C# 获取当前年份的周期及周期所在日期范围(推荐)

最近有一个项目要用到年份周期，用于数据统计图表展示使用，当中用到年份周期,以及年份周期所在的日期范围。当初设想通过已知数据来换算年份周期，经过搜索资料发现通过数据库SQL语句来做，反而更加复杂。现在改变思路通过C#后台代码来算出两段日期范围中年份周期，在依据年份周期所对应的日期范围进行数据库查询进行统计。需要解决以下两个点问题，

第一点：依据日期查找所在年份的第几周；

第二点：依据年份所在的周期计算出周期所在的日期范围。

using System;
using System.Collections.Generic;
using System.Globalization;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace ConsoleApplication6
{
  class Program
  {
    static void Main(string[] args)
    {
      GregorianCalendar gc = new GregorianCalendar();
      int weekOfYear = gc.GetWeekOfYear(DateTime.Now, CalendarWeekRule.FirstDay, DayOfWeek.Monday);
      Console.WriteLine("当前第{0}周", weekOfYear);
      DateTime startDate, lastDate;
      for (int i = 1; i <= 53; i++)
      {
 GetDaysOfWeeks(DateTime.Now.Year, i, out startDate, out lastDate);
 Console.WriteLine("第{0}周", i);
 Console.WriteLine(startDate);
 Console.WriteLine(lastDate);
      }
      Console.ReadLine();
    }
    public static bool GetDaysOfWeeks(int year, int index, out DateTime first, out DateTime last)
    {
      first = DateTime.MinValue;
      last = DateTime.MinValue;
      if (year < 1700 || year > 9999)
      {
 //"年份超限"
 return false;
      }
      if (index < 1 || index > 53)
      {
 //"周数错误"
 return false;
      }
      DateTime startDay = new DateTime(year, 1, 1); //该年第一天
      DateTime endDay = new DateTime(year + 1, 1, 1).AddMilliseconds(-1);
      int dayOfWeek = 0;
      if (Convert.ToInt32(startDay.DayOfWeek.ToString("d")) > 0)
 dayOfWeek = Convert.ToInt32(startDay.DayOfWeek.ToString("d")); //该年第一天为星期几
      if (dayOfWeek == 0) { dayOfWeek = 7; }
      if (index == 1)
      {
 first = startDay.AddDays(7 - dayOfWeek - 6);
 if (dayOfWeek == 6)
 {
   last = first;
 }
 else
 {
   last = startDay.AddDays((7 - dayOfWeek));
 }
      }
      else
      {
 first = startDay.AddDays((8 - dayOfWeek) + (index - 2) * 7); //index周的起始日期
 last = first.AddDays(6);
 //if (last > endDay)
 //{
 //  last = endDay;
 //}
      }
      if (first > endDay) //startDayOfWeeks不在该年范围内
      {
 //"输入周数大于本年最大周数";
 return false;
      }
      return true;
    }
  }
}

　　执行结果　

总结

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