在MySQL中计算年龄（InnoDb）
如果我将一个人的出生date以dd-mm-yyyy的forms存储在表格中，并且从当前date中减去它，那么返回的date格式是什么？
我怎样才能用这个返回的格式来计算某个人的年龄？
如果该值存储为DATETIME数据types：
SELECT YEAR(CURRENT_TIMESTAMP) - YEAR(dob) - (RIGHT(CURRENT_TIMESTAMP, 5) < RIGHT(dob, 5)) as age FROM YOUR_TABLE
当你考虑闰年时不太确切：
SELECT DATEDIFF(CURRENT_DATE, STR_TO_DATE(t.birthday, '%d-%m-%Y'))/365 AS ageInYears FROM YOUR_TABLE t
您可以使用TIMESTAMPDIFF(unit, datetime_expr1, datetime_expr2)函数：
SELECT TIMESTAMPDIFF(YEAR, '1970-02-01', CURDATE()) AS age
演示
select *,year(curdate())-year(dob) - (right(curdate(),5) < right(dob,5)) as age from your_table
以这种方式，你甚至可以考虑出生的月份和date，以便有一个更准确的年龄计算。
SELECT TIMESTAMPDIFF (YEAR, YOUR_COLUMN, CURDATE()) FROM YOUR_TABLE AS AGE
点击进行演示
简单但优雅..
select floor(datediff (now(), birthday)/365) as age
只是：
DATE_FORMAT(FROM_DAYS(TO_DAYS(NOW())-TO_DAYS(`birthDate`)), '%Y')+0 AS age
尝试这个：
SET @birthday = CAST('1980-05-01' AS DATE); SET @today = CURRENT_DATE(); SELECT YEAR(@today) - YEAR(@birthday) - (CASE WHEN MONTH(@birthday) > MONTH(@today) OR (MONTH(@birthday) = MONTH(@today) AND DAY(@birthday) > DAY(@today)) THEN 1 ELSE 0 END);
今年 – 出生年度（生日之后今年是多less岁），并且根据今年是否有过生日的情况调整。
它不会受到这里介绍的其他方法的舍入误差。
从这里自由改编
这是如何计算MySQL中的年龄：
select date_format(now(), '%Y') - date_format(date_of_birth, '%Y') - (date_format(now(), '00-%m-%d') < date_format(date_of_birth, '00-%m-%d')) as age from table
你可以做一个function来做到这一点：
drop function if exists getIdade; delimiter | create function getIdade( data_nascimento datetime ) returns int begin declare idade int; declare ano_atual int; declare mes_atual int; declare dia_atual int; declare ano int; declare mes int; declare dia int; set ano_atual = year(curdate()); set mes_atual = month( curdate()); set dia_atual = day( curdate()); set ano = year( data_nascimento ); set mes = month( data_nascimento ); set dia = day( data_nascimento ); set idade = ano_atual - ano; if( mes > mes_atual ) then set idade = idade - 1; end if; if( mes = mes_atual and dia > dia_atual ) then set idade = idade - 1; end if; return idade; end| delimiter ;
现在，你可以从一个date的年龄：
select getIdade('1983-09-16');
如果你的date格式是Ymd H：i：s，你可以这样做：
select getIdade(substring_index('1983-09-16 23:43:01', ' ', 1));
你可以在任何地方重用这个function;）
我更喜欢用这种方式来使用函数。
DELIMITER $$ DROP FUNCTION IF EXISTS `db`.`F_AGE` $$ CREATE FUNCTION `F_AGE`(in_dob datetime) RETURNS int(11) NO SQL BEGIN DECLARE l_age INT; IF DATE_FORMAT(NOW( ),'00-%m-%d') >= DATE_FORMAT(in_dob,'00-%m-%d') THEN -- This person has had a birthday this year SET l_age=DATE_FORMAT(NOW( ),'%Y')-DATE_FORMAT(in_dob,'%Y'); ELSE -- Yet to have a birthday this year SET l_age=DATE_FORMAT(NOW( ),'%Y')-DATE_FORMAT(in_dob,'%Y')-1; END IF; RETURN(l_age); END $$ DELIMITER ;
现在使用
SELECT F_AGE('1979-02-11') AS AGE;
要么
SELECT F_AGE(date) AS age FROM table;
由于这个问题被标记为mysql ，我有以下的实现对我有用，我希望类似的替代scheme将在那里的其他RDBMS的。 这里的SQL：
select YEAR(now()) - YEAR(dob) - ( DAYOFYEAR(now()) < DAYOFYEAR(dob) ) as age from table where ...
简单地做
SELECT birthdate, (YEAR(CURDATE())-YEAR(birthdate)) AS age FROM `member`
birthdate是保存birthdate名称的字段名称将CURDATE（）转换为YEAR（）命令减去年份（）中的YEAR（）
有两个简单的方法来做到这一点：
1-
select("users.birthdate", DB::raw("FLOOR(DATEDIFF(CURRENT_DATE, STR_TO_DATE(users.birthdate, '%Y-%m-%d'))/365) AS age_way_one"),
2-
select("users.birthdate",DB::raw("(YEAR(CURDATE())-YEAR(users.birthdate)) AS age_way_two"))
