具有多个日期字段的每个日期的 SQL COUNT答案

【问题标题】：SQL COUNT by each date with multiple date fields具有多个日期字段的每个日期的 SQL COUNT
【发布时间】：2019-05-18 14:35:35
【问题描述】：

我有下表和查询，但没有给出正确的结果。

JOB 表有 Open、Finish 和 Closed 日期。

现在我需要按每个日期和位置提取所选日期组之间的打开、完成和关闭作业的计数。

请帮助我得到如下预期结果中的结果

+-------+-----------+------------+-----------+----------+
| JOB_id| DateOpen  | DateFinish | DateClose | Location |
+-------+-----------+------------+-----------+----------+
|   100 | 16-Dec-18 | 18-Dec-18  | 19-Dec-18 | A        |
|   101 | 16-Dec-18 | 18-Dec-18  | 19-Dec-18 | A        |
|   102 | 17-Dec-18 | 19-Dec-18  | 20-Dec-18 | C        |
|   103 | 10-Dec-18 | 11-Dec-18  | 16-Dec-18 | D        |
|   104 | 17-Dec-18 | 19-Dec-18  | 18-Dec-18 | E        |
+-------+-----------+------------+-----------+----------+

查询：

SELECT count(DateOpen) as Opened,
       count(DateFinish) as Finised,
      count(DateClose) as Closed,
      (DateOpen) as Date 
FROM JOBS
WHERE DateOpen BETWEEN '12/16/2018' AND DATEADD(DAY, 1, '12/17/2018')
group by DateOpen

预期结果：

+-----------+------+----------+--------+----------+
|   Date    | Open | Finished | Closed | Location |
+-----------+------+----------+--------+----------+
| 16-Dec-18 |    2 |        0 |      0 | A        |
| 16-Dec-18 |    0 |        0 |      1 | D        |
| 17-Dec-18 |    1 |        0 |      0 | C        |
| 17-Dec-18 |    1 |        0 |      0 | E        |
+-----------+------+----------+--------+----------+

【问题讨论】：

当前输出是多少？
能否粘贴查询生成数据
如何通过状态区分 JOB_id 101 和 102 ？两者在 DateFinished 和 DateClosed 中都有相同的日期
为什么您的日期列还不是 DATE 类型？
@SalmanA BETWEEN

标签： sql sql-server datetime sql-server-2012 sql-server-2014

【解决方案1】：

您可以将所有打开、完成和关闭日期放在一个列中，然后将其加入您的工作表：

DECLARE @date1 AS DATE = '2018-12-16';
DECLARE @date2 AS DATE = '2018-12-17';

WITH dates(date) AS (
    SELECT DateOpen FROM jobs
    UNION
    SELECT DateFinish FROM jobs
    UNION
    SELECT DateClose FROM jobs
)
SELECT dates.date
     , Location
     , COUNT(CASE WHEN dates.date = DateOpen THEN 1 END) AS Opened
     , COUNT(CASE WHEN dates.date = DateFinish THEN 1 END) AS Finished
     , COUNT(CASE WHEN dates.date = DateClose THEN 1 END) AS Closed
FROM dates
LEFT JOIN jobs ON dates.date IN (DateOpen, DateFinish, DateClose)
WHERE dates.date BETWEEN @date1 AND @date2
GROUP BY dates.date
       , Location

结果：

| date       | Location | Opened | Finished | Closed |
|------------|----------|--------|----------|--------|
| 16/12/2018 | A        | 2      | 0        | 0      |
| 16/12/2018 | D        | 0      | 0        | 1      |
| 17/12/2018 | C        | 1      | 0        | 0      |
| 17/12/2018 | E        | 1      | 0        | 0      |

Demo on DB Fiddle

【讨论】：

您的查询正确提取了数据，但加载时间很长，因为我有一百万行
你有在表上创建的索引吗？
我感觉发生了隐式转换

【解决方案2】：

您可以使用以下查询来获取所需的结果集。

select coalesce(t1.date, t2.date, t3.date) as date, t1.Opened, t2.Finished, t3.Closed,
coalesce(t1.location, t2.location, t3.location) as location
from
(SELECT Convert(date,DateOpen) as Date ,count(JobID) as Opened, location      
FROM JOBS
WHERE DateOpen BETWEEN '12/16/2018' AND DATEADD(DAY, 1, '12/17/2018')
group by Convert(date,DateOpen), location 
) t1
Full join 
(SELECT Convert(date,DateFinish) as Date ,count(JobID) as Finished, location      
FROM JOBS
WHERE DateFinish BETWEEN '12/16/2018' AND DATEADD(DAY, 1, '12/17/2018')
group by Convert(date,DateFinish), location 
) t2  ON(t1.date = t2.date and t1.location = t2.location)
FULL JOIN
(SELECT Convert(date,DateClose) as Date ,count(JobID) as Closed, location      
FROM JOBS
WHERE DateClose BETWEEN '12/16/2018' AND DATEADD(DAY, 1, '12/17/2018')
group by Convert(date,DateClose), location 
) t3  ON(t2.date = t3.date and t2.location = t3.location)

