Question

在mysql数据库中我存储了5个表。我使用PHP来存储和检索数据。我想以链接的方式检索/回显/显示这5个表的信息。我在每个表上都有academy_id作为外键。每个学院都有一个联系人。有些学院可能会共用同一个联系人。但不确定如何以独特的信息展示每个学院。如何通过php / mysql select查询显示这些值？

Academy Name | MOU_ID | Academy_ID | STATUS | Academy Created | Course Name | Course Start Date | Instructor First Name | Contact First Name

表

CREATE TABLE IF NOT EXISTS `academies` (
  `id` int(11) NOT NULL auto_increment,
  `name` varchar(25) NOT NULL,
  `mou_id` int(11) default NULL,
  `academy_id` int(11) NOT NULL,
  `status` enum('INACTIVE','ACTIVE') NOT NULL default 'ACTIVE',
  `created_date` date NOT NULL
  PRIMARY KEY  (`id`)
);

INSERT INTO `courses` (`course_id`, `course_name`) VALUES
(1, 'MATH'),
(2, 'ENGLISH'),
(3, 'BIOLOGY'),

CREATE TABLE IF NOT EXISTS `academy_courses` (
  `unique_id` int(11) NOT NULL auto_increment,
  `academy_id` int(11) NOT NULL,
  `course_id` int(11) NOT NULL,
  `start_date` date default NULL,
  PRIMARY KEY  (`unique_id`),
  KEY `course_id` (`academy_id`,`course_id`)
);


CREATE TABLE IF NOT EXISTS `instructors` (
  `instructor_id` int(11) NOT NULL auto_increment,
  `academy_id` int(11) NOT NULL,
  `instructor_fname` varchar(50) NOT NULL
  PRIMARY KEY  (`instructor_id`),
  KEY `academy_id` (`academy_id`)
);


CREATE TABLE IF NOT EXISTS `main_contact` (
  `contact_id` int(11) NOT NULL auto_increment,
  `academy_id` int(11) NOT NULL,
  `contact_fname` varchar(50) NOT NULL,
  PRIMARY KEY  (`contact_id`),
  KEY `academy_id` (`academy_id`)
);

CREATE TABLE IF NOT EXISTS `main_contact_bridge` (
  `academy_id` int(11) NOT NULL,
  `contact_id` int(11) NOT NULL,
  PRIMARY KEY  (`contact_id`,`academy_id`),
  KEY `academy_id` (`academy_id`)
);

Answer 1

我认为这是你想要得到的：

SELECT a.name, a.mou_id, a.academy_id, a.status, a.created_date, ac.course_id, ac.start_date, i.instructor_fname, mc.contact_fname
FROM academy_courses ac 
LEFT JOIN academies a USING (academy_id) 
LEFT JOIN instructors i USING (academy_id) 
LEFT JOIN main_contact mc USING (academy_id);

Answer 2

我会将instructor_id添加到academy_courses中，否则如果您有多个教师，您将获得每个教师的所有重复数据。然后你会这样做：

SELECT a.name AS 'Academy Name', 
       a.mou_id AS 
       a.academy_id
       a.status
       a.created_date AS 'Academy Created',
       c.course_name AS 'Course Name',
       ac.start_date AS 'Course Start Date',
       i.instructor_fname AS 'Instructor First Name',
       co.contact_fname AS 'Contact First Name'
FROM academies a
INNER JOIN academy_courses ac ON a.id = ac.academy_id
INNER JOIN courses c ON c.id = ac.course_id
INNER JOIN instructors i ON i.instructor_id = ac.instructor_id
INNER JOIN main_contact co ON co.academy_id = a.id

如果您想从Academy获取信息，即使课程/ academy_courses / etc不存在，您也可以写一个LEFT JOIN。

Answer 3

您正在寻找的是加入表格。有不同类型的连接。请查看此解释http://www.codinghorror.com/blog/2007/10/a-visual-explanation-of-sql-joins.html

显示共享一个外键的多个表的值

3 个答案: