当日期可能丢失时，以“...的周”组获取数据

Question

我在一个带日期的表中有数据，并且想按“周”（例如“2017-05-01 周”）计算行数，其中结果具有周的日期（从星期一开始）和匹配行数——即使那周没有行。 （这都将在一个日期范围内。）

我可以通过在DATEPART(wk, D)（其中D 是日期列）上分组来轻松地将事情划分为几周，但我正在努力解决：

1. 如何获取“星期”日期并填写，以及

2. 如何在数据中没有匹配行的情况下连续一周

这里按周分组：

SET DATEFORMAT ymd;
SET DATEFIRST 1; -- Monday is first day of week

DECLARE @startDate DATETIME = '2017-05-01';
DECLARE @endDate DATETIME = '2017-07-01';

SELECT      DATEPART(wk, D) AS [Week Number], COUNT(*) AS [Count]
FROM        #temp
GROUP BY    DATEPART(wk, D)
ORDER BY    DATEPART(wk, D);

这给了我：

+−−−−−−−−−−−−−−+−−−−−−−−−+ |周数 |计数 | +−−−−−−−−−−−−−−+−−−−−−−−−+ | 19 | 5 | | 20 | 19 | | 22 | 8 | | 23 | 10 | | 24 | 5 | | 26 | 4 | +−−−−−−−−−−−−−−+−−−−−−−−−+

但理想情况下我想要：

+−−−−−−−−−−−−−+−−−−−−−−+ |周 |计数 | +−−−−−−−−−−−−−+−−−−−−−−+ | 2017-05-01 | 5 | | 2017-05-08 | 19 | | 2017-05-15 | 0 | | 2017-05-22 | 8 | | 2017-05-29 | 10 | | 2017-06-05 | 5 | | 2017-06-12 | 0 | | 2017-06-19 | 4 | | 2017-06-26 | 0 | +−−−−−−−−−−−−−+−−−−−−−−+

我该怎么做？

---

设置测试信息：

SET DATEFIRST 1;
SET DATEFORMAT ymd;

CREATE TABLE #temp (
    D DATETIME
);
GO

INSERT INTO #temp (D)
VALUES      -- Week of 2017-05-01 (#19)
            ('2017-05-01'),('2017-05-01'),('2017-05-01'),
            ('2017-05-06'),('2017-05-06'),
            -- Week of 2017-05-08 (#20) - note no data actually on the 8th
            ('2017-05-10'),
            ('2017-05-11'),('2017-05-11'),('2017-05-11'),('2017-05-11'),('2017-05-11'),('2017-05-11'),
            ('2017-05-12'),('2017-05-12'),('2017-05-12'),('2017-05-12'),
            ('2017-05-13'),('2017-05-13'),('2017-05-13'),('2017-05-13'),('2017-05-13'),('2017-05-13'),('2017-05-13'),
            ('2017-05-14'),
            -- Week of 2017-05-15 (#21)
            -- (note we have no data for this week)
            -- Week of 2017-05-22 (#22)
            ('2017-05-22'),('2017-05-22'),('2017-05-22'),
            ('2017-05-23'),('2017-05-23'),('2017-05-23'),('2017-05-23'),('2017-05-23'),
            -- Week of 2017-05-29 (#23)
            ('2017-05-29'),('2017-05-29'),('2017-05-29'),
            ('2017-06-02'),('2017-06-02'),
            ('2017-06-03'),
            ('2017-06-04'),('2017-06-04'),('2017-06-04'),('2017-06-04'),
            -- Week of 2017-06-05 (#24) - note no data actually on the 5th
            ('2017-06-08'),('2017-06-08'),('2017-06-08'),
            ('2017-06-11'),('2017-06-11'),
            -- Week of 2017-06-12 (#25)
            -- (note we have no data for this week)
            -- Week of 2017-06-19 (#26)
            ('2017-06-19'),('2017-06-19'),('2017-06-19'),
            ('2017-06-20');
GO

Answer 1

为此，您必须生成一个包含星期一日期及其周数的表格或 CTE（如 this answer 所示，为我们需要在下面执行的操作稍作修改），然后是 LEFT JOIN 或 OUTER APPLY使用周数将您的数据按周分组：

