问题描述
我想要一个Java程序来计算两个日期之间的天数。
I want a Java program that calculates days between two dates.
- 输入第一个日期(德语表示法;带空格:dd mm yyyy)
- 输入第二个日期。
- 程序应计算两个日期之间的天数。
我如何包含闰年和夏季?
How can I include leap years and summertime?
我的代码:
import java.util.Calendar;
import java.util.Date;
import java.util.Scanner;
public class NewDateDifference {
public static void main(String[] args) {
System.out.print("Insert first date: ");
Scanner s = new Scanner(System.in);
String[] eingabe1 = new String[3];
while (s.hasNext()) {
int i = 0;
insert1[i] = s.next();
if (!s.hasNext()) {
s.close();
break;
}
i++;
}
System.out.print("Insert second date: ");
Scanner t = new Scanner(System.in);
String[] insert2 = new String[3];
while (t.hasNext()) {
int i = 0;
insert2[i] = t.next();
if (!t.hasNext()) {
t.close();
break;
}
i++;
}
Calendar cal = Calendar.getInstance();
cal.set(Calendar.DAY_OF_MONTH, Integer.parseInt(insert1[0]));
cal.set(Calendar.MONTH, Integer.parseInt(insert1[1]));
cal.set(Calendar.YEAR, Integer.parseInt(insert1[2]));
Date firstDate = cal.getTime();
cal.set(Calendar.DAY_OF_MONTH, Integer.parseInt(insert2[0]));
cal.set(Calendar.MONTH, Integer.parseInt(insert2[1]));
cal.set(Calendar.YEAR, Integer.parseInt(insert2[2]));
Date secondDate = cal.getTime();
long diff = secondDate.getTime() - firstDate.getTime();
System.out.println ("Days: " + diff / 1000 / 60 / 60 / 24);
}
}
推荐答案
你正在与你的字符串进行一些不必要的转换。有一个等级 - 试试这个:
You are making some conversions with your Strings that are not necessary. There is a SimpleDateFormat
class for it - try this:
SimpleDateFormat myFormat = new SimpleDateFormat("dd MM yyyy");
String inputString1 = "23 01 1997";
String inputString2 = "27 04 1997";
try {
Date date1 = myFormat.parse(inputString1);
Date date2 = myFormat.parse(inputString2);
long diff = date2.getTime() - date1.getTime();
System.out.println ("Days: " + TimeUnit.DAYS.convert(diff, TimeUnit.MILLISECONDS));
} catch (ParseException e) {
e.printStackTrace();
}
编辑:由于有一些关于这段代码的正确性:它确实照顾了闰年。但是,函数失去精度,因为毫秒转换为天数(有关详细信息,请参阅链接的文档)。如果这是一个问题, diff
也可以手动转换:
Since there have been some discussions regarding the correctness of this code: it does indeed take care of leap years. However, the TimeUnit.DAYS.convert
function loses precision since milliseconds are converted to days (see the linked doc for more info). If this is a problem, diff
can also be converted by hand:
float days = (diff / (1000*60*60*24));
请注意,这是 float
值,不一定是 int
。
Note that this is a float
value, not necessarily an int
.
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