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';

执行输出结果：

    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。