MySQL SELECT value before MAX - mysql

Question

How to select 1st, 2nd or 3rd value before MAX ?
usually we do it with order by and limit
SELECT * FROM table1
ORDER BY field1 DESC
LIMIT 2,1
but with my current query I don't know how to make it...
Sample table
+----+------+------+-------+
| id | name | type | count |
+----+------+------+-------+
| 1 | a | 1 | 2 |
| 2 | ab | 1 | 3 |
| 3 | abc | 1 | 1 |
| 4 | b | 2 | 7 |
| 5 | ba | 2 | 1 |
| 6 | cab | 3 | 9 |
+----+------+------+-------+
I'm taking name for each type with max count with this query
SELECT
`table1b`.`name`
FROM
(SELECT
`table1a`.`type`, MAX(`table1a`.`count`) AS `Count`
FROM
`table1` AS `table1a`
GROUP BY `table1a`.`type`) AS `table1a`
INNER JOIN
`table1` AS `table1b` ON (`table1b`.`type` = `table1a`.`type` AND `table1b`.`count` = `table1a`.`Count`)
and I want one more column additional to name with value before max(count)
so result should be
+------+------------+
| name | before_max |
+------+------------+
| ab | 2 |
| b | 1 |
| cab | NULL |
+------+------------+
Please ask if something isn't clear ;)

Answer 1

AS per your given table(test) structure, the query has to be as follows :
select max_name.name,before_max.count
from
(SELECT type,max(count) as max
FROM `test`
group by type) as type_max
join
(select type,name,count
from test
) as max_name on (type_max.type = max_name.type and count = type_max.max )
left join
(select type,count
from test as t1
where count != (select max(count) from test as t2 where t1.type = t2.type)
group by type
order by count desc) as before_max on(type_max.type = before_max .type)