Java 如何根据现在的时间毫秒来计算时间的公式(不想用原有的类)
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Java 如何根据现在的时间毫秒来计算时间的公式(不想用原有的类)
java.sql.Timestamp time ;
long tt = -1 ;
time = new java.sql.Timestamp(new java.util.Date().getTime());
tt = time.getTime()
来获得信息.
我希望可以根据一些数学公司来确定现在的年、月、日、时、分、秒、毫秒的信息.
不要用现有的类
java.sql.Timestamp time ;
long tt = -1 ;
time = new java.sql.Timestamp(new java.util.Date().getTime());
tt = time.getTime()
来获得信息.
我希望可以根据一些数学公司来确定现在的年、月、日、时、分、秒、毫秒的信息.
不要用现有的类
特意写了一个 你看看是不是这个意思public static void main(String[] args) {
int timezone = 8;
long timeMillis = new Date().getTime();//1970
long totalSeconds = timeMillis / 1000;
totalSeconds += 60 * 60 * timezone;
int second = (int) (totalSeconds % 60);// 秒
long totalMinutes = totalSeconds / 60;
int minute = (int) (totalMinutes % 60);// 分
long totalHours = totalMinutes / 60;
int hour = (int) (totalHours % 24);// 时
int totalDays = (int) (totalHours / 24);
int _year = 1970;
int year = _year + totalDays / 366;
int month = 1;
int day = 1;
int diffDays;
boolean leapYear;
while (true) {
int diff = (year - _year) * 365;
diff += (year - 1) / 4 - (_year - 1) / 4;
diff -= ((year - 1) / 100 - (_year - 1) / 100);
diff += (year - 1) / 400 - (_year - 1) / 400;
diffDays = totalDays - diff;
leapYear = (year % 4 == 0) && (year % 100 != 0) || (year % 400 == 0);
if (!leapYear && diffDays < 365 || leapYear && diffDays < 366) {
break;
} else {
year++;
}
}
int[] monthDays;
if (diffDays >= 59 && leapYear) {
monthDays = new int[] { -1, 0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335 };
} else {
monthDays = new int[] { -1, 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334 };
}
for (int i = monthDays.length - 1; i >= 1; i--) {
if (diffDays >= monthDays[i]) {
month = i;
day = diffDays - monthDays[i] + 1;
break;
}
}
System.out.println(year);
System.out.println(month);
System.out.println(day);
System.out.println(hour);
System.out.println(minute);
System.out.println(second);
}
int timezone = 8;
long timeMillis = new Date().getTime();//1970
long totalSeconds = timeMillis / 1000;
totalSeconds += 60 * 60 * timezone;
int second = (int) (totalSeconds % 60);// 秒
long totalMinutes = totalSeconds / 60;
int minute = (int) (totalMinutes % 60);// 分
long totalHours = totalMinutes / 60;
int hour = (int) (totalHours % 24);// 时
int totalDays = (int) (totalHours / 24);
int _year = 1970;
int year = _year + totalDays / 366;
int month = 1;
int day = 1;
int diffDays;
boolean leapYear;
while (true) {
int diff = (year - _year) * 365;
diff += (year - 1) / 4 - (_year - 1) / 4;
diff -= ((year - 1) / 100 - (_year - 1) / 100);
diff += (year - 1) / 400 - (_year - 1) / 400;
diffDays = totalDays - diff;
leapYear = (year % 4 == 0) && (year % 100 != 0) || (year % 400 == 0);
if (!leapYear && diffDays < 365 || leapYear && diffDays < 366) {
break;
} else {
year++;
}
}
int[] monthDays;
if (diffDays >= 59 && leapYear) {
monthDays = new int[] { -1, 0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335 };
} else {
monthDays = new int[] { -1, 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334 };
}
for (int i = monthDays.length - 1; i >= 1; i--) {
if (diffDays >= monthDays[i]) {
month = i;
day = diffDays - monthDays[i] + 1;
break;
}
}
System.out.println(year);
System.out.println(month);
System.out.println(day);
System.out.println(hour);
System.out.println(minute);
System.out.println(second);
}
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