使用外键连接表

Question

我的结果在10公里以内的研究所.但我想让学院的课程名称为GATE.我怎样才能做到这一点？course_records在研究所有一个外国机构.我无法加入这些表格.任何形式的帮助将不胜感激.

 $result=$conn->query("SELECT *, ( 6371 * acos( cos( radians($user_latitude) ) * cos( radians( latitude ) ) * cos( radians( longitude ) - radians($user_longitude) ) + sin( radians($user_latitude) ) * sin( radians( latitude ) ) ) ) AS distance FROM institutes  HAVING distance < 10 ORDER BY distance  LIMIT 0 , 10 ");

mysql> select*from institutes;

+--------------+-------------------+---------------------+----------------+----------------------------------+-------------+--------+----------+-----------+
| institute_id | name              | email               | contact_number | address                          | telephone   | tut_id | latitude | longitude |
+--------------+-------------------+---------------------+----------------+----------------------------------+-------------+--------+----------+-----------+
|           23 | Dhananjay Classes | dhananjay@gmail.com | 9999888877     | Palam dabri Road,Mahavir Enclave | 011-1234567 |     11 |  28.5892 |   77.0858 |
|           24 | ffe               | rr@rere             | 323232         |                                  |             |     11 |  28.5667 |   77.2833 |
+--------------+-------------------+---------------------+----------------+----------------------------------+-------------+--------+----------+-----------+

mysql> select*from course_records;

+-----------+------+--------------+-------+--------------+--------------------------+--------------------+---------------+-----------+---------+--------------+
| course_id | name | subject      | fees  | num_students | num_students_per_teacher | month_of_admission | num_of_trials | commision | created | institute_id |
+-----------+------+--------------+-------+--------------+--------------------------+--------------------+---------------+-----------+---------+--------------+
|         1 | GATE | CSE          | 10000 | 110          | 20                       | January            | 3             | yes       | NULL    |           23 |
|         2 | NDA  | all_subjects | 7000  | 50           | 20                       | April              | 3             | yes       | NULL    |           23 |
|         3 | 12th | Math         | 2     | 90           | 20                       |                    | 2             |           | NULL    |           23 |
+-----------+------+--------------+-------+--------------+--------------------------+--------------------+---------------+-----------+---------+--------------+

Answer 1

尝试一下：

$result=$conn->query("SELECT
i.name as inst_name, cr.name as course_name,
( 6371 * acos( cos( radians($user_latitude) ) * cos( radians( i.latitude ) ) * cos( radians( i.longitude ) - radians($user_longitude) ) + sin( radians($user_latitude) ) * sin( radians( i.latitude ) ) ) ) AS distance
FROM
institutes i
join course_records cr on i.institute_id = cr.institute_id
where cr.name = 'GATE'
HAVING
distance < 10 ORDER BY distance  LIMIT 0 , 10 ");

您可以使用表的别名（例如表cr.subject中的课程主题）从两个表中添加更多字段作为选择字段course_records。