I have two tables which I want to connect.
TABLE_A:
+-----------+-----------+---------+
| row_id | category | val_1 |
+-----------+-----------+---------+
| 1067 | cat1 | 6.5 |
| 2666 | test | 6.5 |
| 2710 | cat1 | 2.1 |
| 2710 | test | 7.1 |
| 2767 | test | 3 |
| 71142 | cat1 | 5 |
| 50666 | other | 6.5 |
| 71142 | other | 1 |
| 345342 | cat1 | 6.5 |
| 345342 | test | 2.8 |
+-----------+-----------+---------+
TABLE_B:
+-----------+-----------+
| row_id | val_2 |
+-----------+-----------+
| 1067 | 2.0 |
| 2666 | 9 |
| 2701 | 2.2 |
| 2708 | 1 |
| 2709 | 6.5 |
| 2710 | 5.2 |
| 2765 | 6.5 |
| 2766 | 15 |
| 2767 | 8 |
| 71142 | 5 |
| 2783 | 4.5 |
| 50666 | 6.5 |
| 101588 | 9 |
| 101588 | 3 |
| 3452 | 8.0 |
| 23422 | 5 |
| 345342 | 6.5 |
+-----------+-----------+
RESULT_TABLE:
+-----------+-----------+-----------+------------+
| row_id | val_2 | val_1 | category |
+-----------+-----------+-----------+------------+
| 1067 | 2.0 | 6.5 | cat1 |
| 2666 | 9 | 6.5 | test |
| 2701 | 2.2 | 2.2 | NULL |
| 2708 | 1 | 1 | NULL |
| 2709 | 6.5 | 1 | NULL |
| 2710 | 5.2 | 2.1 | cat1 |
| 2710 | 5.2 | 7.1 | test |
| 2765 | 6.5 | 1 | NULL |
| 2766 | 15 | 1 | NULL |
| 2767 | 8 | 3 | test |
| 71142 | 5 | 5 | cat1 |
| 71142 | 5 | 1 | other |
| 2783 | 4.5 | 1 | NULL |
| 50666 | 6.5 | 6.5 | other |
| 101588 | 9 | 1 | NULL |
| 101588 | 3 | 1 | NULL |
| 3452 | 8.0 | 1 | NULL |
| 23422 | 5 | 1 | NULL |
| 345342 | 6.5 | 6.5 | cat1 |
| 345342 | 6.5 | 2.8 | test |
+-----------+-----------+-----------+------------+
I tried to use something like this:
SELECT TABLE_A.row_id, TABLE_A.category, TABLE_A.val_1, TABLE_B.val_2
FROM TABLE_A
INNER JOIN TABLE_B ON TABLE_B.row_id = TABLE_A.row_id
ORDER BY row_id;
However, the result included only rows where the row_id column exists in TABLE_A.
Is there way to connect TABLE_A and TABLE_B to produce the result shown in RESULT_TABLE?
If you want all the results, you need an outer join, not an inner one. (Inner only returns the rows where there is a match; outer returns all rows, with the matching rows “stitched together”)