获取给定间隔和步长的日期范围 [英] Get a range of dates given an interval and step

问题描述

修改 看起来创建一个表以分钟为单位保存DateTimes以进行连接将是最有意义的. 100年的分钟记录约有5200万行.通过Ticks进行索引应该可以使查询运行得非常快.现在变成

Edit It looks like creating a table that holds the DateTimes by minutes to join against would make the most sense. 100 years worth of minutes is ~52M rows. Indexed by the Ticks should make the query run pretty fast. It now becomes

感谢大家的反馈！

我有一个名为Recurrence的类，如下所示:

I have a class called Recurrence that looks like this:

public class Recurrence
{
    public int Id { get; protected set; }
    public DateTime StartDate { get; protected set; }
    public DateTime? EndDate { get; protected set; }
    public long? RecurrenceInterval { get; protected set; }

}

这是一个实体框架POCO类.我想对此类使用标准查询运算符做两件事. (以便查询完全在服务器端运行.)

It is an entity framework POCO class. There are two things I want to do with this class all with standard query operators. (So that the query runs entirely server side).

首先，我想创建一个查询，该查询返回从开始日期到结束日期(包括给定的重复间隔)的所有日期.迭代函数很简单

First I want to create a query that returns all the dates from the start date to the end date inclusive with the given recurrence interval. The iterative function is simple

for(i=StartDate.Ticks; i<=EndDate.Ticks; i+=RecurrenceInterval)
{
  yield return new DateTime(i);
}

Enumerable.Range()将是一个选项，但没有Range的长版本.我想我在这里唯一的选择是Aggregate，但是我对该功能仍然不太满意.

Enumerable.Range() would be an option but there is no long version of Range. I'm thinking my only option here is Aggregate but I'm still not very strong with that function.

最后，当我使该查询正常工作后，我想从那里返回时间范围内(即在不同的开始日期和结束日期之间)的值.使用SkipWhile/TakeWhile做到这一点很容易.

Finally once I have that query working, I want to return the values from there that are within a time window i.e. between a different start and end date. That is easy enough to do using SkipWhile/TakeWhile.

如果DateTime.Ticks为int，这就是我的处理方法

Here's how I could do it if DateTime.Ticks was an int

from recurrence in Recurrences
let range =
Enumerable
  .Range(
    (int)recurrence.StartDate.Ticks,
    recurrence.EndDate.HasValue ? (int)recurrence.EndDate.Value.Ticks : (int)end.Ticks)
  .Where(i=>i-(int)recurrence.StartDate.Ticks%(int)recurrence.RecurrenceLength.Value==0)
  .SkipWhile(d => d < start.Ticks)
  .TakeWhile(d => d <= end.Ticks)
from date in range
select new ScheduledEvent { Date = new DateTime(date) };

我想我需要的是可以在EF查询上执行的LongRange实现.

I guess what I need is an implementation of LongRange that could execute over an EF Query.

推荐答案

以下是产生递归点和指定子间隔的交点的函数:

Here's the function that yields the intersection of the Recurrence points and a specified subinterval:

public class Recurrence
{
    public int Id { get; protected set; }
    public DateTime StartDate { get; protected set; }
    public DateTime? EndDate { get; protected set; }
    public long? RecurrenceInterval { get; protected set; }

    // returns the set of DateTimes within [subStart, subEnd] that are
    // of the form StartDate + k*RecurrenceInterval, where k is an Integer
    public IEnumerable<DateTime> GetBetween(DateTime subStart, DateTime subEnd)
    {            
        long stride = RecurrenceInterval ?? 1;
        if (stride < 1) 
            throw new ArgumentException("Need a positive recurrence stride");

        long realStart, realEnd;

        // figure out where we really need to start
        if (StartDate >= subStart)
            realStart = StartDate.Ticks;
        else
        {
            long rem = subStart.Ticks % stride;
            if (rem == 0)
                realStart = subStart.Ticks;
            else
                // break off the incomplete stride and add a full one
                realStart = subStart.Ticks - rem + stride;
        }
        // figure out where we really need to stop
        if (EndDate <= subEnd)
            // we know EndDate has a value. Null can't be "less than" something
            realEnd = EndDate.Value.Ticks; 
        else
        {
            long rem = subEnd.Ticks % stride;
            // break off any incomplete stride
            realEnd = subEnd.Ticks - rem;
        }
        if (realEnd < realStart)
            yield break; // the intersection is empty

        // now yield all the results in the intersection of the sets
        for (long t = realStart; t <= realEnd; t += stride)
            yield return new DateTime(t);
    }

}

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