days

VB.NET - counting days between two dates with exclusions

由 限于喜欢 提交于 2019-12-24 02:28:25
问题 I'm trying to count the days between two dates, excluding Saturdays and Sundays . I've written this code so far Dim startDay As Integer Dim endDay As Integer Dim days As Integer Dim count As Integer startDay = dtpStartDate.Value.DayOfWeek endDay = dtpEndDate.Value.DayOfWeek For days = startDay To endDay If days = 0 Or days = 6 Then 'Sunday = 0, Saturday = 6 count += 1 End If Next lblNoOfDays.Text = count It works fine if you choose the two dates within the same week. (ex: 23rd Jan to 27th Jan

C++ Calendar problems

由 风格不统一 提交于 2019-12-24 01:25:56
问题 I'm trying to write a program that outputs a calendar. The user has to input the day in which the month starts (Monday-0, Tuesday-1, etc.), and how many days are in the month. Depending on what day the month starts, the calendar dates will start under that specific day. The issues I'm having is, I'm not sure how to get the calendar to start under a specific day, and I'm not sure how to get the dates to go to a new line after the 7 days. Any help would be appreciated. We haven't learnt much so

如何使用PHP计算两个日期之间的差异？

由 雨燕双飞 提交于 2019-12-23 20:02:37
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> 我有两个日期格式： Start Date: 2007-03-24 End Date: 2009-06-26 现在，我需要以以下形式查找这两者之间的区别： 2 years, 3 months and 2 days 如何在PHP中做到这一点？ #1楼 前段时间，我编写了 format_date 函数，因为它为 您 提供 了如何选择日期的许多选项 ： function format_date($date, $type, $seperator="-") { if($date) { $day = date("j", strtotime($date)); $month = date("n", strtotime($date)); $year = date("Y", strtotime($date)); $hour = date("H", strtotime($date)); $min = date("i", strtotime($date)); $sec = date("s", strtotime($date)); switch($type) { case 0: $date = date("Y".$seperator."m".$seperator."d",mktime($hour, $min, $sec, $month,

How to integrate :missed days with :committed days in habits.rb?

由 依然范特西╮ 提交于 2019-12-23 02:58:12
问题 How can we integrate t.integer :missed with t.text :committed so that when a User checks off he :missed 3 :committed days in a :level he has to restart the :level ? for each :missed day he checks off, an additional :committed day is added back into the :level so that he must make it up before advancing? Each habit has 5 levels before "Mastery" is achieved! class Habit < ActiveRecord::Base belongs_to :user before_save :set_level acts_as_taggable serialize :committed, Array def self.comitted

ASP.NET的路由系统：路由映射

由 允我心安 提交于 2019-12-21 16:41:08
总的来说，我们可以通过RouteTable的静态属性Routes得到一个基于应用的全局路由表，通过上面的介绍我们知道这是一个类型的RouteCollection的集合对象，我们可以通过调用它的MapPageRoute进行路由映射，即注册URL模板与某个物理文件的匹配关系。路由注册的核心就是在全局路由表中添加一个Route对象，该对象的绝大部分属性都可以通过MapPageRoute方法的相关参数来指定。接下来我们通过实现演示的方式来说明路由注册的一些细节问题。 目录 一、变量默认值 二、约束 三、对现成文件的路由 四、注册路由忽略地址 五、直接添加路由对象 我们已前面介绍的关于获取天气预报信息的路由地址，我们在创建的ASP.NET Web应用中创建一个Weather.aspx页面，不过我们并不打算在该页面中呈现任何天气信息，而是将基于该页面的路由信息打印出来。该页面主体部分的HTML如下所示，我们不仅将基于当前页面的RouteData对象的Route和RouteHandler属性类型输出来，还将存储于Values和DataTokens字典的变量显示出来。 1: <body> 2: <form id="form1" runat="server"> 3: <div> 4: <table> 5: <tr> 6: <td>Route:</td> 7: <td><%=RouteData

