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我有一個使用計數表插入指定月份的缺失行的查詢。該查詢測試一個月是否缺少一天，然後從列中獲取不同的值（在該月），並將這些值和缺失的日期值插入表中。從不同的表中獲取價值並在不同的查詢中使用價值
該表用於監控工作人員的工作日期，時間以及他們工作的部分。問題是中的某一列如果工作人員移動部分，Distinct語句會爲該特定日期創建兩組行。
這是工作的是獲得意外的結果查詢，（@Zohar佩萊德幫助我做到這一點）
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]
這是結果，正如你可以看到它會爲工作人員Bill Price兩天由於2016-06-08由於他將Section從Level 1更改爲Level 2。因爲Distinct聲明將從Section列獲得兩個值。
（運行代碼，你會得到所有的結果，這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員目前部分中的另一表，看代碼下面。
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
結果如下，得到該職員的數據爲Section。現在
Level 2
，如果我能得到這個值，並把它作爲在Section列，將解決我的問題的默認值。
此代碼顯然是不正確，但而不是獲取Section列的distinct值，替換與來自其他表@T3， 像這樣的東西的價值是價值，
@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
任何幫助將不勝感激。
一名員工會在一天多於一節的工作中工作嗎？ – Squirrel
@Squirrel，不，他們可能會在月中改變，就是這樣。 – KyloRen
是否有任何你存儲員工變更部分的日期？ – Squirrel