SET DATEFORMAT ymd;
SET DATEFIRST 1;

DECLARE @startDate DATETIME = '2017-05-01';
DECLARE @endDate DATETIME = '2017-07-01';

;WITH Mondays AS (
    SELECT  @startDate AS D, DATEPART(WK, @startDate) AS W
    UNION ALL
    SELECT  DATEADD(DAY, 7, D), DATEPART(WK, DATEADD(DAY, 7, D))
    FROM    Mondays m
    WHERE   DATEADD(DAY, 7, D) < @endDate
)
SELECT      LEFT(CONVERT(NVARCHAR(MAX), Mondays.D, 120), 10) AS [Week Of], d.Count
FROM        Mondays
OUTER APPLY (
            SELECT  COUNT(*) AS [Count]
            FROM    #temp
            WHERE   DATEPART(WK, D) = W
            AND     D >= @startDate
            AND     D < @endDate
) d
ORDER BY    Mondays.D;

对此有两个注意事项：

1. 我假设我们可以确保 @startDate 是星期一，这很容易在查询之外完成，或者如果需要可以通过 T-SQL 中的简单循环完成（备份到 WEEKPART(WEEKDAY, @startDate) 是1)。 （或者最坏的情况，我们可以生成所有日期，然后使用 WEEKPART(WEEKDAY, ...) 过滤它们。）

2. 我假设日期范围总是一年或更短；否则，我们会有重复的周数。如果日期范围可能超过一年，请将周数与我们仅使用上述周数的任何地方的年份结合起来（例如，DATEPART(YEAR, D) * 100 + DATEPART(wk, D)）。

Answer 2

你可以用这个。

SET DATEFORMAT ymd;
SET DATEFIRST 1; -- Monday is first day of week

DECLARE @startDate DATETIME = '2017-05-01';
DECLARE @endDate DATETIME = '2017-07-01';

;WITH OrgResult AS ( -- Grouping result with missing week. Answer of the first question
    SELECT 
        DATEADD(DAY, 1 - DATEPART (WEEKDAY, D), D) [Week] -- Fist Day Of the Week
        , COUNT(*) [Count]
    FROM #temp
        WHERE D BETWEEN @startDate AND @endDate
    GROUP BY 
        DATEADD(DAY, 1 - DATEPART (WEEKDAY, D), D)
)
, Result AS -- Adds only missing weeks. Answer of the second question
(
    SELECT * FROM OrgResult
    UNION ALL
    SELECT DATEADD( DAY, 7, R.[Week] ), 0 [Count] 
    FROM Result R 
    WHERE NOT EXISTS( SELECT * FROM OrgResult O WHERE [Week] = DATEADD( DAY, 7, R.[Week] ) )
            AND DATEADD( DAY, 7, R.[Week] ) <= @endDate
)
SELECT * FROM Result
ORDER BY [Week]

结果：

Week        Count
----------- -----------
2017-05-01  5
2017-05-08  19
2017-05-15  0
2017-05-22  8
2017-05-29  10
2017-06-05  5
2017-06-12  0
2017-06-19  4
2017-06-26  0

Answer 3

这是另一种方法。我将其包含在内，因为它比递归 CTE 解决方案生成的读取更少，而且速度会快很多

WITH E(N) AS (SELECT 1 FROM (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1))x(x)),
iTally(N) AS 
(
  SELECT TOP (((DATEDIFF(day,@startdate, @endDate))/7)+1)
    (ROW_NUMBER() OVER (ORDER BY (SELECT 1))-1)
  FROM E a, E b, E c
)
SELECT WeekOf = DATEADD(WEEK,N,@startDate), [count] = COUNT(t.D)
FROM iTally i
LEFT JOIN #temp t ON t.D >= DATEADD(WEEK,N,@startDate) AND t.D < DATEADD(WEEK,N+1,@startDate)
GROUP BY DATEADD(WEEK,N,@startDate)
ORDER BY DATEADD(WEEK,N,@startDate); -- not required

结果：

WeekOf     count
---------- -----------
2017-05-01 5
2017-05-08 19
2017-05-15 0
2017-05-22 8
2017-05-29 10
2017-06-05 5
2017-06-12 0
2017-06-19 4
2017-06-26 0