Bagaimana untuk Mengira Nilai Purata untuk Nilai Lulus Berbeza dalam Pertanyaan SQL?

+------------+--------------+----------------+
| id         | pass         | val            |
+------------+--------------+----------------+
| DA02959106 | 5.0000000000 |  44.4007000000 |
| 08A5969201 | 1.0000000000 | 182.4100000000 |
| 08A5969201 | 2.0000000000 | 138.7880000000 |
...

id, AVG of 'val' for 'pass' = 1, AVG of 'val' for 'pass' = 2, etc

+------------+---------+---------+---------+---------+---------+---------+---------+
| id         | val_1   | val_2   | val_3   | val_4   | val_5   | val_6   | val_7   |
+------------+---------+---------+---------+---------+---------+---------+---------+
| DA02959106 | 186.147 | 148.266 | 111.905 | 76.3985 | 44.4007 | 0       | 0       |
+------------+---------+---------+---------+---------+---------+---------+---------+

SELECT id, pass, AVG(val) AS val_1 
FROM data_r1 
GROUP BY id, pass;

SELECT d1.id,
    (SELECT IFNULL(ROUND(AVG(d2.val), 4) ,0) FROM data_r1 d2 
     WHERE d2.id = d1.id AND pass = 1) as val_1,
    (SELECT IFNULL(ROUND(AVG(d2.val), 4) ,0) FROM data_r1 d2 
     WHERE d2.id = d1.id AND pass = 2) as val_2,
    ...
from data_r1 d1
GROUP BY d1.id