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我有一个使用计数表插入指定月份的缺失行的查询。该查询测试一个月是否缺少一天，然后从列中获取不同的值（在该月），并将这些值和缺失的日期值插入表中。从不同的表中获取价值并在不同的查询中使用价值
该表用于监控工作人员的工作日期，时间以及他们工作的部分。问题是中的某一列如果工作人员移动部分，Distinct语句会为该特定日期创建两组行。
这是工作的是获得意外的结果查询，（@Zohar佩莱德帮助我做到这一点）
DECLARE @startdt DATETIME = '2016/6/1';
DECLARE @enddt DATETIME = '2016/7/1';
DECLARE @value nvarchar(50);
DECLARE @T as TABLE
(
Staff_ID int,
[Date] date,
[Year] int,
Mon int,
[Day] int,
First_Name varchar(10),
Last_Name varchar(10),
Section varchar(10),
Time_Worked datetime
)
INSERT INTO @T VALUES
(1001, '2016-06-01', 2016, 6, 1, 'Bill', 'Price', 'Level 1', '2016-06-01 8:30:00.000'),
(1001, '2016-06-02', 2016, 6, 2, 'Bill', 'Price', 'Level 1', '2016-06-02 8:30:00.000'),
(1001, '2016-06-03', 2016, 6, 3, 'Bill', 'Price', 'Level 1', '2016-06-03 8:30:00.000'),
(1001, '2016-06-04', 2016, 6, 4, 'Bill', 'Price', 'Level 1', '2016-06-04 8:30:00.000'),
(1001, '2016-06-05', 2016, 6, 5, 'Bill', 'Price', 'Level 1', '2016-06-05 8:30:00.000'),
(1001, '2016-06-06', 2016, 6, 6, 'Bill', 'Price', 'Level 1', '2016-06-06 8:30:00.000'),
(1001, '2016-06-07', 2016, 6, 7, 'Bill', 'Price', 'Level 2', '2016-06-07 8:30:00.000'),-- Different section
(1001, '2016-07-05', 2016, 7, 5, 'Bill', 'Price', 'Level 2', '2016-07-5 8:30:00.000'),
(1002, '2016-06-01', 2016, 6, 1, 'Mary', 'Somers', 'Level 1', '2016-06-01 8:30:00.000'),
(1002, '2016-06-05', 2016, 6, 5, 'Mary', 'Somers', 'Level 1', '2016-06-05 8:30:00.000'),
(1002, '2016-06-08', 2016, 6, 8, 'Mary', 'Somers', 'Level 1', '2016-06-08 8:30:00.000'),
(1003, '2016-06-03', 2016, 6, 3, 'Mark', 'Jones', 'Level 1', '2016-06-03 8:30:00.000'),
(1003, '2016-06-04', 2016, 6, 4, 'Mark', 'Jones', 'Level 1', '2016-06-05 8:30:00.000')
[email protected] = SELECT Section_Data FROM Staff_Manager.dbo.Staff_Data_TBL WHERE Staff_No = 1001
SET NOCOUNT ON;
IF object_id('dbo.Tally') is not null drop table dbo.tally
SELECT TOP 30000 IDENTITY(int,1,1) as ID
INTO dbo.Tally FROM master.dbo.SysColumns
ALTER table dbo.Tally
add constraint PK_ID primary key clustered(ID)
; WITH Calendar AS
(
SELECT dateadd(DD, ID-1, @startdt) as [Date]
FROM dbo.Tally
WHERE dateadd(DD, ID-1, @startdt) < @enddt
)
INSERT INTO @T(Staff_ID, [Date], [Year], Mon, [Day], First_Name, Last_Name, Section)
SELECT DISTINCT Staff_ID, C.[Date], Year(C.[Date]), MONTH(C.[Date]), DAY(C.[Date]), First_Name, Last_Name, Section
FROM @T T
CROSS APPLY
(
SELECT Cal.[Date]
FROM Calendar Cal
WHERE MONTH(Cal.[Date]) = MONTH(T.[Date])
AND YEAR(Cal.[Date]) = YEAR(T.[Date])
AND NOT EXISTS
(
SELECT 1
FROM @T T2
WHERE T.Staff_ID = T2.Staff_ID
AND T2.[Date] = Cal.[Date]
