如何在 MySQL 中匹配一个字符来代替 %?

mysqlmysqli database

要在 MySQL 中仅匹配一个字符,可以使用下划线 (_) 代替 %。语法如下:

SELECT *FROM yourTableName WHERE yourColumnName LIKE ‘yourString_’;

为了理解上述语法,让我们创建一个表。创建表的查询如下:

mysql> create table OneCharactermatchDemo
   -> (
   -> Id int NOT NULL AUTO_INCREMENT,
   -> PassoutYear year,
   -> PRIMARY KEY(Id)
   -> );
Query OK, 0 rows affected (0.76 sec)

使用 insert 命令在表中插入一些记录。 查询语句如下:

mysql> insert into OneCharactermatchDemo(PassoutYear) values('2008');
Query OK, 1 row affected (0.14 sec)
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2007');
Query OK, 1 row affected (0.18 sec)
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2010');
Query OK, 1 row affected (0.10 sec)
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2011');
Query OK, 1 row affected (0.13 sec)
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2012');
Query OK, 1 row affected (0.12 sec)
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2006');
Query OK, 1 row affected (0.10 sec)
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2019');
Query OK, 1 row affected (0.14 sec)
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2016');
Query OK, 1 row affected (0.13 sec)
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2017');
Query OK, 1 row affected (0.34 sec)
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2015');
Query OK, 1 row affected (0.17 sec)
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2020');
Query OK, 1 row affected (0.13 sec)

使用 select 语句显示表中的所有记录。查询如下:

mysql> select *from OneCharactermatchDemo;

输出结果如下:

+----+-------------+
| Id | PassoutYear |
+----+-------------+
|  1 |        2008 |
|  2 |        2007 |
|  3 |        2010 |
|  4 |        2011 |
|  5 |        2012 |
|  6 |        2006 |
|  7 |        2019 |
|  8 |        2016 |
|  9 |        2017 |
| 10 |        2015 |
| 11 |        2020 |
+----+-------------+
11 rows in set (0.00 sec)

以下是使用下划线匹配一个字符的查询。查询如下:

mysql> select *from OneCharactermatchDemo where PassoutYear Like '200_';

输出结果如下:

+----+-------------+
| Id | PassoutYear |
+----+-------------+
|  1 |        2008 |
|  2 |        2007 |
|  6 |        2006 |
+----+-------------+
3 rows in set (0.00 sec)

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