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yEAr DAy

int month_day(int year, int yearday, int *pmonth, int *pday) /* year是年, yearday是天数, 若year和yearday合理, 则*pmonth和*pday是计算得出的月和日,函数返回1; 否则,函数返回0。 */ { int mon_day[]={31,28,31,30,31,30,31,31,30,3...

YEAR(TODAY())——今天对应的年份 MONTH(TODAY())+1——今天对应月份+1 DATE(YEAR(TODAY()),MONTH(TODAY())+1,0)——合起来就是今天所在月份的最后一天, 因为DATE(YEAR(TODAY()),MONTH(TODAY())+1,1)对应的是今天所在月份的下个月的第一天 其他应该不...

scanf("%d%d%d", &date.year, &date.month, &date.day); 改成scanf("%d,%d,%d", &date.year, &date.month, &date.day);输入多个数据要加分隔符。

汗,这样的问题,你一点悬赏都没,虽然我乐于助人,也不太想回答啊,这个一看就是课程设计的,建议楼主不要偷懒,自己好好写一下吧,也不难,而且对你理解OOP的思想很重要啊 呵呵,楼主向我求助了,既然是妹妹,我就帮个忙吧,写的非常细,而且...

select * from ( select * from table where to_date(year || month || day)

给你简化了一下。易懂,不出错。 #include int month_day(int year, int yearday, int *pmonth, int *pday) { int *a,*b,i,k=1; int p[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}, r[13]={0,31,29,31,30,31,30,31,31,30,31,30,31}; if(yearda...

class Date { int year=0,month=0,day=0; public void setY(int year){ this.year=year; } public void setM(int month){ this.month=month; } public void setD(int day){ this.day=day; } public int getY(){ return year; } public int getM(...

今天的前一天/当年/当月的后一个月

#includeint main(void){ int day1,day=0,month,year,result=0;//day初始化为0int day_of_year(int year,int month,int day);printf("Enter year:");scanf("%d",&year);//去掉\nprintf("Enter month:");scanf("%d",&month);//去掉\nprintf("Ente...

倒数第二个“)”靠后了,改进:=DATE(YEAR(TODAY()),MONTH(TODAY()),DAY(TODAY())-1)

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