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第几周计算代码和C#函数方法

周计算方法C#

/// 当前月有多少天
        /// </summary>
        /// <param name="y"></param>
        /// <param name="m"></param>
        /// <returns></returns>
        public static int HowMonthDay(int y, int m)
        {
            int mnext;
            int ynext;
            if (m < 12)
            {
                mnext = m + 1;
                ynext = y;
            }
            else
            {
                mnext = 1;
                ynext = y + 1;
            }
            DateTime dt1 = System.Convert.ToDateTime(y + "-" + m + "-1");
            DateTime dt2 = System.Convert.ToDateTime(ynext + "-" + mnext + "-1");
            TimeSpan diff = dt2 - dt1;
            return diff.Days;
        }

        /**//// <summary>
        /// 得到一年中的某周的起始日和截止日
        /// 年 nYear
        /// 周数 nNumWeek
        /// 周始 out dtWeekStart
        /// 周终 out dtWeekeEnd
        /// </summary>
        /// <param name="nYear"></param>
        /// <param name="nNumWeek"></param>
        /// <param name="dtWeekStart"></param>
        /// <param name="dtWeekeEnd"></param>
        public static void GetWeek(int nYear, int nNumWeek, out   DateTime dtWeekStart, out   DateTime dtWeekeEnd)
        {
            DateTime dt = new DateTime(nYear, 1, 1);
            dt = dt + new TimeSpan((nNumWeek - 1) * 7, 0, 0, 0);
            dtWeekStart = dt.AddDays(-(int)dt.DayOfWeek + (int)DayOfWeek.Monday);
            dtWeekeEnd = dt.AddDays((int)DayOfWeek.Saturday - (int)dt.DayOfWeek + 1);
        }

        /**//// <summary>
        /// 求某年有多少周
        /// 返回 int
        /// </summary>
        /// <param name="strYear"></param>
        /// <returns>int</returns>
        public static int GetYearWeekCount(int strYear)
        {
            string returnStr = "";

            System.DateTime fDt = DateTime.Parse(strYear.ToString() + "-01-01");
            int k = Convert.ToInt32(fDt.DayOfWeek);//得到该年的第一天是周几
            if (k == 1)
            {
                int countDay = fDt.AddYears(1).AddDays(-1).DayOfYear;
                int countWeek = countDay / 7 + 1;
                return countWeek;

            }
            else
            {
                int countDay = fDt.AddYears(1).AddDays(-1).DayOfYear;
                int countWeek = countDay / 7 + 2;
                return countWeek;
            }

        }

        /**//// <summary>
        /// 求当前日期是一年的中第几周
        /// </summary>
        /// <param name="date"></param>
        /// <returns></returns>
        public static int WeekOfYear(DateTime curDay)
        {
            int firstdayofweek = Convert.ToInt32(Convert.ToDateTime(curDay.Year.ToString() + "- " + "1-1 ").DayOfWeek);

            int days = curDay.DayOfYear;
            int daysOutOneWeek = days - (7 - firstdayofweek);

            if (daysOutOneWeek <= 0)
            {
                return 1;
            }
            else
            {
                int weeks = daysOutOneWeek / 7;
                if (daysOutOneWeek % 7 != 0)
                    weeks++;

                return weeks + 1;

            }

  }

=========================================================================

C 语言中求某一天是日历上第几周怎么计算?
用c语言编程,只知道年月日,星期几,那么怎么计算某一天是第几周呢?
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按票数排序 2 个回答

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1
反对,不会显示你的姓名
skydiver,C++ is a terrible language.
文闻 赞同
参考 Stackoverflow:c++ - How do I calculate the week number given a date?
使用 C 标准库 time.h 就可以完成这个计算。
示例代码如下(来自上面链接)
#include <stdio.h>
#include <string.h>
#include <time.h>

int
main(void)
{
  struct tm tm;
  char timebuf[64];

  // Zero out struct tm
  memset(&tm, 0, sizeof(struct tm));

  // November 4, 2008 11:00 pm
  tm.tm_sec = 0;
  tm.tm_min = 0;
  tm.tm_hour = 23;
  tm.tm_mday = 4;
  tm.tm_mon = 10;
  tm.tm_year = 108;

  // Call mktime to recompute tm.tm_wday and tm.tm_yday
  mktime(&tm);

  if (strftime(timebuf, sizeof(timebuf), "%W", &tm) != 0) {
    printf("Week number is: %s\n", timebuf);
  }

  return 0;
}
2013-05-27 添加评论 感谢 分享 收藏 • 没有帮助 • 举报

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反对,不会显示你的姓名
汪建,茫茫人海一小研
lena zhao 赞同
1、知道年月日,可以计算这个日期是这一年的第几天m,也就是减去1月1日,例如:2月2日是这一年的第33天
2、将得到的天数m除以7,取整,得到过去了多少周n
3、将得到的天数m对7取余,与给定日期是星期几做对比,可以判断出是否出现了跨周的情况,如果出现了,再将2中得到的n加1

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