Heim > Datenbank > MySQL-Tutorial > [MySQL]--)查询5天之内过生日的同事中的跨年问题的解决过程_MySQL

[MySQL]--)查询5天之内过生日的同事中的跨年问题的解决过程_MySQL

WBOY
Freigeben: 2016-06-01 13:02:26
Original
1288 Leute haben es durchsucht

前言:

遇到朋友提问,如下:

SELECT * FROM ali_users WHERE DATEDIFF(CAST(CONCAT(DATE_FORMAT(NOW(),'%y'),DATE_FORMAT(birthday,'-%m-%d'))AS DATE),CAST(DATE_FORMAT(NOW(),'%y-%m-%d') AS DATE))
1,准备测试数据,需要包含跨年的数据

1.1,准备测试数据的SQL

USE test;
DROP TABLE IF EXISTS ali_users;
CREATE TABLE ali_users (username VARCHAR(10),birthday DATE NOT NULL,iphone VARCHAR(16));
INSERT INTO ali_users SELECT \'MaoYi\',\'1985-09-04\',\'13998786543\' UNION ALL
SELECT \'LiuEr\',\'1985-08-30\',\'13998786543\' UNION ALL
SELECT \'ZhangSan\',\'1981-01-01\',\'13998786543\' UNION ALL
SELECT \'LiSi\',\'1983-01-02\',\'13998786543\' UNION ALL
SELECT \'WangWu\',\'1984-11-01\',\'13998786543\' UNION ALL
SELECT \'ZhaoLiu\',\'1984-11-01\',\'13998786543\' UNION ALL
SELECT \'SongQi\',\'1986-08-31\',\'13998786543\' UNION ALL
SELECT \'HuangBa\',\'1989-09-01\',\'13998786543\' UNION ALL
SELECT \'ZengJiu\',\'1989-09-02\',\'13998786543\' UNION ALL
SELECT \'LuoShi\',\'1985-09-03\',\'13998786543\' UNION ALL
SELECT \'Tom\',\'1995-09-05\',\'13998786543\' UNION ALL
SELECT \'Licy\',\'1991-12-30\',\'13998286543\' UNION ALL
SELECT \'Cari\',\'1992-12-31\',\'13998286543\' UNION ALL
SELECT \'Mark\',\'1992-01-03\',\'13998286543\' UNION ALL
SELECT \'Ruby\',\'1992-01-04\',\'13998286547\';
1.2,在数据库命令行执行SQL

mysql> USE test;
DATABASE CHANGED
mysql> DROP TABLE IF EXISTS ali_users;
QUERY OK, 0 ROWS affected (0.00 sec)

mysql> CREATE TABLE ali_users (username VARCHAR(10),birthday DATE NOT NULL,iphone VARCHAR(16));
QUERY OK, 0 ROWS affected (0.01 sec)

mysql> INSERT INTO ali_users SELECT \'MaoYi\',\'1985-09-04\',\'13998786543\' UNION ALL
-> SELECT \'LiuEr\',\'1985-08-30\',\'13998786543\' UNION ALL
-> SELECT \'ZhangSan\',\'1981-01-01\',\'13998786543\' UNION ALL
-> SELECT \'LiSi\',\'1983-01-02\',\'13998786543\' UNION ALL
-> SELECT \'WangWu\',\'1984-11-01\',\'13998786543\' UNION ALL
-> SELECT \'ZhaoLiu\',\'1984-11-01\',\'13998786543\' UNION ALL
-> SELECT \'SongQi\',\'1986-08-31\',\'13998786543\' UNION ALL
-> SELECT \'HuangBa\',\'1989-09-01\',\'13998786543\' UNION ALL
-> SELECT \'ZengJiu\',\'1989-09-02\',\'13998786543\' UNION ALL
-> SELECT \'LuoShi\',\'1985-09-03\',\'13998786543\' UNION ALL
-> SELECT \'Tom\',\'1995-09-05\',\'13998786543\' UNION ALL
-> SELECT \'Licy\',\'1991-12-30\',\'13998286543\' UNION ALL
-> SELECT \'Cari\',\'1992-12-31\',\'13998286543\' UNION ALL
-> SELECT \'Mark\',\'1992-01-03\',\'13998286543\' UNION ALL
-> SELECT \'Ruby\',\'1992-01-04\',\'13998286547\';
QUERY OK, 15 ROWS affected (0.01 sec)
Records: 15 Duplicates: 0 WARNINGS: 0

mysql> SELECT * FROM ali_users;
+----------+------------+-------------+
| username | birthday | iphone |
+----------+------------+-------------+
| MaoYi | 1985-09-04 | 13998786543 |
| LiuEr | 1985-08-30 | 13998786543 |
| ZhangSan | 1981-01-01 | 13998786543 |
| LiSi | 1983-01-02 | 13998786543 |
| WangWu | 1984-11-01 | 13998786543 |
| ZhaoLiu | 1984-11-01 | 13998786543 |
| SongQi | 1986-08-31 | 13998786543 |
| HuangBa | 1989-09-01 | 13998786543 |
| ZengJiu | 1989-09-02 | 13998786543 |
| LuoShi | 1985-09-03 | 13998786543 |
| Tom | 1995-09-05 | 13998786543 |
| Licy | 1991-12-30 | 13998286543 |
| Cari | 1992-12-31 | 13998286543 |
| Mark | 1992-01-03 | 13998286543 |
| Ruby | 1992-01-04 | 13998286547 |
+----------+------------+-------------+
15 ROWS IN SET (0.00 sec)

