Php 如果是这样的话（我故意让每个人都出生在1973年），你是如何得到5月和6月生日的所有输出的。我在年份中添加了curdate减去他们的出生年份，如果他们的生日在这个日历上过了，那就再多过一年year@ZombieHunter这将在年底失败-它在明年年初找不到

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如果是这样的话（我故意让每个人都出生在1973年），你是如何得到5月和6月生日的所有输出的。我在年份中添加了curdate减去他们的出生年份，如果他们的生日在这个日历上过了，那就再多过一年year@ZombieHunter这将在年底失败-它在明年年初找不到生日。@因此，当现在是12月25日时，你是说他的将不起作用？+1它起作用，有了索引，速度会非常快。你能把这些问题编辑成你的答案吗？（如果SQLfiddle死亡）。如果能够解决这个问题，而不需要另外记录下一个生日，那将是一件好事。是的，那只是一个要删除的报告表

$results = mysql_query("SELECT * FROM Club WHERE Birthday < DATE_SUB(NOW(), INTERVAL 30 DAY)");

SELECT * FROM Club WHERE DAYOFYEAR(Birthday) BETWEEN DAYOFYEAR(NOW()) AND DAYOFYEAR(NOW()) + 30;

create table peeps
( id int not null auto_increment,
  lastname varchar(40) not null,
birthdate datetime not null,
 primary key (id)
);

insert peeps (lastname,birthdate) values ('sam', '1973-01-01');
insert peeps (lastname,birthdate) values ('julie', '1973-02-01');
insert peeps (lastname,birthdate) values ('kim', '1973-03-01');
insert peeps (lastname,birthdate) values ('fred', '1973-04-01');
insert peeps (lastname,birthdate) values ('oscar1', '1973-05-01');
insert peeps (lastname,birthdate) values ('oscar2', '1973-05-02');
insert peeps (lastname,birthdate) values ('oscar3', '1973-05-04');
insert peeps (lastname,birthdate) values ('oscar4', '1973-05-06');
insert peeps (lastname,birthdate) values ('oscar5', '1973-05-08');
insert peeps (lastname,birthdate) values ('oscar6', '1973-05-10');
insert peeps (lastname,birthdate) values ('oscar7', '1973-05-12');
insert peeps (lastname,birthdate) values ('oscar8', '1973-05-14');
insert peeps (lastname,birthdate) values ('oscar9', '1973-05-16');
insert peeps (lastname,birthdate) values ('oscar10', '1973-05-18');
insert peeps (lastname,birthdate) values ('oscar11', '1973-05-20');
insert peeps (lastname,birthdate) values ('oscar12', '1973-05-22');
insert peeps (lastname,birthdate) values ('oscar13', '1973-05-24');
insert peeps (lastname,birthdate) values ('Felix the Cat1', '1973-06-01');
insert peeps (lastname,birthdate) values ('Felix the Cat2', '1973-06-05');
insert peeps (lastname,birthdate) values ('Felix the Cat3', '1973-06-07');
insert peeps (lastname,birthdate) values ('Bonehead7', '1973-07-01');
insert peeps (lastname,birthdate) values ('Bonehead8', '1973-08-01');
insert peeps (lastname,birthdate) values ('Bonehead9', '1973-09-01');
insert peeps (lastname,birthdate) values ('Bonehead10', '1973-10-01');
insert peeps (lastname,birthdate) values ('Bonehead11', '1973-11-01');
insert peeps (lastname,birthdate) values ('Bonehead12', '1973-12-01');

create table peeps_next_birthday
(id int not null,
 next_birthdate datetime not null
 );

insert into peeps_next_birthday (id,next_birthdate) select id,birthdate from peeps;


UPDATE peeps_next_birthday set next_birthdate=date_add(next_birthdate, interval (year(curdate())-year(next_birthdate)) year);

update peeps_next_birthday set next_birthdate=date_add(next_birthdate,interval 1 year)
where curdate()>next_birthdate;

   ///  *************************** now show the birthdays in the coming 30 days

select t2.id,t2.lastname,t1.next_birthdate
from peeps_next_birthday t1
join peeps t2
on t2.id=t1.id
where datediff(t1.next_birthdate,curdate())<=30

SELECT
    birthday,
    -- Shift the birthday into the current year
    CAST(CONCAT(YEAR(NOW()), '-', MONTH(birthday), '-', DAY(birthday)) AS DATE) AS BirthdayInCurrentYear
FROM Club
WHERE CAST(CONCAT(YEAR(NOW()), '-', MONTH(birthday), '-', DAY(birthday)) AS DATE)
    BETWEEN NOW() AND NOW() + INTERVAL 30 DAY;