)
) C
SELECT Staff_ID, [Date], [Year], Mon, [Day], First_Name, Last_Name, Section, Time_Worked
FROM @T
ORDER BY Staff_ID, [Date]
这是结果，正如你可以看到它会为工作人员Bill Price两天由于2016-06-08由于他将Section从Level 1更改为Level 2。因为Distinct声明将从Section列获得两个值。
（运行代码，你会得到所有的结果，这只是展示一下我的意思是SNIPPIT。）
Staff_ID Date Year Mon Day First_Name Last_Name Section Time_Worked
1001 2016-06-01 2016 6 1 Bill Price Level 1 2016-06-01 08:30:00.000
1001 2016-06-02 2016 6 2 Bill Price Level 1 2016-06-02 08:30:00.000
1001 2016-06-03 2016 6 3 Bill Price Level 1 2016-06-03 08:30:00.000
1001 2016-06-04 2016 6 4 Bill Price Level 1 2016-06-04 08:30:00.000
1001 2016-06-05 2016 6 5 Bill Price Level 1 2016-06-05 08:30:00.000
1001 2016-06-06 2016 6 6 Bill Price Level 1 2016-06-06 08:30:00.000
1001 2016-06-07 2016 6 7 Bill Price Level 2 2016-06-07 08:30:00.000
1001 2016-06-08 2016 6 8 Bill Price Level 1 NULL
1001 2016-06-08 2016 6 8 Bill Price Level 2 NULL
1001 2016-06-09 2016 6 9 Bill Price Level 1 NULL
1001 2016-06-09 2016 6 9 Bill Price Level 2 NULL
1001 2016-06-10 2016 6 10 Bill Price Level 1 NULL
1001 2016-06-10 2016 6 10 Bill Price Level 2 NULL
1001 2016-06-11 2016 6 11 Bill Price Level 1 NULL
1001 2016-06-11 2016 6 11 Bill Price Level 2 NULL
1001 2016-06-12 2016 6 12 Bill Price Level 1 NULL
1001 2016-06-12 2016 6 12 Bill Price Level 2 NULL
1001 2016-06-13 2016 6 13 Bill Price Level 1 NULL
1001 2016-06-13 2016 6 13 Bill Price Level 2 NULL
所以我保存的工作人员目前部分中的另一表，看代码下面。
DECLARE @value nvarchar(50);
DECLARE @T3 as TABLE
(
Staff_ID int,
First_Name varchar(10),
Last_Name varchar(10),
Section varchar(10)
)
INSERT INTO @T3 VALUES
(1001, 'Bill', 'Price', 'Level 2'),
(1002, 'Mary', 'Somers', 'Level 1'),
(1003, 'Mark', 'Jones', 'Level 1')
SELECT Section FROM @T3 WHERE Staff_ID = 1001
结果如下，得到该职员的数据为Section。现在
Level 2
，如果我能得到这个值，并把它作为在Section列，将解决我的问题的默认值。
此代码显然是不正确，但而不是获取Section列的distinct值，替换与来自其他表@T3， 像这样的东西的价值是价值，
@value = SELECT Section FROM @T3 WHERE Staff_ID = 1001
再有值插入到新行中。
DECLARE @value nvarchar(50);
@value = SELECT Section FROM @T3 WHERE Staff_ID = 1001
INSERT INTO @T(Staff_ID, [Date], [Year], Mon, [Day], First_Name, Last_Name, Section)
SELECT DISTINCT Staff_ID, C.[Date], Year(C.[Date]), MONTH(C.[Date]), DAY(C.[Date]), First_Name, Last_Name, @value
FROM @T T
CROSS APPLY
(
SELECT Cal.[Date]
FROM Calendar Cal
WHERE MONTH(Cal.[Date]) = MONTH(T.[Date])
AND YEAR(Cal.[Date]) = YEAR(T.[Date])
AND Section = @value
AND NOT EXISTS
(
SELECT 1
FROM @T T2
WHERE T.Staff_ID = T2.Staff_ID
AND T2.[Date] = Cal.[Date]
)
) C
任何帮助将不胜感激。
一名员工会在一天多于一节的工作中工作吗？ – Squirrel
@Squirrel，不，他们可能会在月中改变，就是这样。 – KyloRen
是否有任何你存储员工变更部分的日期？ – Squirrel