Question

如何计算SQL中两个日期之间的年、月、日的持续时间/年龄？ instituitiondate 和 nextdate 具有“日期”数据类型？

这是我试过的：

SELECT
    instituitiondate, 
    MAX(NextDate) AS LastDate,
    DATEDIFF (YEAR, InstituitionDate, MAX(NextDate)) Years, 
    DATEDIFF (MONTH, InstituitionDate, MAX(NextDate)) Months, 
    DATEDIFF (DAY, InstituitionDate, MAX(NextDate)) Days 
FROM
    Main_Cause_List 
WHERE
    CaseNo = '372/4/2020' 
GROUP BY
    InstituitionDate;

Answer 1

我不知道您的问题是否有确切答案，但这应该是一个很好的近似：

SELECT
    instituitiondate, 
    MAX(NextDate) AS LastDate,
    floor(datediff(day, InstituitionDate, MAX(NextDate)) / 365.25) as Years, 
    floor(datediff(day, InstituitionDate, MAX(NextDate)) / 30.4375) % 12 as Months, 
    floor(datediff(day, InstituitionDate, MAX(NextDate)) % 30.4375) as Days 
FROM
    Main_Cause_List 
WHERE
    CaseNo = '372/4/2020' 
GROUP BY
    InstituitionDate;

它计算年龄时考虑了平均每年 365.25 天和每月 30.44 天。

已编辑：将每月天数从 30.44 更改为 30.4375，因为 30.4375 * 12 = 365.25，消除了计算年份与相应月份和剩余天数之间的非常小的不一致。< /p>

Answer 2

SELECT
    instituitiondate, 
    MAX(NextDate) AS LastDate,
    floor(datediff(day, InstituitionDate, MAX(NextDate)) / 365.25) as Years, 
    floor(datediff(day, InstituitionDate, MAX(NextDate)) / 30.44) % 12 as Months, 
    floor(datediff(day, InstituitionDate, MAX(NextDate)) % 30.44) as Days 
FROM
    Main_Cause_List 
WHERE
    CaseNo = '372/4/2020' 
GROUP BY
    InstituitionDate;

如何计算SQL中两个日期之间的年，月，日的持续时间/年龄？ instituitiondate 和 nextdate 具有“日期”数据类型？

2 个答案: