我写了一个简单的代码,将常规日期转换为Julian日期。
以下是需要相同转换的代码:
public int convertToJulian(String unformattedDate)
{
/*Unformatted Date: ddmmyyyy*/
int resultJulian = 0;
if(unformattedDate.length() > 0)
{
/*Days of month*/
int[] monthValues = {31,28,31,30,31,30,31,31,30,31,30,31};
String dayS, monthS, yearS;
dayS = unformattedDate.substring(0,2);
monthS = unformattedDate.substring(3, 5);
yearS = unformattedDate.substring(6, 10);
/*Convert to Integer*/
int day = Integer.valueOf(dayS);
int month = Integer.valueOf(monthS);
int year = Integer.valueOf(yearS);
//Leap year check
if(year % 4 == 0)
{
monthValues[1] = 29;
}
//Start building Julian date
String julianDate = "1";
//last two digit of year: 2012 ==> 12
julianDate += yearS.substring(2,4);
int julianDays = 0;
for (int i=0; i < month-1; i++)
{
julianDays += monthValues[i];
}
julianDays += day;
if(String.valueOf(julianDays).length() < 2)
{
julianDate += "00";
}
if(String.valueOf(julianDays).length() < 3)
{
julianDate += "0";
}
julianDate += String.valueOf(julianDays);
resultJulian = Integer.valueOf(julianDate);
}
return resultJulian;
}
此代码将01.01.2013转换为113001
我想要做的是将Julian日期转换为没有时间细节的常规日期。 例如:Julian日期是113029 ==&gt;定期日期29.01.2013
请告诉我你如何做的想法。
感谢。
答案 0 :(得分:4)
如果你想要113029 ==&gt; 29.01.2013尝试
String j = "113029";
Date date = new SimpleDateFormat("Myydd").parse(j);
String g = new SimpleDateFormat("dd.MM.yyyy").format(date);
System.out.println(g);
输出
29.01.2013
答案 1 :(得分:3)
public static int[] julianToGregorian(double injulian) {
int JGREG= 15 + 31*(10+12*1582);
int jalpha,ja,jb,jc,jd,je,year,month,day;
double julian = injulian + 0.5 / 86400.0;
ja = (int) julian
if (ja>= JGREG) {
jalpha = (int) (((ja - 1867216) - 0.25) / 36524.25);
ja = ja + 1 + jalpha - jalpha / 4;
}
jb = ja + 1524;
jc = (int) (6680.0 + ((jb - 2439870) - 122.1) / 365.25);
jd = 365 * jc + jc / 4;
je = (int) ((jb - jd) / 30.6001);
day = jb - jd - (int) (30.6001 * je);
month = je - 1;
if (month > 12) month = month - 12;
year = jc - 4715;
if (month > 2) year--;
if (year <= 0) year--;
return new int[] {year, month, day};
}