在MySQL中计算年龄（InnoDb）

如果我的出生日期以表格的形式存储在表格中，dd-mm-yyyy并且从当前日期中减去该日期，那么返回的日期格式是什么？

如何使用这种返回的格式来计算某人的年龄？

如果该值存储为DATETIME数据类型：

SELECT YEAR(CURRENT_TIMESTAMP) - YEAR(dob) - (RIGHT(CURRENT_TIMESTAMP, 5) < RIGHT(dob, 5)) as age 
  FROM YOUR_TABLE

考虑leap年时不太精确：

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)功能：

SELECT TIMESTAMPDIFF(YEAR, '1970-02-01', CURDATE()) AS age

演示版

select *,year(curdate())-year(dob) - (right(curdate(),5) < right(dob,5)) as age from your_table

这样，您甚至可以考虑出生的月份和日期，以便更准确地计算年龄。

SELECT TIMESTAMPDIFF (YEAR, YOUR_COLUMN, CURDATE()) FROM YOUR_TABLE AS AGE

点击演示..

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);

它返回今年-出生年份（该人今年生日后的年龄），并根据该人今年是否生日来进行调整。

它不会受到此处介绍的其他方法的舍入误差的影响。

^{从这里自由改编}

由于该问题已被标记mysql，因此我有以下适用于我的实现，希望其他RDBMS可以使用类似的替代方法。这是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是保持生日日期名称通过YEAR（）命令将CURDATE（）转到年份的字段名称，减去生日日期字段中的YEAR（）

这是在MySQL中计算年龄的方法：

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

您可以创建一个函数来做到这一点：

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，您可以从日期获取年龄：

select getIdade('1983-09-16');

如果日期格式为Ymd H：i：s，则可以执行以下操作：

select getIdade(substring_index('1983-09-16 23:43:01', ' ', 1));

您可以在任何地方重用此功能;）

我更喜欢以这种方式使用函数。

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;

有两种简单的方法可以做到这一点：

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"))