以下是该计划：
class DateDiff {
int month[];//stores the number of days in all the months
int d,y,m,n;
DateDiff(int d,int m,int y) {
month=new int[]{31,28,31,30,31,30,31,31,30,31,30,31};
this.d=d;
this.m=m;
this.y=y;
if(isLeap(y))
month[1]=29;//if year is leap, February has 29 days
else
month[1]=28;
n=0;
}
//function for checking for Leap Year
boolean isLeap(int y3) {
if((y3%400==0) || ((y3%100!=0)&&(y3%4==0)))
return true;
else
return false;
}
//function for finding the number of days that have passed from start of the year
int getDayNum(int d1,int m1,int y1) {
int i=0;
if(isLeap(y1))//February leap year correction
month[1]=29;
else
month[1]=28;
for(;i<m1;i++)
if(d<=month[i]&&i==m1-1) {
n+=d1;//add days when month hasn't completed
break;
}
else
n+=month[i]; //add the normal number of days in the month
return n;
}
//finding difference between the dates
int diff(int d2,int m2,int y2) {
int daysLeft=getDayNum(d,m,y);//store the number of days passed since start of the year y
if(isLeap(y))
daysLeft=366-daysLeft;//subtracting gives the number of days left in year y
else
daysLeft=365-daysLeft;
for(int x=y+1;x<y2;x++)//for subsequent years add 366 or 365 days as required
if(isLeap(x))
daysLeft+=366;
else
daysLeft+=365;
daysLeft+=getDayNum(d2,m2,y2);//add the number of days that have passed since start of year y2
return daysLeft;}}
public class DateDifference{
public static int main(int d1,int m1,int y1,int d2,int m2,int y2) {
DateDiff obj=new DateDiff(d1,m1,y1);
return obj.diff(d2,m2,y2);
}
}
我没有使用此程序获得正确的输出。通过d1=20,m2=3,y1=1997和d2=14,m2=2,y2=2015（日期是1997年3月20日和2015年2月14日）。
正确的差异是6540天。我的计划提供6619天。有人可以指出错误吗？
答案 0 :(得分：1)
public DateDiff(int d,int m,int y){
m--;
public int diff(int d2,int m2,int y2) {
m2--;
在某些地方你必须调整人类＆＃34;月份编号1-12到您选择的内部表格0-11。
为所有来电添加公开和私有，具体取决于您是否这样做，并且不会调整月份。
稍后更多修正
删除这些行。他们在一个无用的地方。
if(isLeap(y))
month[1]=29;//if year is leap, February has 29 days
else
month[1]=28;
n=0;
不要使用字段来获取本地金额。 （int n）。
int getDayNum(int d1,int m1,int y1) {
if(isLeap(y1))//February leap year correction
month[1]=29;
else
month[1]=28;
int n = 0;
for(int i = 0; i<m1; i++){
n += month[i];
}
n += d1;//add days when month hasn't completed
return n;
}
循环结构不明智。不，如果！
编码风格的一些改进：
class DateDiff
int month[] = new int[]{31,28,31,30,31,30,31,31,30,31,30,31};
int d,y,m;
DateDiff(int d,int m,int y) {
this.d=d;
this.m=m-1;
this.y=y;
}
boolean isLeap(int y3) {
return (y3%400==0) || ((y3%100!=0)&&(y3%4==0));
}
int getDayNum(int d1,int m1,int y1){
month[1] = isLeap(y1) ? 29 : 28;
int n = 0;
for(int i = 0; i < m1; i++){
n += month[i];
}
n += d1;//add days when month hasn't completed
return n;
}
int diff(int d2,int m2,int y2){
m2--;
int daysLeft = -getDayNum(d,m,y);
for( int x = y; x<y2; x++ ){
daysLeft += isLeap(x) ? 366 : 365;
}
daysLeft += getDayNum(d2,m2,y2);
return daysLeft;
}
}
答案 1 :(得分：0)
import java.util.Calendar;
public class DateExample {
public long diff(int d1, int m1, int y1, int d2, int m2, int y2) {
long days = 0;
try {
Calendar calendar1 = Calendar.getInstance();
calendar1.set(y1, m1, d1);
Calendar calendar2 = Calendar.getInstance();
calendar2.set(y2, m2, d2);
long diff = calendar2.getTimeInMillis()
- calendar1.getTimeInMillis();
days = (diff / (1000 * 60 * 60 * 24));
} catch (Exception e) {
e.printStackTrace();
}
return days;
}
public static void main(String[] args) {
DateExample dateExample = new DateExample();
System.out.println(dateExample.diff(20, 3, 1997, 14, 2, 2015));
}
}
答案 2 :(得分：0)
1）JAVA：
int trhFark(String w_trh_1, String w_trh_2) { //YYYYMMDD
int w_fark = Integer.MAX_VALUE;
if ( w_trh_1.length() == 8 && w_trh_2.length() == 8 ) {
String w_GG = w_trh_1.substring(6, 8);
String w_AA = w_trh_1.substring(4, 6);
String w_YYYY = w_trh_1.substring(0, 4);
Calendar c1 = Calendar.getInstance();
c1.set(Integer.valueOf(w_YYYY).intValue(), Integer.valueOf(w_AA).intValue(), Integer.valueOf(w_GG).intValue());
w_GG = w_trh_2.substring(6, 8);
w_AA = w_trh_2.substring(4, 6);
w_YYYY = w_trh_2.substring(0, 4);
Calendar c2 = Calendar.getInstance();
c2.set(Integer.valueOf(w_YYYY).intValue(), Integer.valueOf(w_AA).intValue(), Integer.valueOf(w_GG).intValue());
w_fark = (int) Math.floor( (c2.getTimeInMillis() - c1.getTimeInMillis()) / (1000 * 60 * 60 * 24 ) );
}
return w_fark;
}
public Main() {
System.out.println("20141230" + ", " + "20150102" + " => " + trhFark("20141230", "20150102"));
System.out.println("19970320" + ", " + "20150214" + " => " + trhFark("19970320", "20150214"));
}
2）JAVASCRIPT：
function trhFark(w_trh_1, w_trh_2) { //YYYYMMDD
var w_fark = 99999;
if ( w_trh_1.length == 8 && w_trh_2.length == 8 ) {
var w_GG = w_trh_1.substr(6, 2);
var w_AA = w_trh_1.substr(4, 2);
var w_YYYY = w_trh_1.substr(0, 4);
var d1 = new Date(); d1.setFullYear(parseInt(w_YYYY, 10), parseInt(w_AA, 10), parseInt(w_GG, 10));
w_GG = w_trh_2.substr(6, 2);
w_AA = w_trh_2.substr(4, 2);
w_YYYY = w_trh_2.substr(0, 4);
var d2 = new Date(); d2.setFullYear(parseInt(w_YYYY, 10), parseInt(w_AA, 10), parseInt(w_GG, 10));
w_fark = Math.floor( (d2.getTime() - d1.getTime()) / (1000 * 60 * 60 * 24 ) );
}
alert(w_fark);
}
答案 3 :(得分：0)
从LocalDateTime包中使用LocalDate或java.time和句点。
LocalDateTime localDateTime = LocalDateTime.now();
LocalDateTime localDateTime1 = LocalDateTime.now().plusDays(10);
Period period = Period.between(localDateTime.toLocalDate(), localDateTime1.toLocalDate());
System.out.println(period.getDays()); // 10