Home > Database > Mysql Tutorial > MYSQL中利用select查询某字段中包含以逗号分隔的字符串的记录方法_MySQL

MYSQL中利用select查询某字段中包含以逗号分隔的字符串的记录方法_MySQL

WBOY
Release: 2016-06-01 13:02:24
Original
1139 people have browsed it

首先我们建立一张带有逗号分隔的字符串。
CREATE TABLE test(id int(6) NOT NULL AUTO_INCREMENT,PRIMARY KEY (id),pname VARCHAR(20) NOT NULL,pnum VARCHAR(50) NOT NULL);

然后插入带有逗号分隔的测试数据
INSERT INTO test(pname,pnum) VALUES('产品1','1,2,4');
INSERT INTO test(pname,pnum) VALUES('产品2','2,4,7');
INSERT INTO test(pname,pnum) VALUES('产品3','3,4');
INSERT INTO test(pname,pnum) VALUES('产品4','1,7,8,9');
INSERT INTO test(pname,pnum) VALUES('产品5','33,4');

查找pnum字段中包含3或者9的记录
mysql> SELECT * FROM test WHERE find_in_set('3',pnum) OR find_in_set('9',pnum);
+----+-------+---------+
| id | pname | pnum    |
+----+-------+---------+
|  3 | 产品3 | 3,4     |
|  4 | 产品4 | 1,7,8,9 |
+----+-------+---------+
2 rows in set (0.03 sec)


使用正则
mysql> SELECT * FROM test WHERE pnum REGEXP '(3|9)';
+----+-------+---------+
| id | pname | pnum    |
+----+-------+---------+
|  3 | 产品3 | 3,4     |
|  4 | 产品4 | 1,7,8,9 |
|  5 | 产品5 | 33,4    |
+----+-------+---------+
3 rows in set (0.02 sec)
这样会产生多条记录,比如33也被查找出来了,不过MYSQL还可以使用正则,挺有意思的

find_in_set()函数返回的所在的位置,如果不存在就返回0
mysql> SELECT find_in_set('e','h,e,l,l,o');
+------------------------------+
| find_in_set('e','h,e,l,l,o') |
+------------------------------+
|                            2 |
+------------------------------+
1 row in set (0.00 sec)

还可以用来排序,如下;
mysql> SELECT * FROM TEST WHERE id in(4,2,3);
+----+-------+---------+
| id | pname | pnum    |
+----+-------+---------+
|  2 | 产品2 | 2,4,7   |
|  3 | 产品3 | 3,4     |
|  4 | 产品4 | 1,7,8,9 |
+----+-------+---------+
3 rows in set (0.03 sec)

如果想要按照ID为4,2,3这样排序呢?
mysql> SELECT * FROM TEST WHERE id in(4,2,3) ORDER BY find_in_set(id,'4,2,3');
+----+-------+---------+
| id | pname | pnum    |
+----+-------+---------+
|  4 | 产品4 | 1,7,8,9 |
|  2 | 产品2 | 2,4,7   |
|  3 | 产品3 | 3,4     |
+----+-------+---------+
3 rows in set (0.03 sec) 
Copy after login

Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template