将日期范围转换为间隔描述

Mic*_*een 11 database-agnostic date-format

最近的一个项目要求报告资源何时被完全消耗。除了用尽日历日期,我还被要求以类似英语的格式显示剩余时间,比如“1 年零 3 个月”。

内置DATEDIFF函数

返回在指定开始日期和结束日期之间跨越的指定日期部分边界的计数 ...。

如果按原样使用,可能会产生误导或混淆的结果。例如,使用 YEAR 间隔将显示 1999-12-31 (YYYY-MM-DD) 和 2000-01-01 相隔一年,而常识会说这些日期仅相隔 1 天。相反,使用 DAY 1999-12-31 和 2010-12-31 间隔 4,018 天,而大多数人会认为“11 年”是更好的描述。

从天数开始计算月和年,很容易出现闰年和月份大小的错误。

我想知道如何在各种 SQL 方言中实现这一点?示例输出包括:

create table TestData(
    FromDate date not null,
    ToDate date not null,
    ExpectedResult varchar(100) not null); -- exact formatting is unimportant

insert TestData (FromDate, ToDate, ExpectedResult)
values ('1999-12-31', '1999-12-31', '0 days'),
       ('1999-12-31', '2000-01-01', '1 day'),
       ('2000-01-01', '2000-02-01', '1 month'),
       ('2000-02-01', '2000-03-01', '1 month'),              -- month length not important
       ('2000-01-28', '2000-02-29', '1 month, 1 day'),       -- leap years to be accounted for
       ('2000-01-01', '2000-12-31', '11 months, 30 days'),
       ('2000-02-28', '2000-03-01', '2 days'),
       ('2001-02-28', '2001-03-01', '1 day'),                -- not a leap year
       ('2000-01-01', '2001-01-01', '1 year'),
       ('2000-01-01', '2011-01-01', '11 years'),
       ('9999-12-30', '9999-12-31', '1 day'),                -- catch overflow in date calculations
       ('1900-01-01', '9999-12-31', '8099 years 11 months 30 days');  -- min(date) to max(date)
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我碰巧在使用 SQL Server 2008R2,但我有兴趣了解其他方言如何处理这个问题。

Pau*_*ite 10

此答案显示了使用 SQL Server (2005+) CLR 函数的实现。

-- Enable CLR (if necessary)
EXECUTE sys.sp_configure 
    @configname = 'clr enabled',
    @configvalue = 1;

RECONFIGURE;
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组装和功能

CREATE ASSEMBLY DBA
AUTHORIZATION dbo
FROM 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WITH PERMISSION_SET = SAFE;
GO
CREATE FUNCTION dbo.IntervalDescription
(
    @From date, 
    @To date
)
RETURNS nvarchar(100)
AS EXTERNAL NAME 
    DBA.UserDefinedFunctions.IntervalDescription;
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用法

SELECT 
    TD.FromDate,
    TD.ToDate,
    TD.ExpectedResult, 
    IntervalDescription = dbo.IntervalDescription(TD.FromDate, TD.ToDate) 
FROM dbo.TestData AS TD;
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结果

Plan

Output

来源

我不是 C# 程序员!

-- Enable CLR (if necessary)
EXECUTE sys.sp_configure 
    @configname = 'clr enabled',
    @configvalue = 1;

RECONFIGURE;
Run Code Online (Sandbox Code Playgroud)


And*_*y M 9

以下解决方案适用于 SQL Server。该方法与Serg 的相似之处在于查询仅使用 DATEADD 和 DATEDIFF 函数。但是,它不考虑负间隔(FromDate > ToDate),它从总月差中得出年和月:

WITH
  MonthDiff AS
  (
    SELECT
      t.FromDate,
      t.ToDate,
      t.ExpectedResult,
      Months = x.Months - CASE WHEN DAY(t.FromDate) > DAY(t.ToDate) THEN 1 ELSE 0 END
    FROM
      dbo.TestData AS t
      CROSS APPLY (SELECT DATEDIFF(MONTH, t.FromDate, t.ToDate)) AS x (Months)
  )
SELECT
  t.FromDate,
  t.ToDate,
  t.ExpectedResult,
  Result = ISNULL(NULLIF(ISNULL(x.Years  + CASE x.Years  WHEN '1' THEN ' year '  ELSE ' years '  END, '')
                       + ISNULL(x.Months + CASE x.Months WHEN '1' THEN ' month ' ELSE ' months ' END, '')
                       + ISNULL(x.Days   + CASE x.Days   WHEN '1' THEN ' day '   ELSE ' days '   END, ''), ''), '0 days')
FROM
  MonthDiff AS t
  CROSS APPLY
  (
    SELECT
      CAST(NULLIF(t.Months / 12, 0) AS varchar(10)),
      CAST(NULLIF(t.Months % 12, 0) AS varchar(10)),
      CAST(NULLIF(DATEDIFF(DAY, DATEADD(MONTH, t.Months, t.FromDate), t.ToDate), 0) AS varchar(10))
  ) AS x (Years, Months, Days)
;
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输出:

WITH
  MonthDiff AS
  (
    SELECT
      t.FromDate,
      t.ToDate,
      t.ExpectedResult,
      Months = x.Months - CASE WHEN DAY(t.FromDate) > DAY(t.ToDate) THEN 1 ELSE 0 END
    FROM
      dbo.TestData AS t
      CROSS APPLY (SELECT DATEDIFF(MONTH, t.FromDate, t.ToDate)) AS x (Months)
  )
SELECT
  t.FromDate,
  t.ToDate,
  t.ExpectedResult,
  Result = ISNULL(NULLIF(ISNULL(x.Years  + CASE x.Years  WHEN '1' THEN ' year '  ELSE ' years '  END, '')
                       + ISNULL(x.Months + CASE x.Months WHEN '1' THEN ' month ' ELSE ' months ' END, '')
                       + ISNULL(x.Days   + CASE x.Days   WHEN '1' THEN ' day '   ELSE ' days '   END, ''), ''), '0 days')
FROM
  MonthDiff AS t
  CROSS APPLY
  (
    SELECT
      CAST(NULLIF(t.Months / 12, 0) AS varchar(10)),
      CAST(NULLIF(t.Months % 12, 0) AS varchar(10)),
      CAST(NULLIF(DATEDIFF(DAY, DATEADD(MONTH, t.Months, t.FromDate), t.ToDate), 0) AS varchar(10))
  ) AS x (Years, Months, Days)
;
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Mic*_*een 8

我的版本,在 SQL Server 2008R2 SP2 中实现。

CREATE FUNCTION dbo.ReadableInterval(
    @FromDate AS date,
    @ToDate AS date
)
RETURNS TABLE AS RETURN 
(
with YearStep as
(
    select
        max(n1.Number) as YearNumber
    from dbo.Numbers as n1
    where n1.Number <= DATEDIFF(YEAR, @FromDate, @ToDate)  -- see comment (A)
    and DATEADD(YEAR, n1.Number, @FromDate) <= @ToDate     -- see comment (B)
)
, MonthStep as
(
    select
        max(n2.Number) as MonthNumber
    from dbo.Numbers as n2
    cross apply YearStep as y1
    where n2.Number <= DATEDIFF(MONTH, DATEADD(YEAR, y1.YearNumber, @FromDate), @ToDate)
    and DATEADD(MONTH, n2.Number, DATEADD(YEAR, y1.YearNumber, @FromDate)) <= @ToDate
)
, DayStep as
(
    select
        DATEDIFF(day, DATEADD(MONTH, m1.MonthNumber, DATEADD(YEAR, y2.YearNumber, @FromDate)), @ToDate) as DayNumber
    from MonthStep as m1
    cross apply YearStep as y2
)
select
    y.YearNumber,
    m.MonthNumber,
    d.DayNumber
from YearStep as y
cross apply MonthStep as m
cross apply DayStep as d
)
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使用给定的测试数据,结果是

select
    td.FromDate,
    td.ToDate,
    td.ExpectedResult,
    ri.YearNumber as Years,
    ri.MonthNumber as Months,
    ri.DayNumber as [Days]
from dbo.TestData as td
cross apply dbo.ReadableInterval(td.FromDate, td.ToDate) as ri;
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CREATE FUNCTION dbo.ReadableInterval(
    @FromDate AS date,
    @ToDate AS date
)
RETURNS TABLE AS RETURN 
(
with YearStep as
(
    select
        max(n1.Number) as YearNumber
    from dbo.Numbers as n1
    where n1.Number <= DATEDIFF(YEAR, @FromDate, @ToDate)  -- see comment (A)
    and DATEADD(YEAR, n1.Number, @FromDate) <= @ToDate     -- see comment (B)
)
, MonthStep as
(
    select
        max(n2.Number) as MonthNumber
    from dbo.Numbers as n2
    cross apply YearStep as y1
    where n2.Number <= DATEDIFF(MONTH, DATEADD(YEAR, y1.YearNumber, @FromDate), @ToDate)
    and DATEADD(MONTH, n2.Number, DATEADD(YEAR, y1.YearNumber, @FromDate)) <= @ToDate
)
, DayStep as
(
    select
        DATEDIFF(day, DATEADD(MONTH, m1.MonthNumber, DATEADD(YEAR, y2.YearNumber, @FromDate)), @ToDate) as DayNumber
    from MonthStep as m1
    cross apply YearStep as y2
)
select
    y.YearNumber,
    m.MonthNumber,
    d.DayNumber
from YearStep as y
cross apply MonthStep as m
cross apply DayStep as d
)
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解释

我的一般方法是从较早的日期向前推进,首先是几年,然后是几个月,然后是几天。在每个粒度级别,目标是尽可能接近结束日期而无需越过它,然后在下一个较低级别继续。

我使用数字表来促进接近但未结束的计算。从这张表中,DATEADD我可以找到ToDate代码中注释 (B)之前的最大年数/月/天数。

由于我正在寻找 MAX 数字并且我的 Numbers 表聚集在它上面,因此优化器正在执行降序扫描,将值提供给 DATEADD。这会导致日期溢出错误,因为 Numbers 包含超过 100,000 行。DATEADD(YEAR, 100000, @FromDate)大于 9999-12-31 并引发错误。谓词 (A) 给出了向后扫描开始的 Number 值的上限,从而避免了日期溢出。因此,即使对于非常大的日期范围,查询计划也会遍历很少的行。

这种方法用于查找年份和月份,除了月份的起点是由我在第一个 CTE 中找到的年份提出的。DAYS 是我的最低粒度级别,所以一个简单的 DATEDIFF 就足够了。

这可以扩展到更细的粒度,如果需要,以小时、分钟和秒为单位返回间隔。


Mic*_*een 7

PostgreSQL 支持age开箱即用的功能:

select
  FromDate,
  ToDate,
  ExpectedResult,
  age(ToDate, FromDate)
from TestData;
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这给出了期望的结果,给予或接受一些额外的时间值。

select
  FromDate,
  ToDate,
  ExpectedResult,
  age(ToDate, FromDate)
from TestData;
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Ser*_*erg 5

不需要number表格或计数的版本。在 Michael Green 的测试数据上给出相同的结果。他们在数据上有所不同 @FromDate > @ToDateReadableInterval2返回与空值相反的负值。

CREATE FUNCTION dbo.ReadableInterval2(
    @FromDate AS date,
    @ToDate AS date
)
RETURNS TABLE AS RETURN 
(with checkData as (
    select 
       fromDate = case when @FromDate > @ToDate then @ToDate else @FromDate end,
       toDate = case when @FromDate <= @ToDate then @ToDate else @FromDate end,
       k = case when @FromDate > @ToDate then -1 else 1 end
), MonthStep as (
    select k, FromDate, ToDate,
        YearNumber = x.months / 12,
        MonthNumber = x.months % 12
    from checkdata
    cross apply(
        select months = DATEDIFF(MONTH, FromDate, ToDate)
            - case when DAY(FromDate) > DAY(ToDate) then 1 else 0 end
        ) x
)
select YearNumber = k*YearNumber, 
      MonthNumber = k*MonthNumber,
      DayNumber = k*DATEDIFF(day, DATEADD(MONTH, MonthNumber, DATEADD(YEAR, YearNumber, FromDate)), ToDate) 
    from MonthStep 
)
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  • @AaronBertrand 我同意它们非常有用。但就在这里,我看不到哪个问题编号表有助于解决。没有递归,没有 XML,纯标量 DATEADD,DATEDIFF 函数。可能有点啰嗦。 (3认同)