【讨论】：

错误：从字符串转换日期和/或时间时转换失败。

【解决方案3】：

首先，我建议您停止将日期存储为文本并使用正确的数据类型。

要执行您需要的操作，请通过选择每个日期列和位置来列出所有日期和位置，并将它们合并为一个结果（第一个 cte - allDates）。然后我们需要一个不同的列表（第二个 cte - 聚合）来选择，并计算表中与当前日期/位置值匹配的行数。

这是整个解决方案：

declare @JOBS table(JOB_ID int, DateOpen varchar(10), DateFinish varchar(10), DateClose varchar(10), Location varchar(5))
insert into @JOBS values
(   100, '16-Dec-18', '18-Dec-18', '19-Dec-18', 'A'),
(   101, '16-Dec-18', '18-Dec-18', '19-Dec-18', 'A'),
(   102, '17-Dec-18', '19-Dec-18', '20-Dec-18', 'C'),
(   103, '10-Dec-18', '11-Dec-18', '16-Dec-18', 'D'),
(   104, '17-Dec-18', '19-Dec-18', '18-Dec-18', 'E')

;with allDates as (
select convert(date, DateOpen) as [Date], Location from @JOBS
union
select convert(date, DateFinish), Location from @JOBS
union
select convert(date, DateClose), Location from @JOBS
),
aggregated as (
    select [Date], Location
    from allDates
    group by [Date], Location
)
select
    a.Date
    , (select count(*) from @JOBS where a.[Date] = DateOpen and a.Location = Location) [Open]
    , (select count(*) from @JOBS where a.[Date] = DateFinish and a.Location = Location) Finished
    , (select count(*) from @JOBS where a.[Date] = DateClose and a.Location = Location) Closed
    , Location
from aggregated a
where a.Date between '20181216' and '20181217'

【讨论】：

那你为什么将它们转换为日期，例如group by Convert(date,DateOpen)?

【解决方案4】：

您可以使用带有 sum 的 case 语句，

    SELECT  Convert(date,DateOpen) as Date ,      
           sum(case when DateOpen =DateOpen then 1 else 0 end) as Opened,
           sum(case when DateFinish=DateOpen then 1 else 0 end) as Finised,
           sum(case when DateClose=DateOpen then 1 else 0 end) as Closed,
           Location

    FROM JOBS
    WHERE DateOpen BETWEEN '12/16/2018' AND DATEADD(DAY, 1, '12/17/2018')
    group by Convert(date,DateOpen),Location

    UNION

    SELECT  Convert(date,DateClose) as Date ,      
           sum(case when DateOpen =DateClose then 1 else 0 end) as Opened,
           sum(case when DateFinish=DateClose then 1 else 0 end) as Finised,
           sum(case when DateClose=DateClose then 1 else 0 end) as Closed,
           Location

    FROM JOBS
    WHERE DateClose BETWEEN '12/16/2018' AND DATEADD(DAY, 1, '12/16/2018')
    group by Convert(date,DateClose),Location

【讨论】：

【解决方案5】：

DECLARE @startDate DATETIME = '12/16/2018'
DECLARE @endDate DATETIME = '12/17/2018'

SELECT 
  count(CASE when DateOpen BETWEEN @startDate AND @endDate THEN 1 end) as Opened,
  count(CASE when DateFinish BETWEEN @startDate AND @endDate THEN 1 end) as Finised,
  count(CASE when DateClose BETWEEN @startDate AND @endDate THEN 1 end) as Closed,
  DateOpen as Date,
  Location
FROM JOBS
WHERE DateOpen BETWEEN @startDate AND @endDate
group by DateOpen, Location

【讨论】：

【解决方案6】：

您可以尝试以下代码，使用 SUM 函数对输入日期的进程状态求和：

SELECT Convert(date1,DateOpen) as Date,
       sum(case
            when DateOpen = Convert(date1,DateOpen) then 1
            else 0
           end) as Open,
       sum(case
            when DateFinish = Convert(date1,DateOpen) then 1
            else 0
           end) as Finished,
      sum(case
            when DateClose = Convert(date1,DateOpen) then 1
            else 0
           end) as Closed,
      Location
FROM JOBS
group by Location
UNION ALL  
SELECT Convert(date2,DateOpen) as Date,
       sum(case
            when DateOpen = Convert(date2,DateOpen) then 1
            else 0
           end) as Open,
       sum(case
            when DateFinish = Convert(date2,DateOpen) then 1
            else 0
           end) as Finished,
      sum(case
            when DateClose = Convert(date2,DateOpen) then 1
            else 0
           end) as Closed,
      Location
FROM JOBS
group by Location;

