使用日期过滤器相关表 SQL Server 计数记录

Question

我正在尝试计算两个日期（日历表）和工作日（agenda_user 表）之间每个用户的报告数量（report_user 表） ）。

这是我的表格的图表：

日历表：

 DATE         Year       Month
---------------------------------
2020-01-01      2020      1
2020-01-02      2020      1
2020-01-03      2020      1
2020-01-04      2020      1

AGENDA_USER 表：

ID_USER        DATE         Value
---------------------------------
1           2020-01-01       1
2           2020-01-01       1
1           2020-01-02       0
2           2020-01-02       1

用户表：

ID_USER        Name     
-------------------------
1              Jack
2              Robert

Report_Result 表：

ID_USER        Date          Result
-----------------------------------
1             2020-01-01      good
1             2020-01-01      good
2             2020-01-01      bad
2             2020-01-01      good
2             2020-01-02      good
2             2020-01-02      good

我正在尝试使用 SQL 查询查找结果

ID_USER        Date          Number of report     Day work      report/work
---------------------------------------------------------------------------
1             2020-01-01            2                1               2/1 = 2
2             2020-01-01            2                1                1
1             2020-01-02            0                0                0
2             2020-01-02            2                1                2

SELECT 
    REPORT_USER.ID_USER, 
    COUNT(ID_USER) AS result
FROM [DB].[dbo].REPORT_USER AS report,
JOIN [DB].[dbo].[USER] AS [USER]
     ON [USER].ID_USER = report.ID_USER
JOIN [DB].[dbo].AGENDA_USER AS agenda
     ON agenda.ID_USER = report.ID_USER
WHERE CAST(agenda.[Date] AS DATE) >= '2020-09-01'
    AND CAST(agenda.[Date] AS DATE) <= '2021-07-28'
    AND [USER].ID_user = 1167
GROUP BY 
    report.ID_VENDEUR;

Answer 1

我不完全确定我理解你的问题，但我认为我已经相当接近了，所以这是一个开始，指出我的无效假设，我们可以改进。更多数据，尤其是议程和报告中的数据，确实会有所帮助。下面是一个解释（加上代码中的注释）。

整个流程是生成您想要报告的日期/人员列表 (cteUserDays)，生成每个用户在每个日期生成的报告数量列表 (cteReps)，生成谁在哪些天工作的列表(cteWork)，然后使用 LEFT OUTER JOIN 将所有 3 个部分连接在一起，以便报告涵盖所有日期的所有工作人员。 编辑：添加 cteRepRaw ，其中 DATETIME 转换为 DATE 并过滤掉“坏”报告。分组和计数仍然发生在 cteReps 中，但加入 cteUserDays 不存在，因为如果有 NULL，它会加 1 来计数。

DECLARE @Cal TABLE (CalDate DATETIME, CalYear int, CalMonth int)
DECLARE @Agenda TABLE (UserID int, CalDate DATE, AgendaVal int)
DECLARE @User TABLE (UserID int, UserName nvarchar(50))
DECLARE @Reps TABLE (UserID int, CalDate DATETIME, RepResult nvarchar(50))

INSERT INTO @Cal(CalDate, CalYear, CalMonth)
VALUES ('2020-01-01', 2020, 1), ('2020-01-02', 2020, 1), ('2020-01-03', 2020, 1), ('2020-01-04', 2020, 1)

INSERT INTO @Agenda(UserID, CalDate, AgendaVal)
VALUES (1, '2020-01-01', 1), (2, '2020-01-01', 1), (1, '2020-01-02', 0), (2, '2020-01-02', 1)

INSERT INTO @User (UserID , UserName )
VALUES (1, 'Jack'), (2, 'Robert')

INSERT INTO @Reps (UserID , CalDate , RepResult )
VALUES (1, '2020-01-01', 'good'), (1, '2020-01-01', 'good')
    , (2, '2020-01-01', 'bad'), (2, '2020-01-01', 'good')
    , (2, '2020-01-02', 'good'), (2, '2020-01-02', 'good')


; with cteUserDays as (
    --First, you want zeros in your table where no reports are, so build a table for that
    SELECT CONVERT(DATE, D.CalDate) as CalDate --EDIT add CONVERT here
      , U.UserID FROM @Cal as D CROSS JOIN @User as U
    WHERE D.CalDate >= '2020-01-01' AND D.CalDate <= '2021-07-28'
    --EDIT Watch the <= date here, a DATE is < DATETIME with hours of the same day
), cteRepRaw as (--EDIT Add this CTE to convert DATETIME to DATE so we can group on it
    --Map the DateTime to a DATE type so we can group reports from any time of day
    SELECT R.UserID 
        , CONVERT(DATE, R.CalDate) as CalDate --EDIT add CONVERT here
        , R.RepResult
    FROM @Reps as R 
    WHERE R.RepResult='good' --EDIT Add this test to only count good ones
), cteReps as (
    --Get the sum of all reports for a given user on a given day, though some might be missing (fill 0)
    SELECT R.UserID , R.CalDate , COUNT(*) as Reports --SUM(COALESCE(R.RepResult, 0)) as Reports
    FROM cteRepRaw as R--cteUserDays as D 
        --Some days may have no reports for that worker, so use a LEFT OUTER JOIN
        --LEFT OUTER JOIN cteRepRaw as R on D.CalDate = R.CalDate AND D.UserID = R.UserID
    GROUP BY R.UserID , R.CalDate 
) , cteWork as (
    --Unclear what values "value" in Agenda can take, but assuming it's some kind of work 
    -- unit, like "hours worked" or "shifts" so add them up 
    SELECT A.UserID , A.CalDate, SUM(A.AgendaVal) as DayWork FROM @Agenda as A 
    WHERE A.CalDate >= '2020-01-01' AND A.CalDate <= '2021-07-28'
    GROUP BY A.CalDate, A.UserID 
) 
SELECT D.UserID , D.CalDate, COALESCE(R.Reports, 0) as Reports, W.DayWork 
    --NOTE: While it's probably a mistake to credit a report to a day a worker had
    --no shifts, it could happen and would throw an error so check
    , CASE WHEN W.DayWork > 0 THEN R.Reports / W.DayWork ELSE 0 END as RepPerWork
FROM cteUserDays as D 
    LEFT OUTER JOIN cteReps as R on D.CalDate=R.CalDate AND R.UserID = D.UserID 
    LEFT OUTER JOIN cteWork as W on D.UserID = W.UserID AND D.CalDate = W.CalDate 
ORDER BY CalDate , UserID 

首先，根据您的 OP“议程”中的 cmets 表示用户何时工作，您没有说明它的结构，所以我假设它可以在给定的一天为给定的人有多个条目（即4 小时轮班和 8 小时轮班），所以我将它们加起来得到总工作量（cteWork）。我还假设如果有人不工作，您就无法为他们提供报告。我对此进行了检查，但通常我希望您的数据验证器预先筛选出来。

其次，我假设每条记录有 1 个报告，给定用户每天可以有多个。您已经在给定的内容中找到了，但这对这个解决方案很重要，所以我重申一下，以防其他人稍后阅读。

第三，我假设您希望为所有用户报告所有天数，我通过在用户和天数 (cteUserDays) 之间生成交叉连接来保证这一点