从整数输入重建日期的最佳方法是什么?


8

我有很多财务报告,我们希望能够将两个输入(年份和季度)作为变量传递给他们。

我这样做,但是我真的不喜欢它:

    declare @quarter int,
    @year int,
    @date date

    set @quarter = 4
    set @year = 2018


    set @date = cast(@year as varchar(4)) + '-01-01'
    set @date = dateadd(quarter, @quarter - 1, @date)


    print @date

问题从整数输入重建日期的最佳方法是什么?

预期结果:

    2018-10-01

Answers:


5

您可以在SQL Server中创建日期维度或日历表并对其进行查询

--demo setup 
drop table if exists #dim
DECLARE @StartDate DATE = '20000101', @NumberOfYears INT = 30;

-- prevent set or regional settings from interfering with 
-- interpretation of dates / literals

SET DATEFIRST 7;
SET DATEFORMAT mdy;
SET LANGUAGE US_ENGLISH;

DECLARE @CutoffDate DATE = DATEADD(YEAR, @NumberOfYears, @StartDate);

-- this is just a holding table for intermediate calculations:

CREATE TABLE #dim
(
  [date]       DATE PRIMARY KEY, 
  [day]        AS DATEPART(DAY,      [date]),
  [month]      AS DATEPART(MONTH,    [date]),
  FirstOfMonth AS CONVERT(DATE, DATEADD(MONTH, DATEDIFF(MONTH, 0, [date]), 0)),
  [MonthName]  AS DATENAME(MONTH,    [date]),
  [week]       AS DATEPART(WEEK,     [date]),
  [ISOweek]    AS DATEPART(ISO_WEEK, [date]),
  [DayOfWeek]  AS DATEPART(WEEKDAY,  [date]),
  [quarter]    AS DATEPART(QUARTER,  [date]),
  [year]       AS DATEPART(YEAR,     [date]),
  FirstOfYear  AS CONVERT(DATE, DATEADD(YEAR,  DATEDIFF(YEAR,  0, [date]), 0)),
  Style112     AS CONVERT(CHAR(8),   [date], 112),
  Style101     AS CONVERT(CHAR(10),  [date], 101)
);

-- use the catalog views to generate as many rows as we need

INSERT #dim([date]) 
SELECT d
FROM
(
  SELECT d = DATEADD(DAY, rn - 1, @StartDate)
  FROM 
  (
    SELECT TOP (DATEDIFF(DAY, @StartDate, @CutoffDate)) 
      rn = ROW_NUMBER() OVER (ORDER BY s1.[object_id])
    FROM sys.all_objects AS s1
    CROSS JOIN sys.all_objects AS s2
    -- on my system this would support > 5 million days
    ORDER BY s1.[object_id]
  ) AS x
) AS y;

drop table if exists dbo.DateDimension

CREATE TABLE dbo.DateDimension
(
  --DateKey           INT         NOT NULL PRIMARY KEY,
  [Date]              DATE        NOT NULL,
  [Day]               TINYINT     NOT NULL,
  DaySuffix           CHAR(2)     NOT NULL,
  [Weekday]           TINYINT     NOT NULL,
  WeekDayName         VARCHAR(10) NOT NULL,
  IsWeekend           BIT         NOT NULL,
  IsHoliday           BIT         NOT NULL,
  HolidayText         VARCHAR(64) SPARSE,
  DOWInMonth          TINYINT     NOT NULL,
  [DayOfYear]         SMALLINT    NOT NULL,
  WeekOfMonth         TINYINT     NOT NULL,
  WeekOfYear          TINYINT     NOT NULL,
  ISOWeekOfYear       TINYINT     NOT NULL,
  [Month]             TINYINT     NOT NULL,
  [MonthName]         VARCHAR(10) NOT NULL,
  [Quarter]           TINYINT     NOT NULL,
  QuarterName         VARCHAR(6)  NOT NULL,
  [Year]              INT         NOT NULL,
  MMYYYY              CHAR(6)     NOT NULL,
  MonthYear           CHAR(7)     NOT NULL,
  FirstDayOfMonth     DATE        NOT NULL,
  LastDayOfMonth      DATE        NOT NULL,
  FirstDayOfQuarter   DATE        NOT NULL,
  LastDayOfQuarter    DATE        NOT NULL,
  FirstDayOfYear      DATE        NOT NULL,
  LastDayOfYear       DATE        NOT NULL,
  FirstDayOfNextMonth DATE        NOT NULL,
  FirstDayOfNextYear  DATE        NOT NULL
);
INSERT dbo.DateDimension WITH (TABLOCKX)
SELECT
  --DateKey     = CONVERT(INT, Style112),
  [Date]        = [date],
  [Day]         = CONVERT(TINYINT, [day]),
  DaySuffix     = CONVERT(CHAR(2), CASE WHEN [day] / 10 = 1 THEN 'th' ELSE 
                  CASE RIGHT([day], 1) WHEN '1' THEN 'st' WHEN '2' THEN 'nd' 
                  WHEN '3' THEN 'rd' ELSE 'th' END END),
  [Weekday]     = CONVERT(TINYINT, [DayOfWeek]),
  [WeekDayName] = CONVERT(VARCHAR(10), DATENAME(WEEKDAY, [date])),
  [IsWeekend]   = CONVERT(BIT, CASE WHEN [DayOfWeek] IN (1,7) THEN 1 ELSE 0 END),
  [IsHoliday]   = CONVERT(BIT, 0),
  HolidayText   = CONVERT(VARCHAR(64), NULL),
  [DOWInMonth]  = CONVERT(TINYINT, ROW_NUMBER() OVER 
                  (PARTITION BY FirstOfMonth, [DayOfWeek] ORDER BY [date])),
  [DayOfYear]   = CONVERT(SMALLINT, DATEPART(DAYOFYEAR, [date])),
  WeekOfMonth   = CONVERT(TINYINT, DENSE_RANK() OVER 
                  (PARTITION BY [year], [month] ORDER BY [week])),
  WeekOfYear    = CONVERT(TINYINT, [week]),
  ISOWeekOfYear = CONVERT(TINYINT, ISOWeek),
  [Month]       = CONVERT(TINYINT, [month]),
  [MonthName]   = CONVERT(VARCHAR(10), [MonthName]),
  [Quarter]     = CONVERT(TINYINT, [quarter]),
  QuarterName   = CONVERT(VARCHAR(6), CASE [quarter] WHEN 1 THEN 'First' 
                  WHEN 2 THEN 'Second' WHEN 3 THEN 'Third' WHEN 4 THEN 'Fourth' END), 
  [Year]        = [year],
  MMYYYY        = CONVERT(CHAR(6), LEFT(Style101, 2)    + LEFT(Style112, 4)),
  MonthYear     = CONVERT(CHAR(7), LEFT([MonthName], 3) + LEFT(Style112, 4)),
  FirstDayOfMonth     = FirstOfMonth,
  LastDayOfMonth      = MAX([date]) OVER (PARTITION BY [year], [month]),
  FirstDayOfQuarter   = MIN([date]) OVER (PARTITION BY [year], [quarter]),
  LastDayOfQuarter    = MAX([date]) OVER (PARTITION BY [year], [quarter]),
  FirstDayOfYear      = FirstOfYear,
  LastDayOfYear       = MAX([date]) OVER (PARTITION BY [year]),
  FirstDayOfNextMonth = DATEADD(MONTH, 1, FirstOfMonth),
  FirstDayOfNextYear  = DATEADD(YEAR,  1, FirstOfYear)
FROM #dim
OPTION (MAXDOP 1);

