计算任意2个日期之间的天数（C语言实现）

计算任意2个日期之间的天数（C语言实现）。

计算任意2个日期之间的天数，除了用SQL，也可以用程序实现，本文是说C语言的实现。

考虑已经周全，具体应用自己根据需要进行调整。另外一篇会介绍java方面的，基本上是大同小异。

/*求所在月份的天数*/

int daysMonth(int year,int month,int day)

{

int days[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

int i,sum=0;

for(i=0;i<month;i++)

sum+=days[i];

if(month>2) /*如果是闰年则2月加一天*/

if((year%4==0)&&(year%100!=0)||(year%400==0))

sum+=1;

sum+=day;

return sum;

}

/*交换位置，避免负值*/

void swap(int x1,int x2)

{

int tmp=x1;

x1=x2;

x2=tmp;

}

long countDate(int y1,int m1,int d1,int y2,int m2,int d2)

{

int daysyear1,daysyear2;

long totalDays=0;

int total_day1;

int tmpYear;

int tmpDays;

printf("The first Date is %ld-%ld-%ld/n",y1,m1,d1);

printf("The second Date is %ld-%ld-%ld/n",y2,m2,d2);

if(y1>y2)

{

swap(y1,y2);

swap(m1,m2);

swap(d1,d2);

}

if(y1==y2)

{

daysyear1=daysMonth(y1,m1,d1);

daysyear2=daysMonth(y2,m2,d2);

totalDays=abs(daysyear1-daysyear2)+1;

printf("totalDays is %ld/n",totalDays-1);

}

else

{

daysyear1=daysMonth(y1,m1,d1);

total_day1=365-daysyear1+1;

if(m1<=2)

if(y1%4==0&&(y1%100!=0)||(y1%400==0))

total_day1+=1;

totalDays+=total_day1;

tmpYear=y1;

while(++tmpYear<y2)

{

tmpDays=365;

if((tmpYear%4==0)&&(tmpYear%100!=0)||(tmpYear%400==0))

tmpDays+=1;

totalDays+=tmpDays;

}

daysyear2=daysMonth(y2,m2,d2);

totalDays+=daysyear2;

}

return totalDays-1;

printf("totalDays is %ld/n",totalDays-1);

}

main()

{

int y1,m1,d1,y2,m2,d2;

printf("Plsase input Date,for example 2008-8-1,2008-8-8:/n");

scanf("%ld-%ld-%ld,%ld-%ld-%ld",&y1,&m1,&d1,&y2,&m2,&d2);

countDate(y1,m1,d1,y2,m2,d2);

getch();

}