SQL速度：返回日期范围内的每个日期，并为每个日期返回count（）

我的目标是返回日期范围内的每个日期，并计数（）每个日期的所有记录

MyTable
-------------------------------
| OrderId |   DateFinalized   |
-------------------------------
|   51    | 2016-1-3 12:50:34 |
|   55    | 2016-1-4 10:01:56 |
|   73    | 2016-1-4 11:52:02 |
|   93    | 2016-1-6 01:35:16 |
|   104   | 2016-1-6 02:40:47 |
-------------------------------

挑战在于也要包括没有订单的日期。使用上面的

MyTable

，如果日期范围介于

2016-1-1

和

2016-1-6

之间，则所需输出为：

---------------------
|  MyDate  | Orders |
---------------------
| 2016-1-1 |   0    |
| 2016-1-2 |   0    |
| 2016-1-3 |   1    |
| 2016-1-4 |   2    |
| 2016-1-5 |   0    |
| 2016-1-6 |   2    |
---------------------

为了实现这一点，我使用此查询仅选择日期，并在1秒之内执行：

declare @startdate datetime = '1/1/2016';
declare @enddate datetime = '1/1/2017';

with [dates] as (
    select convert(date, @startdate) as [date] 
    union all
    select dateadd(day, 1, [date])
    from [dates]
    where [date] < @enddate 
)
select 
[date]
from [dates] 
where [date] between @startdate and @enddate
order by [date] desc
option (maxrecursion 0)

问题是，当我将这两个查询合并到一个SQL语句中时

左连接执行需要20秒（！！！）。我尝试了一个“咯咯笑”的子查询，但在13秒时效果并没有太好：

declare @startdate datetime = '1/1/2016';
declare @enddate datetime = '1/1/2017';

with [dates] as (
    select convert(date, @startdate) as [date] 
    union all
    select dateadd(day, 1, [date])
    from [dates]
    where [date] < @enddate 
)
select 
[date]
from [dates] 
where [date] between @startdate and @enddate
order by [date] desc
option (maxrecursion 0)

如何有效地加入结果数据集？

提前感谢您的时间。
使用递归cte是生成日期范围最糟糕的方法之一。与使用递归cte相比，使用堆叠cte可根据需要生成日期范围

如果要跨多行或长时间使用它，或者要多次运行此类操作，最好只创建一个
日期
或
日历
表

内存只有152kb，表中可以有30年的日期，可以这样使用：

/* dates table */ 
declare @fromdate date = '20000101';
declare @years    int  = 30;
/* 30 years, 19 used data pages ~152kb in memory, ~264kb on disk */
;with n as (select n from (values(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) t(n))
select top (datediff(day, @fromdate,dateadd(year,@years,@fromdate)))
    [Date]=convert(date,dateadd(day,row_number() over(order by (select 1))-1,@fromdate))
into dbo.Dates
from n as deka cross join n as hecto cross join n as kilo 
               cross join n as tenK cross join n as hundredK
order by [Date];

create unique clustered index ix_dbo_Dates_date 
  on dbo.Dates([Date]);

并这样查询：

select
    d.[Date]
  , OrderCount = count(o.OrderID)
from dates d
  left join orders o
    on convert(date,o.OrderDate) = d.[Date]
group by d.[Date]
order by d.[Date] desc

编号和日历表参考：











如果您确实不想要日历表，可以使用堆叠的cte部分：

declare @fromdate date = '20160101';
declare @years    int  = 1;
;with n as (select n from (values(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) t(n))
, dates as (
  select top (datediff(day, @fromdate,dateadd(year,@years,@fromdate)))
      [Date]=convert(date,dateadd(day,row_number() over(order by (select 1))-1,@fromdate))
  from n as deka cross join n as hecto cross join n as kilo 
                /* cross join n as tenK cross join n as hundredK */
   order by [Date]
)
select
    d.[Date]
  , OrderCount = count(o.OrderID)
from dates d
  left join orders o
    on convert(date,o.OrderDate) = d.[Date]
group by d.[Date]
order by d.[Date] desc

使用索引：尝试使用数字表而不是递归cte。我用过