通常我们只是用 % 作为通配符,用来表示任意个字符。
但sql中的通配符还有 下划线 _ ,用来标识任意一个字符
实例
SELECT * FROM Websites
WHERE name LIKE '_oogle';
WHERE name LIKE '_oogle';
执行输出结果:
![](https://www.runoob.com/wp-content/uploads/2013/09/wildcards3.jpg)
问题再延伸,如果我们数据中有包含下划线的字符,那么该怎么处理呢?
--需求,找到name中包含下划线的内容,即 把前两条筛出来 with x as (select '_haha' as name from dual union all select '\_2haha' as name from dual union all select '2haha' as name from dual ) select * from x where name like '%_%';
不对,这里的下划线别理解成了通配符
-- 那么使用右斜杠+下划线的模式来转译他,可以么? with x as (select '_haha' as name from dual union all select '\_2haha' as name from dual union all select '2haha' as name from dual ) select * from x where name like '%\_%';
依然不对,这里斜杠并没有起到转译的作用。
查了下,自己以前不认识这个关键字。ESCAPE
-- 使用ESCAPE 关键字来表示【这个字符后边的东西,该被识别成普通字符。】? with x as (select '_haha' as name from dual union all select '\_2haha' as name from dual union all select '2haha' as name from dual ) select * from x where name like '%\_%' ESCAPE '';
结果对了。那么换个别的字符呢?
with x as (select '_haha' as name from dual union all select '\_2haha' as name from dual union all select '2haha' as name from dual ) select * from x where name like '%?_%' ESCAPE '?'; --- with x as (select '_haha' as name from dual union all select '\_2haha' as name from dual union all select '2haha' as name from dual ) select * from x where name like '%!_%' ESCAPE '!';
结果都是
妥了。