4

クイック アップデート -- SQLFiddle リンク: http://sqlfiddle.com/#!2/d038f/2

比較的わかりやすいと思いますが…

全部で 7 つのテーブル、3 つの「メイン」テーブル、「多対多」関係を管理する別の 3 つ、マスター テーブルである別の 1 つがあります。

mysql> show tables;
+--------------------+
| Tables_in_test     |
+--------------------+
| Equipment          |
| Room               |
| Trainer            |
| Training           |
| training_equipment |
| training_room      |
| training_trainer   |
+--------------------+
7 rows in set (0.00 sec)

さて、スキーマと内容:

    mysql> SELECT * FROM Equipment; 
           SELECT * FROM Room; 
           SELECT * FROM Trainer; 
           SELECT * FROM training_equipment; 
           SELECT * FROM training_room; 
           SELECT * FROM training_trainer; 
           SELECT * FROM Training;   



+----+-------------+
| id | equipment   |
+----+-------------+
|  1 | Equipment_1 |
|  2 | Equipment_2 |
|  3 | Equipment_3 |
|  4 | Equipment_4 |
+----+-------------+
4 rows in set (0.00 sec)

+----+--------+
| id | room   |
+----+--------+
|  1 | Room_1 |
|  2 | Room_2 |
+----+--------+
2 rows in set (0.00 sec)

+----+-------+
| id | name  |
+----+-------+
|  1 | John  |
|  2 | Joe   |
|  3 | Jason |
+----+-------+
3 rows in set (0.00 sec)


+-------------+--------------+
| training_id | equipment_id |
+-------------+--------------+
|           1 |            3 |
|           1 |            4 |
|           2 |            1 |
+-------------+--------------+
3 rows in set (0.00 sec)

+-------------+---------+
| training_id | room_id |
+-------------+---------+
|           1 |       1 |
+-------------+---------+
1 row in set (0.01 sec)

+-------------+------------+
| training_id | trainer_id |
+-------------+------------+
|           1 |          2 |
|           1 |          3 |
+-------------+------------+
2 rows in set (0.00 sec)

+----+------------+------------+---------+------+-----------+---+
| id | from       | to         | trainer | room | equipment | a |
+----+------------+------------+---------+------+-----------+---+
|  1 | 1349578297 | 1350096689 |       1 |    1 |         1 | 0 |
+----+------------+------------+---------+------+-----------+---+
1 row in set (0.02 sec)

次のクエリを思いつきましたが、結果が正しくないことがわかります。

mysql> SELECT t.from, r.room, tra.name, e.equipment
    -> FROM Training t
    -> LEFT JOIN training_room tr ON ( t.room = tr.training_id )
    -> LEFT JOIN Room r ON ( tr.room_id = r.id )
    -> LEFT JOIN training_trainer tt ON ( t.trainer = tt.training_id )
    -> LEFT JOIN Trainer tra ON ( tt.trainer_id = tra.id )
    -> LEFT JOIN training_equipment te ON ( t.equipment = te.training_id)
    -> LEFT JOIN Equipment e ON ( te.equipment_id = e.id )
    -> WHERE t.id =1;
+------------+--------+------+-------------+
| from       | room   | name | equipment   |
+------------+--------+------+-------------+
| 1349578297 | Room_1 | Joe  | Equipment_3 |
| 1349578297 | Room_1 | Joe  | Equipment_4 |
| 1349578297 | Room_1 | Jason| Equipment_3 |
| 1349578297 | Room_1 | Jason| Equipment_4 |
+------------+--------+------+-------------+
4 rows in set (0.02 sec)

重複した結果を見たくありません。見たいのは次のとおりです。

+------------+--------+------+-------------+
| from       | room   | name | equipment   |
+------------+--------+------+-------------+
| 1349578297 | Room_1 | Joe  | Equipment_3 |
| 1349578297 | Room_1 | Jason| Equipment_4 |
+------------+--------+------+-------------+

DISTINCT問題を解決しないGROUP BY tra.name, e.equipment

ありがとうございました。

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1 に答える 1

4

申し訳ありませんが、現在のデータベース構造では、要件に一致するクエリがないと思います。ただし、 group_concat() 関数を確認すると、機器とトレーナーの名前を文字列にグループ化できます。次に、php などのサーバー コードを使用して値を抽出できます。 

 SELECT t.from, r.room, group_concat(distinct tra.name), group_concat( distinct e.equipment)
 FROM Training t
 LEFT JOIN training_room tr ON ( t.room = tr.training_id )
 LEFT JOIN Room r ON ( tr.room_id = r.id )
 LEFT JOIN training_trainer tt ON ( t.trainer = tt.training_id )
 LEFT JOIN Trainer tra ON ( tt.trainer_id = tra.id )
 LEFT JOIN training_equipment te ON ( t.equipment = te.training_id)
 LEFT JOIN Equipment e ON ( te.equipment_id = e.id )
 WHERE t.id =1

テーブル「training_equipment」の構造を変更できる場合は、別の列「trainer_id」を追加してから、以下のクエリを使用できます。

SELECT t.from, r.room, tra.name, e.equipment
FROM Training t
LEFT JOIN training_room tr ON ( t.room = tr.training_id )
LEFT JOIN Room r ON ( tr.room_id = r.id )     
LEFT JOIN training_equipment te ON ( t.equipment = te.training_id)
LEFT JOIN Equipment e ON ( te.equipment_id = e.id )
LEFT JOIN Trainer tra ON ( te.trainer_id = tra.id )
WHERE t.id =1
于 2012-10-13T06:11:35.917 に答える