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我正在尝试显示工作时间/天数的间隔,它应该如下所示:

开放时间
(来源:clip2net.com

我有一张桌子,我在其中存储每天的天数、开放时间和关闭时间

桌子
(来源:clip2net.com

然后我创建了查询=>

var groups = from s in this.OpenTimes
orderby s.Day
group s by new { s.Till, s.Start } into gr
select new
{
    Time = gr.Key.Start + "-" + gr.Key.Till,
    Days = this.OpenTimes
        .Where(o => o.Start == gr.Key.Start && o.Till == gr.Key.Till)
        .OrderBy(d => d.Day).Select(d => d.Day).ToArray()
};

此查询提供了包含在此时间范围内的所有分组时间间隔和天但我遇到了问题 - 我创建了代表这些组的后半部分,但它无法正常工作。也许有人可以向我解释需要的观点或显示开放时间的基本逻辑。

谢谢指教...

4

1 回答 1

1

下一个方法对我有用:

结果屏幕

  public string OpeningTimesString
      {
         get
         {
            if (!this.OpeningTimes.IsLoaded)
               this.OpeningTimes.Load();
            var groups = (from s in this.OpeningTimes
                       orderby s.Day, s.Start, s.Stop
                       group s by new { Stop = formatTime(s.Stop), Start = formatTime(s.Start), s.Day } into gr
                       select new
                       {
                          Time = gr.Key.Start + "-" + gr.Key.Stop,
                          Day = gr.Key.Day
                       }).ToList();
            string result = "";
            int tmp = 1;
            for (int i = 0; i < groups.Count(); i++)
            {


               //One one = new One();
               bool exit = false;
               tmp = i;
               while (exit == false)
               {
                  if (i + 1 < groups.Count && groups[i].Time.Equals(groups[i + 1].Time))
                  {
                     i++;
                  }
                  else
                  {
                     if (tmp != i)
                        result += (NormalDayOfWeek)(groups[tmp].Day - 1) + "-" + (NormalDayOfWeek)(groups[i].Day - 1) + " : " + groups[i].Time + "<br />";
                     else
                        result += (NormalDayOfWeek)(groups[i].Day - 1) + " : " + groups[i].Time + "<br />";
                     exit = true;
                  }
               }
            }

            if (result.IsNotNull())
               return result;
            else
               return "[%Not yet defined]";
         }
      }
于 2009-07-19T07:34:15.757 回答