在mysql函数之间使用忽略年份 [英] Using between mysql function ignoring year

问题描述

好的，我有下面的用户表:

id |  name  |  birthday 

1  |  nam1  |  1980-06-29  
2  |  nam2  |  1997-07-08  
3  |  nam3  |  1997-07-20  

假设今天是2012-06-29

Assuming that today is 2012-06-29

我如何获得将在未来15天内庆祝您生日的用户?我已经尝试过了:

How can I get the users who will celebrate your birthdays in the next 15 days? I have tried with this:

select * from users where birthday between '%-06-29' and '%-07-14';

但看起来不是有效的查询.

but looks like is not a valid query.

推荐答案

我认为最安全的方法是将15天的年龄与今天的年龄进行比较:

I think the safest way is to compare their age in 15 days with their age today:

SELECT *
FROM   Users
WHERE  TIMESTAMPDIFF(YEAR, birthday, CURRENT_DATE + INTERVAL 15 DAY)
     > TIMESTAMPDIFF(YEAR, birthday, CURRENT_DATE)

在 sqlfiddle 上查看.

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