--solution
SELECT min(Date)
  FROM [Test].[dbo].[DateDimension]
  where [year] = 2018 and [Quarter]=4

| Date       |
|------------|
| 2018-10-01 |

1
这是正确的方法-通过这种方式我们可以为日期部分建立索引,因此,与将函数放在where子句中的列相比,加入DateDimension的速度将快万千倍。
詹姆斯(James)

12

怎么样

declare @quarter int = 4
declare @year int = 2018

select datefromparts(@year,(@quarter-1)*3+1,1)

或者,如果您仍在使用SQL 2008:

select dateadd(month,(@quarter-1)*3,dateadd(year, @year-2018,'20180101'))

10

我建议不要使用“-”或“ /”之类的日期分隔符,它取决于区域设置,请使用YYYYMMDD格式。

declare @quarter int,
    @year int,
    @date date

    set @quarter = 4
    set @year = 2018


    set @date = cast(@year as varchar(4)) + '0101'
    set @date = dateadd(quarter, 1 - 1, @date)
    print @date

    set @date = cast(@year as varchar(4)) + '0101'
    set @date = dateadd(quarter, 2 - 1, @date)
    print @date

    set @date = cast(@year as varchar(4)) + '0101'
    set @date = dateadd(quarter, 3 - 1, @date)
    print @date

    set @date = cast(@year as varchar(4)) + '0101'
    set @date = dateadd(quarter, 4 - 1, @date)
    print @date
2018-01-01
2018-04-01
2018-07-01
2018-10-01

db <> 在这里拨弄


3
我的理解是,它'2018-10-01'独立于区域设置,并且始终会被SQL Server正确解析。那不是真的吗
a_horse_with_no_name

@a_horse_with_no_name应该是ISO 8601,但至少在我使用SQL Server混合语言安装(英语,西班牙语)的公司中,我结束了使用不太明确的格式YYYYMMDD HH:MM:SS以避免日期转换问题。
McNets

5
@a_horse_with_no_name-对于较新的时间数据类型(例如datedatetime2),这是正确的,但对于旧datetime类型则不正确。因此,作为提问者被铸造date的格式是罚款在这种情况下
马丁·史密斯

1
@MartinSmith:感谢您的澄清
a_horse_with_no_name19年

“ 2001-08-12”并非总是以您期望的方式进行解释。我从不知道如何做,但是在我们的开发环境中,它将解释为yyyy-dd-mm,并抛出“ 2001-08-13”超出范围的错误。我尝试完全避免从字符串到日期的转换,并且在必要时确保使用月份名称-“ 2001年8月12日”是明确的。
IanF1

4

我会完全避免使用字符串,而是将日期算术与已知(甚至未知!)零纪元结合使用。

DECLARE @epoch DATE = CONVERT(DATETIME, 0);
/* for some reason SQL Server let's you cast int to datetime but not to date, the above casts via datetime (second cast implicit) */

SET @date = DATEADD(MONTH, (@quarter-1)*3, DATEADD(YEAR, @year - YEAR(@epoch), @epoch));

这避免了字符串到日期的比较,这比较麻烦,依赖于文化并且昂贵。

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