如何为该条件创建mysql查询？

我必须写一个类似这样的查询

select * from table where title like '% select from tableb %'

select * from table where title like (select column from tableb);

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这个从表B中选择是一个查询，而不是一个字符串您基本上已经得到了它：

您可以这样做：

SELECT * FROM `table` 
where `title` like (SELECT column FROM tableb WHERE id=case)

我找到了

select * from table where title like '%'+ select from tableb +'%'

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你应该这样用

select * from table where title like '% select from tableb %'

select * from table where title like (select column from tableb);