找出两个日期之间的差异C

Question

我试图在C中将一个日期与另一个日期分开，以找出几天之间的差异。 但是，这比我初次看到的要复杂得多，因为由于leap年，我显然不得不考虑不同年份的不同天数，并且取决于它所在的月份，因此需要不同的天数。 我确实尝试使用time.h，但我认为它不会允许allows年。

目前，我的数据以整数形式存储在数组中，例如{2010, 5, 1, 2011, 6, 1} 。

所以有人可以发布或指出我可以帮助我完成此任务的算法吗？ 非常感谢你。

Answer 1

标准库C时间库包含所需的结构和功能。

该头文件包含用于获取和操作日期和时间信息的函数的定义。

还要检查C日期和时间函数以及两个日期之间的C程序天数

编辑：-

尝试这个：

int main()
{
    int day1,mon1,year1,day2,mon2,year2;
    int ref,dd1,dd2,i;
    clrscr();
    printf("Enter first date  day, month, year\n");
    scanf("%d%d%d",&day1,&mon1,&year1);
    printf("Enter second date day, month, year\n");
    scanf("%d%d%d",&day2,&mon2,&year2);
    ref = year1;
    if(year2<year1)
    ref = year2;
    dd1=0;
    dd1=dater(mon1);
    for(i=ref;i<year1;i++)
    {
        if(i%4==0)
        dd1+=1;
    }
    dd1=dd1+day1+(year1-ref)*365;
    dd2=0;
    for(i=ref;i<year2;i++)
    {
        if(i%4==0)
        dd2+=1;
    }
    dd2=dater(mon2)+dd2+day2+((year2-ref)*365);
    printf("\n\n Difference between the two dates is %d days",abs(dd2-dd1));

    getch();
}

int dater(x)
{ int y=0;
    switch(x)
    {
        case 1: y=0; break;
        case 2: y=31; break;
        case 3: y=59; break;
        case 4: y=90; break;
        case 5: y=120;break;
        case 6: y=151; break;
        case 7: y=181; break;
        case 8: y=212; break;
        case 9: y=243; break;
        case 10:y=273; break;
        case 11:y=304; break;
        case 12:y=334; break;
        default: printf("Invalid Input\n\n\n\n"); exit(1);
    }
    return(y);
}

或使用time.h像这样尝试：

#include <stdio.h>  
#include <time.h>       
int main ()
{
  struct tm start_date;
  struct tm end_date;
  time_t start_time, end_time;
  double seconds;

  start_date.tm_hour = 0;  start_date.tm_min = 0;  start_date.tm_sec = 0;
  start_date.tm_mon = 10; start_date.tm_mday = 15; start_date.tm_year = 113;

  end_date.tm_hour = 0;  end_date.tm_min = 0;  end_date.tm_sec = 0;
  end_date.tm_mon = 10; end_date.tm_mday = 20; end_date.tm_year = 113;

  start_time = mktime(&start_date);
  end_time = mktime(&end_date);

  seconds = difftime(end_time, start_time);

  printf ("%.f seconds difference\n", seconds);

  return 0;
}

Answer 2

算法：

1. 将日期转换为struct tm 。
2. 将struct tm转换为time_t 。
3. 取time_t s之差。 它以秒为单位产生日期之间的差。
4. 将秒数差异除以86400，将其转换为天数。

Answer 3

/* This function calculates the differance between two dates, passes 6 parameters date-month-year of first date and date-month-year of second date and prints the difference in date-weeks-year  format */
void Date::subtract(int firstDate, int firstMonth, int firstYear,int secondDate, int secondMonth, int secondYear)
{
/*check the dates are valid or not  */
if(isDateValid(firstDate,firstMonth,firstYear) && isDateValid(secondDate,secondMonth,secondYear) )
{

    firstMonth = (firstMonth + 9) % 12;
    firstYear = firstYear - firstMonth / 10;
    FirstNoOfDays = 365 * firstYear + firstYear/4 - firstYear/100 + firstYear/400 + (firstMonth * 306 + 5) /10 + ( firstDate - 1 );

    secondMonth = (secondMonth + 9) % 12;
    secondYear = secondYear - secondMonth / 10;
    SecondNoOfDays = 365 * secondYear + secondYear/4 - secondYear/100 + secondYear/400 + (secondMonth * 306 + 5) /10 + ( secondDate - 1 );

    dayDifference = abs(FirstNoOfDays - SecondNoOfDays); /* uses absolute if the first date is smaller so it wont give negative number */
    years = dayDifference / 365;
    weeks = (dayDifference % 365)/7;
    days = (dayDifference % 365) % 7;
    cout<<years<<" years "<<weeks<<" weeks "<<days<< " days"<<endl;

}
else
{
    cout<<"Invalid Date"<<endl;
}

}