
/**
Gets the month
@return the month (1...12)
*/
public int getMonth()
{ return month;
}
/**
Gets the year
@return the year (counting from 0, <i> not </i> from 1900)
*/
public int getYear()
{ return year;
}
/**
Gets the weekday
@return the weekday ({@link Day#SUNDAY}, ...,
{@link Day#SATURDAY})
*/
public int weekday()
{ return (toJulian() + 1) % 7 + 1;
}
/**
The number of days between this and day parameter
@param b any date
@return the number of days between this and day parameter
and b (> 0 if this day comes after b)
*/
public int daysBetween(Day b)
{ return toJulian() - b.toJulian();
}
/**
A string representation of the day
@return a string representation of the day
*/
public String toString()
{ return "Day[ " + year + ", " + month + ", " + day + "] ";
}
/**
Makes a bitwise copy of a Day object
@return a bitwise copy of a Day object
*/
public Object clone()
{ try
{ return super.clone();
} catch (CloneNotSupportedException e)
{ // this shouldn 't happen, since we are Cloneable
return null;
}
}
/**
Compares this Day against another object
@param obj another object
@return true if the other object is identical to this Day object
*/
public boolean equals(Object obj)
{ if (!getClass().equals(obj.getClass())) return false;
Day b = (Day)obj;
return day == b.day && month == b.month && year == b.year;
}
/**
Computes the number of days between two dates
@return true iff this is a valid date
*/
private boolean isValid()
{ Day t = new Day();
t.fromJulian(this.toJulian());
return t.day == day && t.month == month
&& t.year == year;
}
/**
@return The Julian day number that begins at noon of
this day
Positive year signifies A.D., negative year B.C.
Remember that the year after 1 B.C. was 1 A.D.
A convenient reference point is that May 23, 1968 noon
is Julian day 2440000.
Julian day 0 is a Monday.
This algorithm is from Press et al., Numerical Recipes
in C, 2nd ed., Cambridge University Press 1992
*/
private int toJulian()
{ int jy = year;
if (year < 0) jy++;
int jm = month;
if (month > 2) jm++;
else
{ jy--;
jm += 13;
}
int jul = (int) (java.lang.Math.floor(365.25 * jy)
+ java.lang.Math.floor(30.6001*jm) + day + 1720995.0);
int IGREG = 15 + 31*(10+12*1582);
// Gregorian Calendar adopted Oct. 15, 1582
if (day + 31 * (month + 12 * year) > = IGREG)
// change over to Gregorian calendar
{ int ja = (int)(0.01 * jy);
jul += 2 - ja + (int)(0.25 * ja);
}
return jul;
}
/**
Converts a Julian day to a calendar date
This algorithm is from Press et al., Numerical Recipes
in C, 2nd ed., Cambridge University Press 1992
@param j the Julian date
*/
private void fromJulian(int j)
{ int ja = j;
int JGREG = 2299161;
/* the Julian date of the adoption of the Gregorian
calendar
*/
if (j > = JGREG)
/* cross-over to Gregorian Calendar produces this
correction
*/
{ int jalpha = (int)(((float)(j - 1867216) - 0.25)
/ 36524.25);
ja += 1 + jalpha - (int)(0.25 * jalpha);
}
int jb = ja + 1524;
int jc = (int)(6680.0 + ((float)(jb-2439870) - 122.1)
/365.25);
int jd = (int)(365 * jc + (0.25 * jc));
int je = (int)((jb - jd)/30.6001);
day = jb - jd - (int)(30.6001 * je);
month = je - 1;
if (month > 12) month -= 12;
year = jc - 4715;
if (month > 2) --year;
if (year <= 0) --year;
}
public static int SUNDAY = 1;
public static int MONDAY = 2;
public static int TUESDAY = 3;
public static int WEDNESDAY = 4;
public static int THURSDAY = 5;
public static int FRIDAY = 6;
public static int SATURDAY = 7;
/** @serial */
private int day;
/** @serial */
private int month;
/** @serial */
private int year;
}
[解决办法]
用得着那么复杂吗？试试下面的代码：
public static void main(String[] args) throws ParseException {
SimpleDateFormat sim = new SimpleDateFormat( "yyyy-MM-dd ");
Date d1 = sim.parse( "2007-06-25 ");
Date d2 = sim.parse( "2007-06-22 ");
System.out.println( (d1.getTime() - d2.getTime())/(3600L*1000*24));
}
如果有时分秒的话，需要改进一下。
[解决办法]
支持
[解决办法]
yyyy-MM-dd格式的用bao110908(bao)(bao) 的就行了,
yyyy-MM-dd hh:mm:ss的用下面这个比较准确
public static int getDaysBetween (String date1, String date2, String pattern) {
Calendar d1 = StringToCalendar(date1, pattern);
Calendar d2 = StringToCalendar(date2, pattern);
if (d1.after(d2)) { // swap dates so that d1 is start and d2 is end
java.util.Calendar swap = d1;
d1 = d2;
d2 = swap;
}
int days = d2.get(java.util.Calendar.DAY_OF_YEAR) -
d1.get(java.util.Calendar.DAY_OF_YEAR);
int y2 = d2.get(java.util.Calendar.YEAR);
if (d1.get(java.util.Calendar.YEAR) != y2) {
d1 = (java.util.Calendar) d1.clone();
do {
days += d1.getActualMaximum(java.util.Calendar.DAY_OF_YEAR);
d1.add(java.util.Calendar.YEAR, 1);
} while (d1.get(java.util.Calendar.YEAR) != y2);
}
return days;
}
