获取ID日期到正负日期之间的日期

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');

[这里我正在寻找date的ID = 4和1,2，.... n天的正负之间的日期。

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;

[这种方法不适用于我寻找50天或更多天数差异的情况。

尝试2：

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

获得单个日期。

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：

消息4104，级别16，状态1，第2行，多部分标识符无法绑定“ t.Dates”。讯息4104，第16级，状态1，第2行多部分标识符“ t.Dates”无法绑定。

预期输出：

给出：ID = 4，天= + 1

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

给出：ID = 4，天= + 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

给出：ID = 4，天= -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

并选择这样的结果

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

我得到了您想要的结果

dt
2019-12-09
2019-12-10

Answer 2

此SQL将返回两个范围之间的正确日期。只需根据您的需求进行调整即可。

    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 3

尝试此查询

FIDDLE DEMO

功能

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 <= DATEDIFF(DAY, @StartDate, DATEADD(day,@Range,@StartDate))

Qurey用法1

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

Qurey用法2

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