MySQL：LIKE子句中的Select语句

Question

I want to use " like '%(selected result)%' " phrase. Below code is mine

 where note like '%(SELECT ds_word.ds_en FROM ds_word
                     WHERE ds_co LIKE '%콜레라%' LIMIT 0 ,1)%'

But '%~~~%' use make an error, that is,

"#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(SELECT code_co.code, code_co.disease_co, code_en.disease_en FROM code_co LEFT' at line 44".

How can i use that expression? Help me, please!

And, below answer outputs the result picture. This is the captured picture on phpmyadmin.

Answer 1

WHERE note LIKE
CONCAT ('%',(SELECT ds_word.ds_en FROM ds_word WHERE ds_co LIKE '%콜레라%' LIMIT 0 ,1), '%')

Answer 2

What I think is, Since you are trying to use a value inside your LIKE statement, you will need to modify it to something like:

where note like CONCAT('%',SELECT ds_word.ds_en FROM ds_word WHERE ds_co LIKE '%콜레라%' LIMIT 0 ,1, '%')

Hope this works for you.

Answer 3

Use \\ to escape the ' which is the escape character

where note 
like '%(SELECT ds_word.ds_en FROM ds_word WHERE ds_co LIKE \'%콜레라%\' LIMIT 0 ,1)%';