MySQL get the 2 last of each version in one table

Question

I would like to retrieve the 2 last version of a process:

| id | name  | version | souce_id |
| 1  | test1 | 1.00    | 0        |
| 2  | test1 | 1.01    | 1        |
| 3  | test1 | 1.02    | 1        |
| 4  | test1 | 1.03    | 1        |
| 5  | test2 | 1.00    | 0        |
| 6  | test2 | 1.01    | 4        |
| 7  | test2 | 1.02    | 4        |
| 8  | test3 | 1.00    | 0        |

and the expected result would be:

| id | name  | version | souce_id |
| 3  | test1 | 1.02    | 1        |
| 4  | test1 | 1.03    | 1        |
| 6  | test2 | 1.01    | 4        |
| 7  | test2 | 1.02    | 4        |
| 8  | test3 | 1.00    | 0        |

I have written this query but I don't like it... and I cannot include it in as a subquery in a WHERE id IN (...) . SELECT (SELECT SUBSTRING_INDEX(GROUP_CONCAT( id ORDER BY process_version DESC), ',', 2) FROM process WHERE process_name = parent . process_name ) AS ids FROM process AS parent WHERE 1 GROUP BY process_name

Answer 1

Eg

SELECT x.*
  FROM my_table x
  JOIN my_table y
    ON y.name = x.name
   AND y.id <= x.id
 GROUP
    BY x.id
 HAVING COUNT(*) <=2;

Answer 2

Thanks Strawberry! With just a small correction I have my request (yours was giving the 2 1st versions):

SELECT `x`.*, `y`.`id`
  FROM `my_table` AS `x`
  JOIN `my_table` AS `y`
    ON `y`.`name` = `x`.`name`
    AND `y`.`id` >= `x`.`id`
GROUP BY `x`.`id`
HAVING COUNT(*) <=2  
ORDER BY `x`.`name` ASC, `x`.`version` ASC