求任意两天的天数差

由 女生的网名这么多〃 提交于 2019-12-20 10:16:51
** 求任意两天的天数差 ** 第二个函数调用了第一个函数，Dev C++编译过不了，应该怎么办，哪位哥帮忙解答一下 # include <stdio.h> # include <stdlib.h> typedef struct D { int year ; int month ; int day ; } Date ; int GetYear ( int year ) { if ( year % 4 == 0 && year % 100 != 0 || year % 400 == 0 ) return 29 ; else return 28 ; } int Getmonth ( int month ) { switch ( month ) { case 1 : case 3 : case 5 : case 7 : case 8 : case 10 : case 12 : return 31 ; case 2 : return GetYear ( year ) ; case 4 : case 6 : case 9 : case 11 : return 30 ; default : return - 1 ; } } int Getday ( Date date1 , Date date2 ) { int day = 0 ; int days ; if ( date1 . year ==

python之print刷新输出技巧(输出进度条)

由 拜拜、爱过 提交于 2019-12-20 05:06:07
import time for i in range(5): print("#",end="",flush=True) time.sleep(1) 这样可以逐个刷新输出“#” import time days=365 for i in range(365): print("进步百分比:{0}%".format(round(i+1)*100/days),end="",flush=True) time.sleep(1) 这个与上面相同，会把字符串里的全部刷新输出，包括“进步百分比”这几个中文字，所以，修改代码如下： import time days=365 for i in range(365): print("\r","进步百分比:{0}%".format(round(i+1)*100/days),end="",flush=True) time.sleep(1) 添加一个转义字符"\r"，使每次输出都回到最开始的位置，这样每次就可以在原处刷新了。 当然，python中也有相应的进度条模块库，比如 tqdm 和 progressbar ，代码如下所示： import time from tqdm import tqdm days=365 for i in tqdm(range(days)): print("\r""进步百分比:{0}%".format(round(i+1)*100

Linux下密码过期时间设置

由 浪尽此生 提交于 2019-12-20 00:17:43
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> Linux下对于新添加的用户，用户密码过期时间是从/etc/login.defs中PASS_MAX_DAYS提取的，普通系统默认就是99999，而有些安全操作系统是90。更改此处，只是让新建的用户默认密码过期时间变化，已有用户密码过期时间仍然不变。 [root @linuxidc ~]# chage --help Usage: chage [options] user Options: -d, --lastday LAST_DAY set last password change to LAST_DAY -E, --expiredate EXPIRE_DATE set account expiration date to EXPIRE_DATE -h, --help display this help message and exit -I, --inactive INACTIVE set password inactive after expiration to INACTIVE -l, --list show account aging information -m, --mindays MIN_DAYS set minimum number of days before password change

HDU 1555 How many days？

由 断了今生、忘了曾经 提交于 2019-12-18 23:47:31
Problem Description 8600的手机每天消费1元，每消费K元就可以获赠1元，一开始8600有M元，问最多可以用多少天？ Input 输入包括多个测试实例.每个测试实例包括2个整数M, k，（2 <= k <= M <= 1000).M = 0, k = 0代表输入结束. Output 对于每个测试实例输出一个整数，表示M元可以用的天数。 Sample Input 2 2 4 3 0 0 Sample Output 3 5 题意：中文题目，就不多说了，略。 分析：水题，看代码吧。 AC源代码（C语言）： 1 #include<stdio.h> 2 3 int main() 4 { 5 int M,k,i,Sumday; 6 while(scanf("%d%d",&M,&k)==2) 7 { 8 if(M==0&&k==0) break; 9 Sumday=0; 10 i=0; 11 while(M) 12 { 13 M-=1; 14 i+=1; 15 Sumday+=1; 16 if(i==k) 17 { 18 M+=1; 19 i=0; /*i要在这里重新初始化*/ 20 } 21 } 22 printf("%d\n",Sumday); 23 } 24 return 0; 25 } 2013-05-03 来源： https://www.cnblogs.com

java.util.Date Calculate difference in days

由 佐手、 提交于 2019-12-17 16:39:25
问题 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
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