mysql>
2,写出查询SQL
SELECT * FROM ali_users WHERE
2,1,跨年问题分析
因为跨年的时候生日字段通常月份比较小是1月,所以如果利用DATEDIFF来判断要与月份比较大12月来比较得到相差天数在N天之内的话,就需要YEAR(NOW())+1,当年年份+1再加上月份才能与NOW()比较得出真实的相差天数。
2.2,5天之内的设定
N天之内,用 BETWEEN 0 AND N 来判断,如果是5天之内(包含今天)那么N值就是4,就是 BETWEEN 0 AND 4
3,验证数据
比如提醒最近5天之内(包括今日)过生日的同事,生日快乐。
3.1,查询的数据都在今年之内的,比如今天是8月30日,那么需要执行的SQL如下:
SELECT * FROM ali_users WHERE
查询的结果应该是从今天8月30日到9月3日之间过生日的同事,包括LiuEr,SongQi,HuangBa,ZengJiu,LuoShi;

mysql> SELECT * FROM ali_users WHERE
-> DATEDIFF(CAST(CONCAT(YEAR(NOW()),DATE_FORMAT(birthday,\'-%m-%d\'))AS DATE),CAST(DATE_FORMAT(NOW(),\'%y-%m-%d\') AS DATE)) BETWEEN 0 AND 4
-> OR/* or后面的是捎带解决跨年问题*/
-> DATEDIFF(CAST(CONCAT(YEAR(NOW())+1,DATE_FORMAT(birthday,\'-%m-%d\'))AS DATE),CAST(DATE_FORMAT(NOW(),\'%y-%m-%d\') AS DATE)) BETWEEN 0 AND 4
-> ;
+----------+------------+-------------+
| username | birthday | iphone |
+----------+------------+-------------+
| LiuEr | 1985-08-30 | 13998786543 |
| SongQi | 1986-08-31 | 13998786543 |
| HuangBa | 1989-09-01 | 13998786543 |
| ZengJiu | 1989-09-02 | 13998786543 |
| LuoShi | 1985-09-03 | 13998786543 |
+----------+------------+-------------+
5 ROWS IN SET (0.00 sec)

mysql>
3.2,查询的生日有跨年的
比如今天是2013年12月30日,要查询5天之内过生日的同事,那么就有2013年的12月30日31日过生日的,也有2014年1月1日2日3日过生日的同事,因为今天是8月30日,所以要把Step#2中的SQL的NOW()改成'2013-12-30 00:10:10'来进行测试,SQL整理如下:
mysql> SELECT * FROM ali_users WHERE
-> DATEDIFF(CAST(CONCAT(YEAR(\'2013-12-30 00:10:10\'),DATE_FORMAT(birthday,\'-%m-%d\'))AS DATE),CAST(DATE_FORMAT(\'2013-12-30 00:10:10\',\'%y-%m-%d\') AS DATE)) BETWEEN 0 AND 4
-> OR/* or后面的是捎带解决跨年问题*/
-> DATEDIFF(CAST(CONCAT(YEAR(\'2013-12-30 00:10:10\')+1,DATE_FORMAT(birthday,\'-%m-%d\'))AS DATE),CAST(DATE_FORMAT(\'2013-12-30 00:10:10\',\'%y-%m-%d\') AS DATE)) BETWEEN 0 AND 4
-> ;
+----------+------------+-------------+
| username | birthday | iphone |
+----------+------------+-------------+
| ZhangSan | 1981-01-01 | 13998786543 |
| LiSi | 1983-01-02 | 13998786543 |
| Licy | 1991-12-30 | 13998286543 |
| Cari | 1992-12-31 | 13998286543 |
| Mark | 1992-01-03 | 13998286543 |
+----------+------------+-------------+
5 ROWS IN SET (0.00 sec)

mysql>
4,总结

Verwandte Etiketten:
Quelle:php.cn
Erklärung dieser Website
Der Inhalt dieses Artikels wird freiwillig von Internetnutzern beigesteuert und das Urheberrecht liegt beim ursprünglichen Autor. Diese Website übernimmt keine entsprechende rechtliche Verantwortung. Wenn Sie Inhalte finden, bei denen der Verdacht eines Plagiats oder einer Rechtsverletzung besteht, wenden Sie sich bitte an admin@php.cn
Beliebte Tutorials
Mehr>
Neueste Downloads
Mehr>
Web-Effekte
Quellcode der Website
Website-Materialien
Frontend-Vorlage