获取ID日期和正负天数之间的日期

Question

我有以下示例数据：

CREATE TABLE tblDates
(
    ID int,
    Dates DATE
);

SELECT * FROM tblDates

INSERT INTO tblDates VALUES(1,'2019-12-01');
INSERT INTO tblDates VALUES(2,'2019-12-05');
INSERT INTO tblDates VALUES(3,'2019-12-02');
INSERT INTO tblDates VALUES(4,'2019-12-09');
INSERT INTO tblDates VALUES(5,'2019-12-11');

我在这里寻找 ID = 4 的 date 和正负第 1,2,....n 天之间的日期。

尝试 1：我尝试使用 UNION ALL。

SELECT Dates FROM tblDates WHERE ID = 4 
UNION ALL
SELECT DATEADD(day,1,Dates) FROM tblDates WHERE ID = 4;

当我寻找 50 天或更多天数差异时，此方法不适用。

尝试 2：

SELECT Dates FROM tblDates WHERE ID = 4 AND Dates between Dates AND DATEADD(day,1,Dates);

有单身约会。

尝试 3：

创建的函数：获取日期的函数

CREATE FUNCTION udf_GetDates(@MinDate DATE,@MaxDate DATE)
RETURNS TABLE
AS
RETURN
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;

查询：

SELECT f.* 
FROM udf_GetDates(t.Dates,DATEADD(day,1,t.Dates)) f        
INNER JOIN tblDates t ON f.[Date] = t.[Dates]
WHERE t.ID = 4

出现错误：

Msg 4104, Level 16, State 1, Line 2 The multi-part identifier "t.Dates" could not be bound. Msg 4104, Level 16, State 1, Line 2 The multi-part identifier "t.Dates" could not be bound.

预期输出：

给定：ID = 4 和 day=+1

Dates
----------- 
2019-12-09
2019-12-10

给定：ID = 4 和 day=+10

Dates
----------- 
2019-12-09
2019-12-10
2019-12-11
2019-12-12
2019-12-13
2019-12-14
2019-12-15
2019-12-16
2019-12-17
2019-12-18

给定：ID = 4 和 day=-5

Dates
----------  
2019-12-05
2019-12-06
2019-12-07
2019-12-08
2019-12-09

Answer 1

我尝试使用其他格式创建函数

create FUNCTION udf_GetDates(@MinDate DATE,@MaxDate DATE)
RETURNS @_result table (dt date)
AS
begin
    insert into @_result
    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;
    return
end

和select这样的结果

declare @_date date=(select Dates from tblDates where ID=4)
select *
from udf_GetDates(@_date,DATEADD(day,1,@_date))

我得到了你想要的结果

dt
2019-12-09
2019-12-10

Answer 2

试试这个查询

FIDDLE DEMO

函数

CREATE FUNCTION udf_GetDates (@StartDate DATE, @Range INT)
RETURNS TABLE
AS
RETURN
    SELECT  
        DATEADD(DAY, nbr - 1, @StartDate) myDate
    FROM    
        (SELECT    
             ROW_NUMBER() OVER (ORDER BY c.object_id) AS Nbr
         FROM      
             sys.columns c) nbrs
    WHERE   
        nbr - 1 <= @Range

查询用法#1：

SELECT f.myDate 
FROM udf_GetDates((SELECT dates FROM tblDates WHERE ID = 4), 2) f

查询用法#2：

SELECT t.*, P.*
FROM tblDates t 
OUTER APPLY udf_GetDates(t.Dates, 5) p
WHERE t.ID = 4

更新的答案：

下一个日期

CREATE FUNCTION udf_GetDates (@StartDate DATE, @Range INT)
RETURNS TABLE
AS
RETURN
    SELECT  
        DATEADD(DAY, nbr - 1, @StartDate) myDate
    FROM    
        (SELECT    
             ROW_NUMBER() OVER (ORDER BY c.object_id) AS Nbr
         FROM      
             sys.columns c) nbrs
    WHERE   
        nbr - 1 <= @Range

以前的日期

CREATE FUNCTION [udf_GetDates_Minuus] (@StartDate DATE, @Range INT)
RETURNS TABLE
AS
RETURN
    SELECT  
        DATEADD(DAY, -(nbr - 1), @StartDate) myDate
    FROM    
        (SELECT    
             ROW_NUMBER() OVER (ORDER BY c.object_id) AS Nbr
         FROM      
             sys.columns c) nbrs
    WHERE   
        nbr - 1 <= @Range

单个函数中的下一个和上一个日期

CREATE FUNCTION udf_GetDatesNextandPrevious(@StartDate DATE, @Range INT)
RETURNS TABLE
AS
RETURN
    SELECT  
        DATEADD(DAY, nbr - 1, @StartDate) myDate
    FROM    
        (SELECT    
             ROW_NUMBER() OVER (ORDER BY c.object_id) AS Nbr
         FROM      
             sys.columns c) nbrs
    WHERE   
        nbr - 1 <= @Range

    UNION

    SELECT  
        DATEADD(DAY, -(nbr - 1), @StartDate) myDate
    FROM    
        (SELECT    
             ROW_NUMBER() OVER (ORDER BY c.object_id) AS Nbr
         FROM      
             sys.columns c) nbrs
    WHERE   
        nbr - 1 <= @Range

Updated Fiddle

Answer 3

这 SQL 将 return 两个范围之间的正确日期。只需根据您的需要进行调整即可。

    DECLARE @From               DATETIME,    
            @To                 DATETIME

    SELECT  @From           =   '2019-11-13',
            @To             =   '2019-11-19'

    ;WITH   Numbers         AS
(
    SELECT  0               AS  Number
UNION ALL
    SELECT  Number + 1      AS  Number
    FROM    Numbers
    WHERE   Number          <   DATEDIFF(d, @From, @To)
)
    SELECT  DATEADD(d, 
            Number, @From)  AS Date
    FROM    Numbers

-- 结果

2019-11-13 00:00:00.000
2019-11-14 00:00:00.000
2019-11-15 00:00:00.000
2019-11-16 00:00:00.000
2019-11-17 00:00:00.000
2019-11-18 00:00:00.000
2019-11-19 00:00:00.000

Answer 4

在 sql 服务器中尝试此代码

DECLARE @selecteddate DATE 
DECLARE @day INT = 10 
DECLARE @id INT = 4; 
DECLARE @count INT = 0; 
DECLARE @table1 TABLE 
  ( 
     date_ DATETIME 
  ) 

SELECT @selecteddate = dates 
FROM   tbldates 
WHERE  id = @id; 

IF( @count <= @day ) 
  BEGIN 
        if(@day > 1)
        begin 
            set @day = @day - 1
        end
      WHILE @count <= @day 
        BEGIN 

            INSERT INTO @table1 
            VALUES      (Dateadd(day, @count, @selecteddate)) 

            SET @count = @count + 1 
        END 
  END 
ELSE 
  BEGIN 
      WHILE @count > @day 
        BEGIN 
            INSERT INTO @table1 
            VALUES      (Dateadd(day, @count, @selecteddate)) 

            SET @count = @count - 1 
        END 
  END 

SELECT * 
FROM   @table1 
ORDER  BY 1

获取ID日期和正负天数之间的日期

Get dates between date of ID and date of plus or minus days

tsql

sql-server

sql-server-2008-r2