View difference between Paste ID: jy0AxuR7 and prbXjbdW

-mysql> describe a;
+mysql> describe a;
-+-----------+-------------+------+-----+---------+----------------+
++-----------+-------------+------+-----+---------+----------------+
-| Field     | Type        | Null | Key | Default | Extra          |
+| Field     | Type        | Null | Key | Default | Extra          |
-+-----------+-------------+------+-----+---------+----------------+
++-----------+-------------+------+-----+---------+----------------+
-| id        | int(11)     | NO   | PRI | NULL    | auto_increment |
+| id        | int(11)     | NO   | PRI | NULL    | auto_increment |
-| parent_id | int(11)     | YES  | MUL | NULL    |                |
+| parent_id | int(11)     | YES  | MUL | NULL    |                |
-| name      | varchar(45) | NO   |     | NULL    |                |
+| name      | varchar(45) | NO   |     | NULL    |                |
-+-----------+-------------+------+-----+---------+----------------+
++-----------+-------------+------+-----+---------+----------------+
-3 rows in set (0.04 sec)
+3 rows in set (0.04 sec)
-
+
-mysql> set @t=5; select distinct b.node_id, a.* from a, (select @t:=if(a.id = @t, a.parent_id, @t) as node_id from a order by a.id DESC) as b where a.id=b.node_id;
+mysql> select * from a;
-Query OK, 0 rows affected (0.00 sec)
++----+-----------+------+
-
+| id | parent_id | name |
-+---------+----+-----------+------+
++----+-----------+------+
-| node_id | id | parent_id | name |
+|  1 |      NULL | ROOT |
-+---------+----+-----------+------+
+|  2 |         1 | L1_0 |
-|       5 |  5 |         4 | L3_0 |
+|  3 |         1 | L1_1 |
-|       4 |  4 |         2 | L2_0 |
+|  4 |         2 | L2_0 |
-|       2 |  2 |         1 | L1_0 |
+|  5 |         4 | L3_0 |
-|       1 |  1 |      NULL | ROOT |
+|  6 |         4 | L3_1 |
-+---------+----+-----------+------+
+|  7 |         3 | L2_1 |
 +----+-----------+------+
 7 rows in set (0.00 sec)
 
 mysql> set @t=5; select distinct b.node_id, a.* from a, (select @t:=if(a.id = @t, a.parent_id, @t) as node_id from a order by a.id DESC) as b where a.id=b.node_id; # ищем путь от ноды с индексом 5 до корня, при условии, что дочерний элемент имеет индекс выше, чем свои родитель
 Query OK, 0 rows affected (0.00 sec)
 
 +---------+----+-----------+------+
 | node_id | id | parent_id | name |
 +---------+----+-----------+------+
 |       5 |  5 |         4 | L3_0 |
 |       4 |  4 |         2 | L2_0 |
 |       2 |  2 |         1 | L1_0 |
 |       1 |  1 |      NULL | ROOT |
 +---------+----+-----------+------+
 4 rows in set (0.02 sec)