3 DEPENDENT SUBQUERY t3 ALL NULL NULL NULL NULL 2 100.00 Using where
Warnings:
Note 1276 Field or reference 't.a' of SELECT #3 was resolved in SELECT #1
Note 1003 select `test`.`t1`.`a` AS `a` from `test`.`t1` where exists(select 28 from `test`.`t2` left join `test`.`t3` on((`test`.`t3`.`a` = 28)) where ('j' < `test`.`t1`.`a`))
SELECT * FROM (SELECT * FROM t1) AS t
WHERE EXISTS (SELECT t2.a FROM t3 RIGHT JOIN t2 ON (t3.a = t2.a)