如何在MySQL中代替％匹配一个字符？
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) -> );
使用insert命令在表中插入一些记录。查询如下：
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2008');
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2007');
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2010');
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2011');
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2012');
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2006');
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2019');
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2016');
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2017');
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2015');
mysql> insert into OneCharactermatchDemo(PassoutYear) values('2020');使用select语句显示表中的所有记录。查询如下：
mysql> select *from OneCharactermatchDemo;
以下是输出：
+----+-------------+ | 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)
这是使用下划线仅匹配一个字符的查询。查询如下：
mysql> select *from OneCharactermatchDemo where PassoutYear Like '200_';
以下是输出：
+----+-------------+ | Id | PassoutYear | +----+-------------+ | 1 | 2008 | | 2 | 2007 | | 6 | 2006 | +----+-------------+ 3 rows in set (0.00 sec)