MySQL查詢以顯示基於三個表的格式

Question

我在下面的三個表格中顯示了學生記錄，學科和有學科的學生。

我想問什么是有效的SQL查詢，以顯示以下結果。 我可以使用JOIN來顯示它，但不能使用以下格式。

+------+-----------+-----------+-----+----------+-----------+--------+
| Name | Address   | Telephone | Sex | Subjects | Teacher   | Active |
+------+-----------+-----------+-----+----------+-----------+--------+
| John | somewhere | 12345     | M   |          | Teacher 1 | YES    |
| John | somewhere | 12345     | M   | Math     |           | YES    |
| John | somewhere | 12345     | M   | Science  |           | YES    |
| John | somewhere | 12345     | M   | English  |           | YES    |
| Matt | somewhere | 123456    | M   |          | Teacher 2 | YES    |
| Matt | somewhere | 23456     | M   | Math     |           | YES    |
| Matt | somewhere | 123456    | M   | Science  |           | YES    |
| Girl | somewhere | 5431      | F   |          | Teacher3  | YES    |
| Girl | somewhere | 5431      | F   | Physics  |           | YES    |
| Girl | somewhere | 5431      | F   | Math     |           | YES    |
+------+-----------+-----------+-----+----------+-----------+--------+

select * from student_record;
+------------+------+-----------------+-----------+-----+----------+--------+
| id_student | name | address         | telephone | sex | teacher  | active |
+------------+------+-----------------+-----------+-----+----------+--------+
|          1 | John | Somewhere       | 12345     | M   | Teacher  | 0      |
|          2 | Matt | Somewhere There | 12345222  | M   | Teacher1 | 0      |
|          3 | Girl | Somewhere here  | 3333      | F   | Teacher2 | 0      |
+------------+------+-----------------+-----------+-----+----------+--------+

select * from subjects;
+------------+--------------+---------------------+
| id_subject | subject_name | subject_description |
+------------+--------------+---------------------+
|          1 | Math         | Math                |
|          2 | Science      | Science             |
|          3 | English      | English             |
|          4 | Physics      | Physics             |
+------------+--------------+---------------------+

select * from with_subjects;
+--------------------+--------------------+------------+
| id_student_subject | student_id_subject | student_id |
+--------------------+--------------------+------------+
|                  1 |                  1 |          1 |
|                  2 |                  2 |          1 |
|                  3 |                  3 |          1 |
|                  4 |                  4 |          1 |
|                  5 |                  4 |          2 |
|                  6 |                  3 |          2 |
|                  8 |                  1 |          2 |
|                  9 |                  1 |          3 |
|                 10 |                  2 |          3 |
|                 11 |                  3 |          3 |
|                 12 |                  4 |          3 |
+--------------------+--------------------+------------+

Answer 1

怎么樣

選擇一個名稱作為“名稱”，一個地址作為“地址”，一個電話作為“電話”，一個性別作為“性別”，空作為“主題”，一個教師作為“教師”，一個活動在student_record中作為“活動”，作為聯合a。名稱作為“名稱”，a。地址作為“地址”，a。電話作為“電話”，a.sex作為“性別”，b.subject_name作為“主題”，空作為“教師”，a。活動為“活動”，來自（student_record作為內部聯接，with_subjects在a.id_student = c.student_id上為c）內部聯接主題為b在c.student_id_subject = b.id_subject

沒有測試過。 它的順序與您的示例相同，但是應該在那里包含所有數據