SQL查询以计算每日总计跳过天数

Question

此查询统计我们一个月内生产的瓶子数量，并按天分组。如果当天没有生产任何瓶子，则从输出中跳过，而不是返回0个瓶子。如果没有生产瓶子，如何返回当天的时间戳？我听说日历表必须用于此目的。

SELECT CONVERT(datetime,CAST(t_stamp AS DATE)), COUNT(bottles) AS 'Good Bottles'
FROM bottles
WHERE t_stamp
    BETWEEN "any date"
    AND "any date"
GROUP BY CAST(t_stamp AS DATE)
ORDER BY CAST(t_stamp AS DATE) ASC

当前输出：

Aug 12, 2019 12:00 am..................4302
Aug 13, 2019 12:00 am..................2302
Aug 17, 2019 12:00 am..................1302
Aug 18, 2019 12:00 am..................4302

所需的输出：

Aug 12, 2019 12:00 am..................4302
Aug 13, 2019 12:00 am..................2302
Aug 14, 2019 12:00 am..................0
Aug 15, 2019 12:00 am..................0
Aug 16, 2019 12:00 am..................0
Aug 17, 2019 12:00 am..................1302
Aug 18, 2019 12:00 am..................4302

Answer 1

您需要生成日期。一个非常简单的方法使用递归CTE：

WITH dates as (
      SELECT CONVERT(date, "any date1") as dte
      UNION ALL
      SELECT DATEADD(day, 1, dte)
      FROM dates
      WHERE dte < "any date2"
     )
SELECT d.dte, COUNT(bottles) AS GoodBottles
FROM dates d LEFT JOIN
     bottles b
     ON CAST(t_stamp as DATE) = d.dte
GROUP BY d.dte
ORDER BY d.dte ASC;

注意：

• 如果您有日历或理货表格，请改用它。
• 如果日期数超过100，则需要添加OPTION (MAXRECURSION 0)。
• COUNT(bottles)看起来可疑。您真的打算SUM(bottles)吗？
• 将列转换为date然后转换为datetime也是可疑的。目前尚不清楚为什么要在第一列使用datetime。