Question

我正在研究ASP .Net上的Gridview。我在SQL Server中有一个表，其中包含名为“surname”和“Date”（DateTime）和持续时间的列。该表用于度假请求。如何构建SQL语句以查看每天将丢失多少人？关键是查询SELECT [Date], COUNT(DISTINCT surname) GROUP BY [Date]不会告诉我实际上8人将在9月2日失踪。例如，给出以下数据：

surname Date       Duration
------- ---------- ---------
Bertram 2011-09-01     3
Coulois 2011-09-01     5
LeBlanc 2011-09-01     6
Fosters 2011-09-01     3
Blanche 2011-09-01     2
Bertram 2011-09-02     6
Gillian 2011-09-02     4
Pikklar 2011-09-02     7
Thierry 2011-09-03     6
Selanne 2011-09-03     6

我想要以下结果：

Date  Count
----- -----
1 Sep     5
2 Sep     8    
3 Sep     10

任何想法如何处理它并使用这些数据生成gridview？感谢你的时间

Answer 1

您可以使用numbers table执行此操作。这里我使用master..spt_values。

declare @T table
(
  surname varchar(20),
  [Date] datetime,
  Duration int
)

insert into @T values
('Bertram', '2011-09-01',     3),
('Coulois', '2011-09-01',     5),
('LeBlanc', '2011-09-01',     6),
('Fosters', '2011-09-01',     3),
('Blanche', '2011-09-01',     2),
('Bertram', '2011-09-02',     6),
('Gillian', '2011-09-02',     4),
('Pikklar', '2011-09-02',     7),
('Thierry', '2011-09-03',     6),
('Selanne', '2011-09-03',     6)


select dateadd(day, N.number, [Date]) as [Date],
       count(*) as [Count]
from @T as T
  inner join master..spt_values as N
    on N.number between 0 and T.Duration
where N.type = 'P'    
group by dateadd(day, N.number, [Date])
order by dateadd(day, N.number, [Date])

结果：

Date                    Count
----------------------- -----------
2011-09-01 00:00:00.000 5
2011-09-02 00:00:00.000 8
2011-09-03 00:00:00.000 10
2011-09-04 00:00:00.000 9
2011-09-05 00:00:00.000 7
2011-09-06 00:00:00.000 7
2011-09-07 00:00:00.000 5
2011-09-08 00:00:00.000 4
2011-09-09 00:00:00.000 3

Answer 2

你可以试试这样的事情

WITH Dates AS
    (SELECT CAST('9/1/2011' AS DATE) AS [DATE]
    UNION SELECT '9/2/2011'
    UNION SELECT '9/3/2011'
    )

SELECT [DATE], SUM(OnVacation) AS COUNT
FROM
    (
    SELECT [DATE], 
        CASE WHEN [DATE] BETWEEN StartDate AND DATEADD(dd, Duration, StartDate) 
             THEN 1 ELSE 0 END AS OnVacation
    FROM Vacations
    CROSS JOIN Dates
    ) x
GROUP BY [DATE]
ORDER BY [DATE]

日期表将是一个包含您要查看的日期列表的表。它是此查询中的公用表表达式（CTE）。

这也假定开始日期是持续时间的第一天。

Answer 3

以下内容应该为您提供从预订的第一天开始到结束的最后一天结束的所有假期的细分（不仅仅是开始日期）。它还应报告零范围内的日期（如果存在）;

2011-09-01  5
2011-09-02  8
2011-09-03  10
2011-09-04  9
2011-09-05  7
2011-09-06  7
2011-09-07  5
2011-09-08  4
2011-09-09  3

代码计算出上次预订的日期，然后计算动态日期范围内每一天的所有预订

DECLARE @MaxDate date
SELECT  @MaxDate = max(dateAdd(day, duration, date)) 
FROM holiday;

WITH HolidayDates (holidayDate)
as
(
    SELECT  MIN(date) holidayDate 
    FROM    holiday 
    UNION ALL 
    SELECT  DateAdd(day, 1, holidayDate)
    FROM    holidayDates 
    WHERE   holidayDate <@MaxDate
)

SELECT      cast(hd.holidayDate as date) holidayDate
            , count(h.surname) PeopleOnHoliday
FROM        HolidayDates hd
LEFT JOIN   holiday h on hd.holidayDate between h.date AND dateAdd(day, duration, date) 
GROUP BY    hd.holidayDate
ORDER BY    hd.holidayDate

希望这会有所帮助...

Answer 4

我喜欢bobs回答Sql statement help。

这是您需要创建一个日期列的CTE，给出一个开始＆amp;结束日期。

DECLARE @start datetime = '2011-01-01'
DECLARE @end datetime = '2011-01-31'

; WITH Dates as
(
    SELECT @start as d
    UNION ALL 
    SELECT DATEADD(DAY, 1, d)FROM dates WHERE d < @end
)
SELECT * FROM Dates

如果你把它插入他的答案，并创建一个接受'start'和'end参数的存储过程，你应该得到答案。

Sql语句帮助

4 个答案: