Question

我有一个包含分钟时间的列。我要做的是创建一个返回以下格式的函数：

0 = NULL
>0 and <60 mins = '##min'
between 60 mins & 1440 mins = '##hr,##min'
>1440mins = '##days,##hr,##min'

我已经设法使用下面的功能完成了一些操作，但是我正在努力解决如何分割天，小时和分钟的问题：

CREATE FUNCTION dbo.udf_Format_Date_Friendly 
(
    -- Add the parameters for the function here
    @minutes int
)
RETURNS nvarchar(100)
AS
BEGIN
    -- Declare the return variable here
    DECLARE @Result nvarchar(100)

    -- Add the T-SQL statements to compute the return value here
    SELECT @Result = CASE WHEN @minutes = 0 THEN ''
            WHEN @minutes <= 60 THEN CAST(@minutes AS nvarchar(100)) + ' min' 
          WHEN @minutes BETWEEN 60 AND 1440 THEN CAST(CAST(@minutes/60 AS INT) AS nvarchar(100)) + ' hr'
           ELSE NULL end

    -- Return the result of the function
    RETURN @Result

END
GO

Answer 1

尝试;

CREATE FUNCTION dbo.udf_Format_Date_Friendly (@minutes int)
RETURNS nvarchar(100)
AS
BEGIN
   return CASE 
    when @minutes < 60 then cast( @minutes as varchar(10)) + 'min'
    when @minutes < 1400 then cast(@minutes/60 as varchar(10)) + 'hr, ' + cast(@minutes%60 as varchar(10)) + 'min'
    else cast(@minutes/(1400) as varchar(10)) + 'days, ' + cast((@minutes%1400)/60 as varchar(10)) + 'hr, ' + cast(((@minutes%1400)%60) as varchar(10)) + 'min'
    end
end
go

<强>演示

;with data(d) as (
    select 0 d union all
    select 45 d union all
    select 80 union all
    select 1800
)
select d, dbo.udf_Format_Date_Friendly(d) val
from data

<强>输出

0   0min
45  45min
80  1hr, 20min
1800    1days, 6hr, 40min

Answer 2

逐件方法可能比当前＆＃34;一个表达式更具可读性，可以做任何事情＆＃34;格式 - 下面的代码将一次构建一个表达式，从最大（天）开始，一直到分钟级别。

CREATE FUNCTION dbo.udf_Format_Date_Friendly 
(
    -- Add the parameters for the function here
    @minutes int
)
RETURNS nvarchar(100)
AS
BEGIN

    -- Declare the return variable here
    DECLARE @Result nvarchar(10) = ''


    IF @minutes > 1440 
      BEGIN
        SET @Result = @Result + CAST(@minutes/1440 AS NVARCHAR(10)) + ' days, '
      END

    IF @minutes > 60 
      BEGIN
        SET @Result = @Result + CAST((@minutes%1440) /60 AS NVARCHAR(10)) + ' hr, '
      END

    IF @minutes > 0
      BEGIN
        SET @Result = @Result + CAST(@minutes % 60 AS NVARCHAR(10)) + ' min'
      END

    -- Return the result of the function
    RETURN @Result

END
GO

日期格式的标量函数

2 个答案: