假如有个表user:

id | name

---------------

1 | tom

2 | ._%\%

3 | Jack

要查name包含下划线_的， 执行sql语句:  select * from user where name like '%_%' ，

竟然会把全部行都查出来了，

要想正确地查，就要  select * from user where name like '%\_%'

要查name包含%的，就要 select * from user where name like '%\%%'

简单的说，要把待查询的字符串里面的_和%替换为\_和\%

还有一种写法，如 select * from user where name like '%/%%' escape '/'

thinkphp中实际应用$where['a']  = ["like","%\_".$catId."\_%"];

打印出sql语句为`a` LIKE '%\\_47\\_%'

亲测可用