Sql server 当日期可能缺失时，以“周…”分组获取数据_Sql Server_Tsql

Sql server 当日期可能缺失时，以“周…”分组获取数据

sql-server tsql

Sql server 当日期可能缺失时，以“周…”分组获取数据,sql-server,tsql,Sql Server,Tsql,我在一个带有日期的表格中有数据，希望按周计算行数，例如2017-05-01的一周，其中结果是星期一开始的一周日期和匹配行数，即使该周没有行。这些都将在一个日期范围内通过对DATEPARTwk进行分组，我可以很容易地将事情划分为几周，其中D是日期列，但我很难做到：如何获取日期和填充的周，以及如何在数据中没有匹配行的情况下保留一周的行以下是按周分组： SET DATEFORMAT ymd; SET DATEFIRST 1; -- Monday is first day of week DE

我在一个带有日期的表格中有数据，希望按周计算行数，例如2017-05-01的一周，其中结果是星期一开始的一周日期和匹配行数，即使该周没有行。这些都将在一个日期范围内

通过对DATEPARTwk进行分组，我可以很容易地将事情划分为几周，其中D是日期列，但我很难做到：

如何获取日期和填充的周，以及

如何在数据中没有匹配行的情况下保留一周的行

以下是按周分组：

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：

+−−−−−−−−−−−−−+−−−−−−−+ | Week Number | Count | +−−−−−−−−−−−−−+−−−−−−−+ | 19 | 5 | | 20 | 19 | | 22 | 8 | | 23 | 10 | | 24 | 5 | | 26 | 4 | +−−−−−−−−−−−−−+−−−−−−−+

为此，您必须生成一个表或CTE，其中包含星期一日期及其周数，如中所示，根据下面需要进行的操作进行了轻微修改，然后使用周数将数据按周分组，左连接或外应用该表或CTE：

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;

关于这一点，有两点需要注意：

我假设我们可以确保@startDate是星期一，这在查询之外很容易完成，或者如果需要备份到WEEKPARTWEEKDAY，@startDate是1，可以使用T-SQL中的简单循环完成。或者最坏的情况是，我们可以生成所有日期，然后用WEEKPARTWEEKDAY过滤它们

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，请将周数与年数组合在一起，我们只使用上面的周数，例如DATEPARTYEAR，D*100+DATEPARTwk，D

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;

关于这一点，有两点需要注意：

这是另一种方法。我包括这一点，因为它将生成比递归CTE解决方案更少的读取，并且将非常快

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

这是另一种方法。我包括这一点，因为它将生成比递归CTE解决方案更少的读取，并且将非常快

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

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]

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]

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