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📄 11.txt

📁 时钟!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
💻 TXT
字号:
时钟代码: 
<html> 
<head><title>VBscript记时</title></head> 
<script language="vbscript"> 
dim minute,sec 
dim str,Date1,NowWeekday 
minute=60 
sec=0 
sub button1_onClick 
if weekday(now()-1)=1 then NowWeekday="星期一" 
if weekday(now()-1)=2 then NowWeekday="星期二" 
if weekday(now()-1)=3 then NowWeekday="星期三" 
if weekday(now()-1)=4 then NowWeekday="星期四" 
if weekday(now()-1)=5 then NowWeekday="星期五" 
if weekday(now()-1)=6 then NowWeekday="星期六" 
if weekday(now()-1)=7 then NowWeekday="星期日" 

Date1=now() & "-" & NowWeekday 
document.form1.text1.value=Date1 
TimeID=setTimeout("button1_onClick",1000) 
End sub 
</script> 
<body> 
<form name=form1> 
<input name=text1 type=text value="" size="50"> 
<input type=button name=button1 value="显示时间"> 
</form> 
</body> 
</html> 
友元函数(求两个日期相差的天数): 
#include<iostream.h> 
#include<stdio.h> 
class Date 
{ 
int year; 
int month; 
int day; 
public: 
Date(int y,int m,int d) 
{ 
year=y;month=m;day=d; 
} 
void disp() 
{ 
printf("%d.%d.%d",year,month,day); 
} 
friend int count_day(Date &d,int); 
friend int leap(int year); 
friend int subs(Date &d1,Date &d2); 
}; 
int count_day(Date &d,int flag) 
{ 
static int day_tab[2][12]={{31,28,31,30,31,30,31,31,30,31,30,31}, 
{31,29,31,30,31,30,31,31,30,31,30,31}}; 
// 使用二维数组存放各月天数,第一行对应非闰年,第二行对应闰年 
int p,i,s; 
if(leap(d.year)) 
p=1; 
else p=0; 
if(flag) 
{ 
s=d.day; 
for(i=1;i<d.month;i++) 
s+=day_tab[p][i-1]; 
} 
else 
{ 
s=day_tab[p][d.month]-d.day; 
for(i=d.month+1; i<=12; i++) 
s+=day_tab[p][i-1]; 
} 
return s; 
} 
int leap(int year) 
{ 
if(year%4==0&&year%100!=0||year%400==0) // 是闰年 
return 1; 
else // 不是闰年 
return 0; 
} 
int subs(Date &d1,Date &d2) 
{ 
int days,day1,day2,y; 
if(d1.year<d2.year) 
{ 
days=count_day(d1,0); 
for(y=d1.year+1; y<d2.year ;y++) 
if(leap(y)) 
days+=366L; 
else 
days+=365L; 
days+=count_day(d2,1); 
} 
else if(d1.year==d2.year) 
{ 
day1=count_day(d1,1); 
day2=count_day(d2,1); 
days=day2-day1; 
} 
else 
days=-1; 
return days; 
} 
void main() 
{ 
Date d1(2000,1,1),d2(2002,10,1); 
int ds=subs(d1,d2); 
printf("输出结果:\n "); 
if(ds>=0) 
{ 
d1.disp(); printf("与"); 
d2.disp(); printf("之间有%d天\n\n",ds); 
} 
else 
printf("时间错误!\n"); 
} 

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