days

1 //定义一个结构体变量，给定年月日确定这一天是一年中的第几天，（考虑闰年问题） 2 #include<stdio.h> 3 struct 4 { 5 int year; 6 int month; 7 int day; 8 }date; 9 10 int main() 11 { 12 int days; 13 printf("input year,month,day:\n"); 14 scanf("%d,%d,%d", &date.year, &date.month, &date.day); 15 switch (date.month) 16 { 17 case 1:days = date.day; break; 18 case 2:days = date.day + 31; break; 19 case 3:days = date.day + 59; break; 20 case 4:days = date.day + 90; break; 21 case 5:days = date.day + 120; break; 22 case 6:days = date.day + 151; break; 23 case 7:days = date.day + 181; break; 24 case 8:days = date.day + 212; break; 25 case 9

1、获取当前时间： datetime.datetime.now() datetime.date.today() time.time()time.localtime() 2、获取前一天、后一天 #获取后一天时间（如 2015-03-07） today=datetime.date.today() temp=datetime.timedelta(days=1) nextDay=today+temp print(nextDay) #获取前一天时间（如 2015-03-05） today=datetime.date.today() temp=datetime.timedelta(days=1) nextDay=today+temp print(nextDay) 3、获取时间间隔(day1和day2大小前后顺序都可以) day1=datetime.date(2019,8,8) day2=datetime.date(2019,8,3) longTimeDay=(day1-day2).days #也可以(day1-day2).seconds 获取秒print(abs(longTimeDay)) 4、格式化日期 import time a = time.time() m, s = divmod(a , 60) h, m = divmod(m, 60) print ("%02d:%02d:%02d"

目录 进阶 类型别名 字符串字面量类型 元祖 例子 越界的元素 枚举 手动赋值 进阶 类型别名 类型别名用来给一个类型起个新名字 type Name = string type NameResolver = () => string type NameOrResolver = Name | NameResolver function getName(n: NameOrResolver): Name { if(typeof n === 'string') { return n } else { return n() } } 上例中， 我们使用 type 创建类型别名。 类型别名常用于联合类型。 字符串字面量类型 字符串字面量类型用来约束取值只能是某几个字符串中的一个。 type EventNames = 'click' | 'scroll' | 'mousemove' function handleEvent(ele: Element, event: EventNames) { // do something } handleEvent(document.getElementById('hello'), 'scoll') handleEvent(document.getElementById('world'), 'dbclick') // err dbclick 未存在于

1 #include <stdio.h> 2 3 4 void Wnl(int y); 5 6 int main(void) 7 { 8 int year; 9 printf("请输入年份(大于1990)："); 10 scanf("%d", &year); 11 Wnl(year); //调用函数Wnl打印当年日历 12 return 0; 13 } 14 15 /* 16 函数功能：打印日历 17 函数形参：年份 18 函数返回值：无 19 备注：无 20 */ 21 void Wnl(int y){ 22 // 1 2 3 4 5 6 7 8 9 10 11 12 23 //31 28 31 30 31 30 31 31 30 31 30 31 24 int mon; 25 int rn = 0; //闰年标志 26 int sum = 0;//存储天数 27 int mon_one_day = 0;//知道每个月第一天的星期几 28 int mon_days = 0;//保存每个月的总天数 29 int i, j = 0; 30 int k = 0; 31 32 int w = (y + (y - 1) / 4 - (y - 1) / 100 + (y - 1) / 400) % 7; //记录所求年份第一天的星期数 33 34 if( (y%4 ==0)&&(y%100

I tried to calculate the difference between two dates and I noticed one thing. When calculating only the days, the start of daylight saving time is included in the interval, so the result will be shorter with 1 day. To obtain accurate results, the value of hours also must be considered. For example: SimpleDateFormat format = new SimpleDateFormat("MM-dd-yyyy"); Date dfrom = format.parse("03-29-2015"); Date dto = format.parse("03-30-2015"); long diff = dto.getTime() - dfrom.getTime(); System.out.println(diff); System.out.println("Days: "+diff / (24 * 60 * 60 * 1000)); System.out.println("Hours:

Possible Duplicate: How can I create a Zerofilled value using JavaScript? I have to output a day number that must always have 3 digits. Instead of 3 it must write 003 , instead of 12 it must write 012 . If it is greater than 100 output it without formatting. I wonder if there's a regex that I could use or some quick in-line script, or I must create a function that should do that and return the result. Thanks! How about: zeroFilled = ('000' + x).substr(-3) For arbitrary width: zeroFilled = (new Array(width).join('0') + x).substr(-width) As per comments, this seems more accurate: lpad = function

I have written a query that counts records hour by hour: select TO_CHAR(copied_timestamp, 'YYYY-MM-DD HH24'),count(*) from req group by TO_CHAR(copied_timestamp, 'YYYY-MM-DD HH24'); the result is: 2012-02-22 13 2280 2012-02-22 15 1250 2012-02-22 16 1245 2012-02-22 19 1258 But I need a result like this: 2012-02-22 13 2280 2012-02-22 14 0 2012-02-22 15 1250 2012-02-22 16 1245 2012-02-22 17 0 2012-02-22 18 0 2012-02-22 19 1258 Also I have these queries that group by day and month too! select TO_CHAR(copied_timestamp, 'YYYY-MM-DD'),count(*) from req group by TO_CHAR(copied_timestamp, 'YYYY-MM-DD')

Heylo, I want to calculate (using the default Perl installation only) the number of days between two dates. The format of both the dates are like so 04-MAY-09. (DD-MMM-YY) I couldn't find any tutorials that discussed that date format. Should I be building a custom date checker for this format? Further reading of the Date::Calc on CPAN it looks unlikely that this format is supported. Thanks. If you care about accuracy, keep in mind that not all days have 86400 seconds. Any solution based on that assumption will not be correct for some cases. Here's a snippet I keep around to calculate and

Series类型的数据，经过 to_datetime 之后就可以用 pandas.Series.dt.day 和 pandas.Series.pd.month。 import pandas as pd a = pd.Series(['2017-1-2','2017-9-1']) print(a) 0 2017-1-2 1 2017-9-1 dtype: object a = pd.to_datetime(a)print(a) 0 2017-01-02 1 2017-09-01 dtype: datetime64[ns] import datetime d = datetime.date(2019,8,19) - a.dt.date print(d) 0 959 days 1 717 days dtype: timedelta64[ns] d = (datetime.date(2019,8,19) - a.dt.date).dt.days print(d) 来源： https://www.cnblogs.com/keye/p/11379734.html

how can I compare two dates return number of days. Ex: Missing X days of the Cup. look my code. NSDateFormatter *df = [[NSDateFormatter alloc]init]; [df setDateFormat:@"d MMMM,yyyy"]; NSDate *date1 = [df dateFromString:@"11-05-2010"]; NSDate *date2 = [df dateFromString:@"11-06-2010"]; NSTimeInterval interval = [date2 timeIntervalSinceDate:date1]; //int days = (int)interval / 30; //int months = (interval - (months/30)) / 30; NSString *timeDiff = [NSString stringWithFormat:@"%dMissing%d days of the Cup",date1,date2, fabs(interval)]; label.text = timeDiff; // output (Missing X days of the Cup)