如何查找缺失的小时数并将数据显示为 0

Question

此查询提取每天每个小时的计数，但忽略没有计数的任何日期和时间。如何填写所选日期之间缺失的日期和小时数？

SELECT CAST(RecordTime AS date) AS Date, DATENAME(dw, RecordTime) AS [Day of the week], DATEPART(hour, RecordTime) AS [Hour of the day], COUNT(*) AS [Hourly Count]
    FROM Counts
    WHERE (RecordTime >= CONVERT(DATETIME, '2022-04-01 00:00:00', 102)) AND (RecordTime < CONVERT(DATETIME, '2022-05-01 00:00:00', 102)) AND (MachineNum = 11) AND (Cavity = 1)
    GROUP BY CAST(RecordTime AS date), DATEPART(hour, RecordTime), DATENAME(dw, RecordTime)
    ORDER BY Date, [Hour of the day]

Answer 1

您需要一个参考 table 来存储所有日期和相应的时间。它将是一种添加了小时的日历 table。

第 1 步：

创造 table 小时。

CREATE TABLE hours
  (
     hour_key INTEGER
  );

INSERT INTO hours
VALUES      (0),
            (1),
            (2),
            (3),
            (4),
            (5),
            (6),
            (7),
            (8),
            (9),
            (10),
            (11),
            (12),
            (13),
            (14),
            (15),
            (16),
            (17),
            (18),
            (19),
            (20),
            (21),
            (22),
            (23);

第 2 步：

创建一个 table，它将具有所需的日期和时间字段。

SELECT Cast(recordtime AS DATE) AS Date_key,
               Datename(dw, recordtime) week_key,
               hour_key
        FROM   hours
               CROSS JOIN counts
        GROUP  BY Cast(recordtime AS DATE),
                  Datename(dw, recordtime),
                  hour_key

注意：使用 Counts table 来导出日期和小时字段是一个糟糕的解决方案，因为它可能是事务性的 table 并且具有巨大的记录数。而是使用日历 table.

最后一步：

使用在步骤 2 中创建的 table 作为主要 table（或用作子查询）和 left join Counts 基于日期和时间。

下面的查询应该会给你想要的输出。

SELECT date_key            AS Date,
       week_key            AS [Day of the week],
       hour_key            AS [Hour of the day],
       Count(c.recordtime) AS [Hourly Count]
FROM   (SELECT Cast(recordtime AS DATE) AS Date_key,
               Datename(dw, recordtime) week_key,
               hour_key
        FROM   hours
               CROSS JOIN counts
        GROUP  BY Cast(recordtime AS DATE),
                  Datename(dw, recordtime),
                  hour_key)cal
       LEFT JOIN counts c
              ON cal.date_key = Cast(c.recordtime AS DATE)
                 AND cal.hour_key = Datepart(hour, c.recordtime)
GROUP  BY date_key,
          week_key,
          hour_key
ORDER  BY date,
          [hour of the day]

SQL Fiddle: Try it here

Answer 2

正如 Larnu 所建议的，您需要生成一个包含范围内所有天+小时组合的完整数据集，以便左连接到。根据我的计算，您需要 30 天 * 24 小时 = 720 行。如果您还没有 a numbers table, or a calendar table, or a sequence generating function，您可以使用递归 CTE 生成它，如下所示：

DECLARE @StartDate     datetime = '20220401',
        @AfterLastDate datetime = '20220501';
        
;WITH days(d) AS
(
  SELECT 0 UNION ALL SELECT d+1 FROM days
  WHERE d < DATEDIFF(DAY, @StartDate, @AfterLastDate) - 1
), hours(h) AS
(
  SELECT 0 UNION ALL SELECT h+1 FROM hours WHERE h<23
),
dates(DayHour, h) AS
(
  SELECT DATEADD(HOUR, hours.h, DATEADD(DAY, days.d, @StartDate)), hours.h
  FROM days CROSS JOIN hours
)
SELECT d.DayHour, DATENAME(WEEKDAY, DayHour), d.h
  FROM dates AS d
  ORDER BY d.DayHour;

输出：

DayHour Day of the week Hour of the day

2022-04-01 00:00:00.000 Friday 0

2022-04-01 01:00:00.000 Friday 1

2022-04-01 02:00:00.000 Friday 2

... 714 more rows ...

