public int[] date(int thisdate) {
        //输入20110928的日期格式,返回输入日期的前7天日期并保存在数组中返回。并且要求剔除节假日和周六日
        }
比如输入20110928,应该返回的是20110927,20110926,20110923,20110922,20110921,20110920,20110919.也就是剔除了24,25号周六日,如果是节假日也要剔除
好像比较难。。节假日和周六日怎么判断

解决方案 »

  1.   

    各个国家的节假日均无法确定
    你 还是自己定义一下吧
    所以最好自建一个表来指定那些日期是节假日。
    然后,可在SQL中根据此表中的数据进行判断
      

  2.   

     if (start.DayOfWeek == DayOfWeek.Saturday || start.DayOfWeek == DayOfWeek.Sunday)
      

  3.   


            public int[] date(int thisdate)
            {
                int[] datet = {0,0,0,0,0,0,0 };
                DateTime today = DateTime.ParseExact(thisdate.ToString(), "yyyyMMdd", null);
                int count = 0;
                int i = -1;
                while (count < 7)
                {
                    DateTime ss = today.AddDays(i);
                    if (ss.DayOfWeek != DayOfWeek.Saturday && ss.DayOfWeek != DayOfWeek.Sunday)
                    {
                        datet[count] =Convert.ToInt32(ss.ToString("yyyyMMdd"));
                        count++;
                    }
                    i--;
                }
                return datet;
            }
      

  4.   


            public int[] date(int thisdate)
            {
                List<string> days = new List<string>();
                days.Add("20111001");
                days.Add("20111002");
                days.Add("20111003");
                days.Add("20111004");
                //算日期麻烦,你自己把11年的假期全加进去            int[] datet = {0,0,0,0,0,0,0 };
                DateTime today = DateTime.ParseExact(thisdate.ToString(), "yyyyMMdd", null);
                int count = 0;
                int i = -1;
                while (count < 7)
                {
                    DateTime ss = today.AddDays(i);
                    if (ss.DayOfWeek != DayOfWeek.Saturday && ss.DayOfWeek != DayOfWeek.Sunday && !days.Contains(ss.ToString("yyyyMMdd")))
                    {
                        datet[count] =Convert.ToInt32(ss.ToString("yyyyMMdd"));
                        count++;
                    }
                    i--;
                }
                return datet;
            }