Question

我正在尝试编写SQL以生成以下数据

Date         Count
2018-09-24   2
2018-09-25   2
2018-09-26   2
2018-09-27   2
2018-09-28   2
2018-09-29   1

我正在使用的基本表的示例是

ID      StartDate   EndDate
187267  2018-09-24  2018-10-01
187270  2018-09-24  2018-09-30

因此，我试图获取两个日期之间的日期列表，然后计算每个日期中有多少基本数据记录。

我开始使用临时表并尝试遍历记录以获取结果，但是我不确定这是否是正确的方法。

到目前为止，我已经有了此代码

WITH ctedaterange 
     AS (SELECT [Dates] = (select ea.StartWork from EngagementAssignment ea where ea.EngagementAssignmentId IN(SELECT ea.EngagementAssignmentId
                                                                                                            FROM EngagementLevel el INNER JOIN
                                                                                                            EngagementAssignment ea ON el.EngagementLevelID = ea.EngagementLevelId
                                                                                                            WHERE el.JobID = 15072 and ea.AssetId IS NOT NULL))
         UNION ALL
         SELECT [dates] + 1 
         FROM   ctedaterange 
         WHERE  [dates] + 1 < = (select ea.EndWork from EngagementAssignment ea where ea.EngagementAssignmentId IN(SELECT ea.EngagementAssignmentId
                                                                                                            FROM EngagementLevel el INNER JOIN
                                                                                                            EngagementAssignment ea ON el.EngagementLevelID = ea.EngagementLevelId
                                                                                                            WHERE el.JobID = 15072 and ea.AssetId IS NOT NULL)))
SELECT [Dates], Count([Dates])
FROM   ctedaterange 
GROUP BY [Dates]

但我收到此错误

子查询返回了多个值。当子查询遵循=，！=，<，<=，>，> =或将子查询用作表达式时，不允许这样做。

当我使用的作业仅在where子句的subselect中生成一条记录时，我得到正确的结果，即：

SELECT ea.EngagementAssignmentId
FROM EngagementLevel el INNER JOIN
EngagementAssignment ea ON el.EngagementLevelID = ea.EngagementLevelId
WHERE el.JobID = 15047 and ea.AssetId IS NOT NULL

产生一条记录。

结果如下：

Dates                   (No column name)
2018-09-24 02:00:00.000 1
2018-09-25 02:00:00.000 1
2018-09-26 02:00:00.000 1
2018-09-27 02:00:00.000 1
2018-09-28 02:00:00.000 1
2018-09-29 02:00:00.000 1
2018-09-30 02:00:00.000 1
2018-10-01 02:00:00.000 1

Answer 1

尝试一下：demo

sdate      total
2018-09-24  2
2018-09-25  2
2018-09-26  2
2018-09-27  2
2018-09-28  2
2018-09-29  2
2018-09-30  1

输出：

from itertools import chain

df = pd.DataFrame({
    'Feature' : df['Feature'].values.repeat(df['Class'].str.len()),
    'Class' : list(chain.from_iterable(df['Class'].values.tolist()))
})
print (df)
  Feature   Class
0   text1  label1
1   text1  label2
2   text2  label2
3   text2  label3

Answer 2

您可以通过更改日期和日期来根据您的范围生成

 DECLARE 
    @DateFrom DATETIME = GETDATE(),
    @DateTo DATETIME = '2018-10-30';

WITH DateGenerate 
AS (
    SELECT @DateFrom as MyDate
    UNION ALL
    SELECT DATEADD(DAY, 1, MyDate)
    FROM DateGenerate
    WHERE MyDate < @DateTo
   )

SELECT 
    MyDate
FROM 
    DateGenerate;

Answer 3

好吧，如果您的日期范围很短，则可以使用递归CTE，如其他答案所示。递归CTE的问题在于范围很大，开始变得无效-因此，我想向您展示一种不同的方法，该方法无需使用递归即可构建日历CTE。

首先，创建并填充示例表（请在您将来的问题中为我们保存此步骤）：

DECLARE @T AS TABLE
(
    ID int,
    StartDate date,
    EndDate date
)
INSERT INTO @T (ID, StartDate, EndDate) VALUES
(187267, '2018-09-24', '2018-10-01'),
(187270, '2018-09-24', '2018-09-30')

然后，在日历cte中获取第一个开始日期和所需的日期数：

DECLARE @DateDiff int, @StartDate Date

SELECT @DateDiff = DATEDIFF(DAY, MIN(StartDate), Max(EndDate)),
       @StartDate = MIN(StartDate)
FROM @T

现在，基于row_number构建日历cte（也就是说，除非您已经有一个可以使用的数字（计数）表）：

;WITH Calendar(TheDate)
AS
(
    SELECT TOP(@DateDiff + 1) DATEADD(DAY, ROW_NUMBER() OVER(ORDER BY @@SPID)-1, @StartDate)
    FROM sys.objects t0
    -- unremark the next row if you don't get enough records...
    -- CROSS JOIN sys.objects t1 
)

请注意，我正在使用row_number() - 1，因此必须选择top(@DateDiff + 1)

最后-查询：

SELECT TheDate, COUNT(ID) As NumberOfRecords
FROM Calendar
JOIN @T AS T
    ON Calendar.TheDate >= T.StartDate
    AND Calendar.TheDate <= T.EndDate
GROUP BY TheDate

结果：

TheDate     |   NumberOfRecords
2018-09-24  |   2
2018-09-25  |   2
2018-09-26  |   2
2018-09-27  |   2
2018-09-28  |   2
2018-09-29  |   2
2018-09-30  |   2
2018-10-01  |   1

You can see a live demo on rextester.

Answer 4

您能在我使用SQL dates table function [dbo]。[DatesTable]的情况下尝试执行SQL CTE查询吗，它在源表中生成介于最小日期和最大日期之间的日期列表

;with boundaries as (
    select
        min(StartDate) minD, max(EndDate) maxD
    from DateRanges
), dates as (
    select
        dates.[date]
    from boundaries
    cross apply [dbo].[DatesTable](minD, maxD) as dates
)
select dates.[date], count(*) as [count]
from dates
inner join DateRanges
on dates.date between DateRanges.StartDate and DateRanges.EndDate
group by dates.[date]
order by dates.[date]

输出符合预期

为两个以上的记录生成一个日期列表

4 个答案: