MySQL complex ORDER BY issue - mysql

Question

I have a complicated ordering issue in my query.
Raw, Unordered Data:
+------+--------+-----------+
| id | job_id | action_id |
+------+--------+-----------+
| 1 | 2 | 1 |
| 2 | 2 | 2 |
| 3 | 1 | 1 |
| 4 | 2 | 3 |
| 5 | 4 | 1 |
| 6 | 1 | 2 |
| 7 | 3 | 1 |
| 8 | 3 | 2 |
| 9 | 4 | 2 |
+------+--------+-----------+
Required Ordering:
+------+--------+-----------+
| id | job_id | action_id |
+------+--------+-----------+
| 7 | 3 | 1 |
| 8 | 3 | 2 |
| | | | * blank lines added for clarity,
| 5 | 4 | 1 | not desired in actual data
| 9 | 4 | 2 |
| | | |
| 3 | 1 | 1 |
| 6 | 1 | 2 |
| | | |
| 1 | 2 | 1 |
| 2 | 2 | 2 |
| 4 | 2 | 3 |
+------+--------+-----------+
The theory behind this ordering:
the largest id is the most recently added entry
the most recent id with action_id of 1
followed by the entries with ascending action_ids that have the same job_id
then the next most recent action_id of 1
ad infinitum
EDIT: I'm not able to add columns to the table in order to aid in sorting, as I've seen in some other solutions to ordering questions.
Any help is greatly appreciated!

Answer 1

My best shot is this:
SELECT * FROM tbl
ORDER BY FIND_IN_SET(job_id,
(SELECT GROUP_CONCAT(job_id ORDER BY ID DESC)
FROM tbl WHERE action_id = 1));

Answer 2

I didn't find a way to do it easily, What do you think of the following code :
select c.id, c.job_id, c.action_id
from (select a.id, a.job_id, a.action_id, min(b.id) as related_id
from myTable a
inner join myTable b
on a.job_id=b.job_id
group by a.job_id) c
group by c.id
order by c.related_id desc, c.action_id