2022-04-30 21:00:00.000 Saturday 21

2022-04-30 22:00:00.000 Saturday 22

2022-04-30 23:00:00.000 Saturday 23

示例db<>fiddle

现在，我们只需要对您现有的 table:

进行左外连接 that

DECLARE @StartDate     datetime = '20220401',
        @AfterLastDate datetime = '20220501';
        
;WITH days(d) AS
(
  SELECT 0 UNION ALL SELECT d+1 FROM days
  WHERE d < DATEDIFF(DAY, @StartDate, @AfterLastDate) - 1
), hours(h) AS
(
  SELECT 0 UNION ALL SELECT h+1 FROM hours WHERE h<23
),
dates(DayHour, h) AS
(
  SELECT DATEADD(HOUR, hours.h, DATEADD(DAY, days.d, @StartDate)), 
    hours.h FROM days CROSS JOIN hours
)
SELECT [Date] = CONVERT(date, d.DayHour), 
  [Day of the week] = DATENAME(WEEKDAY, d.DayHour), 
  [Hour of the day] = d.h,
  [Hourly Count] = COUNT(c.RecordTime)
FROM dates AS d
LEFT OUTER JOIN dbo.Counts AS c
ON c.RecordTime >= d.DayHour
  AND c.RecordTime < DATEADD(HOUR, 1, d.DayHour)
  AND c.MachineNum = 11
  AND c.Cavity = 1
GROUP BY CONVERT(date, d.DayHour), DATENAME(WEEKDAY, DayHour), d.h
ORDER BY [Date], [Hour of the day];

示例db<>fiddle

如果您有数字 table，则生成日期会更容易一些。这是一个简单的示例，仅包含 1,000 行作为您期望的最大日期范围，并使用递归 CTE - 有多种方法可以最初填充数字 table 并且性能并不重要。

CREATE TABLE dbo.Numbers(n int PRIMARY KEY);

;WITH x(x) AS 
(
  SELECT 0 UNION ALL SELECT x+1 FROM x 
  WHERE x < 1000
)
INSERT dbo.Numbers(n) 
  SELECT x FROM x OPTION (MAXRECURSION 0);

现在查询获取范围内的所有日期：

DECLARE @StartDate datetime = '20220401',
        @AfterLastDate datetime = '20220501';
       
;WITH dates(d) AS 
(
  SELECT TOP (DATEDIFF(DAY, @StartDate, @AfterLastDate)) n 
  FROM dbo.Numbers ORDER BY n
),
hours(h) AS 
(
  SELECT TOP (24) n FROM dbo.Numbers ORDER BY n
)
SELECT DayHour = DATEADD(HOUR, hours.h, 
    DATEADD(DAY, dates.d, @StartDate))
  FROM dates CROSS JOIN hours
  ORDER BY DayHour;

然后您可以将其用作核心数据集以左连接，就像上面的示例一样。

如何查找缺失的小时数并将数据显示为 0

How to find missing hours and show data as 0

sql

sql-server

sql-server-2016

第 1 步：

第 2 步：

最后一步：

DayHour	Day of the week	Hour of the day
2022-04-01 00:00:00.000	Friday	0
2022-04-01 01:00:00.000	Friday	1
2022-04-01 02:00:00.000	Friday	2
... 714 more rows ...
2022-04-30 21:00:00.000	Saturday	21
2022-04-30 22:00:00.000	Saturday	22
2022-04-30 23:00:00.000	Saturday	23