已编辑：date1 和 date2 是输入参数。

【讨论】：

【解决方案7】：

你可以试试用 unpivot 看看性能如何

with cols_to_rows
  as (
      select *
        from t
      unpivot(col for val in (dateopen,datefinish,dateclose))m 
      )
select col
       ,location 
       ,count(case when val='dateopen' then 1 end)   as open1
       ,count(case when val='datefinish' then 1 end) as finish
       ,count(case when val='dateclose' then 1 end)  as close1
  from cols_to_rows
where col between cast('2018-12-16' as date)
  and cast('2018-12-17' as date)
group by col,location,val
order by col,location


| 10/12/2018 00:00:00 | D | 1 | 0 | 0 |
| 11/12/2018 00:00:00 | D | 0 | 1 | 0 |
| 16/12/2018 00:00:00 | A | 2 | 0 | 0 |
| 16/12/2018 00:00:00 | D | 0 | 0 | 1 |
| 17/12/2018 00:00:00 | C | 1 | 0 | 0 |
| 17/12/2018 00:00:00 | E | 1 | 0 | 0 |
| 18/12/2018 00:00:00 | A | 0 | 2 | 0 |
| 18/12/2018 00:00:00 | E | 0 | 0 | 1 |
| 19/12/2018 00:00:00 | A | 0 | 0 | 2 |
| 19/12/2018 00:00:00 | C | 0 | 1 | 0 |
| 19/12/2018 00:00:00 | E | 0 | 1 | 0 |
| 20/12/2018 00:00:00 | C | 0 | 0 | 1 |

这是一个 dbfiddle 链接

https://dbfiddle.uk/?rdbms=sqlserver_2012&fiddle=1ca0c0180a4d31d6e112e9f3b1b99715

【讨论】：

【解决方案8】：

试试这个：

    DECLARE @MinDate DATE = '12/16/2018',
            @MaxDate DATE = '12/17/2018'


    DECLARE @DateTable TABLE (DateOpen DATETIME)

    INSERT INTO @DateTable
    SELECT  TOP (DATEDIFF(DAY, @MinDate, @MaxDate) + 1)
            Date = DATEADD(DAY, ROW_NUMBER() OVER(ORDER BY a.object_id) - 1, @MinDate)
    FROM    sys.all_objects a
            CROSS JOIN sys.all_objects b;

    ;with cte
    As
    (
        SELECT DISTINCT d.DateOpen, 
        SUM(CASE WHEN j.DateOpen =d.DateOpen THEN 1 ELSE 0 END) OVER(partition by d.DateOpen,Location) As [Open]
        ,SUM(CASE WHEN j.DateFinish =d.DateOpen THEN 1 ELSE 0 END) OVER(partition by d.DateOpen,Location) As [Finished]
        ,SUM(CASE WHEN j.DateClose =d.DateOpen THEN 1 ELSE 0 END) OVER(partition by d.DateOpen,Location) As [Closed]
        ,Location
        FROM @DateTable d
        CROSS JOIN JOBS j   
    )

    Select * from cte where [Open]>0 or Finished>0 or Closed>0
    Order by DateOpen,Location

【讨论】：

【解决方案9】：

试试这个

WITH CTE AS
(
SELECT *
FROM
(
SELECT Location
FROM T
GROUP BY Location
) L CROSS APPLY
(
  SELECT DateOpen
  FROM T
  WHERE DateOpen BETWEEN '2018-12-16' AND '2018-12-18'
) D
GROUP BY Location, DateOpen
),
 F AS
(
SELECT *,
       (SELECT COUNT(1) FROM T WHERE Location = CTE.Location AND DateOpen = CTE.DateOpen) [Open],
       (SELECT COUNT(1) FROM T WHERE Location = CTE.Location AND DateFinish = CTE.DateOpen)[Finish],
       (SELECT COUNT(1) FROM T WHERE Location = CTE.Location AND DateClose = CTE.DateOpen) [Close]
FROM CTE
)
SELECT DateOpen,
       [Open],
       [Finish],
       [Close],
       Location
FROM F
WHERE [Open] > 0
      OR
      [Finish] > 0
      OR
      [Close] > 0
ORDER BY DateOpen

返回：

+---------------------+------+--------+-------+----------+
|      DateOpen       | Open | Finish | Close | Location |
+---------------------+------+--------+-------+----------+
| 16/12/2018 00:00:00 |    2 |      0 |     0 | A        |
| 16/12/2018 00:00:00 |    0 |      0 |     1 | D        |
| 17/12/2018 00:00:00 |    1 |      0 |     0 | C        |
| 17/12/2018 00:00:00 |    1 |      0 |     0 | E        |
+---------------------+------+--------+-------+----------+

Demo

【讨论】：