我有下表和没有给出正确结果的查询。
JOB表具有打开，结束和关闭日期。
现在，我需要按每个日期和位置拉动所选日期组之间的未完成，已完成和已关闭作业的数量。
请帮助我获得低于预期结果的结果
+-------+-----------+------------+-----------+----------+
| 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 |
+-----------+------+----------+--------+----------+
答案 0 :(得分：1)
您可以将所有开放日期，结束日期和结束日期放在一列中，并使其与工作表一起加入：
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 |
答案 1 :(得分：0)
您可以在下面的查询中使用所需的结果集。
tf.train.Saver()答案 2 :(得分：0)
首先，我建议您停止将日期存储为文本，并使用适当的数据类型。
要执行所需的操作，请选择每个日期列和位置，并将所有日期和位置合并为一个结果（第一个cte-allDates），以列出所有日期和位置。然后，我们需要一个不同的列表（第二个cte-聚合的）进行选择，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：
.humanize()答案 3 :(得分：0)
您可以将case语句与sum一起使用
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
答案 4 :(得分：0)
您可以使用SUM函数尝试以下代码，该函数求和输入日期的进程状态：
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。
答案 5 :(得分：0)
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
答案 6 :(得分：0)
您可以尝试使用unpivot来查看效果如何
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dbfiddle链接
https://dbfiddle.uk/?rdbms=sqlserver_2012&fiddle=1ca0c0180a4d31d6e112e9f3b1b99715
答案 7 :(得分：0)
尝试一下：
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
答案 8 :(得分：0)
尝试这个
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 |
+---------------------+------+--------+-------+----------+