需要考虑18岁以下的2个月的18岁的人的年龄 [英] Need to consider age of person as 18 who is 2months less than 18

问题描述

选择datediff（yy，'1986-12-06'，'2003-10-12 14：55：49.480'）作为年份

这给了我17岁的年龄我现在需要考虑是否只剩下两个月完成18年我将如何做到这一点。请建议

select datediff(yy,'1986-12-06','2003-10-12 14:55:49.480')as years
this gives me age as 17 years now i need to consider if it is only two months remaining in completing 18 years how would i do this.Please suggest

推荐答案

好吧，想想几个月的年龄：谁会去在两个月内18岁，是18 * 12-2 = 214个月。

Well, think on the age in months: who is going to be 18 in two months, is 18*12-2=214 months old.

...datediff(mm,...

年龄与日期不完全相同，真正的问题是您正在访问年数 - 这是一个整数值，并且只考虑计算中的年份数。所以

Ages are not quite the same as dates, and the real problem is that you are accessing the number of years - which is an integer value, and only considers the year number in the calculation. So

SELECT datediff(yy,'1986-01-01','2003-12-31') AS years

和

SELECT datediff（yy，'1986-01- 01' ， 2003-01-01'）AS年

And
SELECT datediff(yy,'1986-01-01','2003-01-01') AS years

也将返回17.



解决方案是使用月数：

Will also both return 17.

A solution is to use the number of months:

SELECT datediff(mm,'1986-01-01','2003-12-31') AS months

哪个好一点 - 你可以比较一下对18 * 12的月数，但如果您需要18年的确切截止日期，它仍然没有帮助：

which is a little better - you can compare the number of months against 18 * 12, but it's still not that helpfull if you need to have an exact cut off date at 18 years:

SELECT datediff(mm,'2003-11-30','2003-12-01')AS months

例如返回1.



更好的解决方案是将年份添加到开始日期：

returns 1 for example.

A better solution is to add the years to the start date:

SELECT DATEDIFF(dd, DATEADD(yy, 18, '1986-12-06'), '2013-10-12') as DAYS

如果结果是负数，则它们不到18，如果它大于零则超过18，如果它是零则是18岁生日！

If the result is negative they are under 18, if it's greater than zero they are over 18, and if it is zero it's their 18th birthday!

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