获取年、月、日的日期差 SQL

Question

有没有一种方法可以根据今天的日期和生日计算某人的年龄，然后以以下方式显示：

If a user is less than (<) 1 year old THEN show their age in MM & days.
Example:  10 months & 2 days old 

If a user is more than 1 year old AND less than 6 years old THEN show their age in YY & MM & days.
Example:  5 years & 3 months & 10 days old

If a user is more than 6 years old THEN display their age in YY.
Example:  12 years

Answer 1

可能不是最有效的方法，但我是这样做的：

我必须首先获得今天的日期和人的生日之间的日期差异。我用它与 ABS() 和 Remainder (%) 函数结合来获取年、月、日等。

declare @year int = 365
declare @month int = 30
declare @sixYears int = 2190

select 
--CAST(DATEDIFF(mm, a.BirthDateTime,  getdate()) AS VARCHAR) as GetMonth,
--CAST(DATEDIFF(dd, DATEADD(mm, DATEDIFF(mm, a.BirthDateTime, getdate()), a.BirthDateTime), getdate()) AS VARCHAR) as GetDays,

CASE 
    WHEN 
        DATEDIFF(dd,a.BirthDateTime,getdate())  < @year 
    THEN 
        cast((DATEDIFF(dd,a.BirthDateTime,getdate()) / (@month)) as varchar) +' Months & ' +
        CAST(ABS(DATEDIFF(dd, DATEADD(mm, DATEDIFF(mm, a.BirthDateTime, getdate()), a.BirthDateTime), getdate())) AS VARCHAR)
        + ' Days'
    WHEN
        DATEDIFF(dd,a.BirthDateTime,getdate()) between @year and @sixYears
    THEN
        cast((DATEDIFF(dd,a.BirthDateTime,getdate()) / (@year)) as varchar) +' Years & ' +
        CAST((DATEDIFF(mm, a.BirthDateTime,  getdate()) % (12)) AS VARCHAR) + ' Months'
    WHEN DATEDIFF(dd,a.BirthDateTime,getdate()) > @sixYears
    THEN cast(a.Age as varchar) + ' Years'

    end as FinalAGE,