php – 如何使用3个多对多关系表从MySQL连接数组

我在本地服务器上用PHPMyAdmin创建了一个MysqL 数据库.在这个数据库中,我存储了我的朋友的名字和最喜欢的NBA球队.这显然是一个多对多的关系.出于这个原因,我在 MySQL中运行以下脚本来为该数据库创建适当的表：

CREATE TABLE `friends` (
  `id` int(4) NOT NULL AUTO_INCREMENT,`name` varchar(30) NOT NULL,PRIMARY KEY (`id`)
)

CREATE TABLE `teams` (
  `id` int(4) NOT NULL AUTO_INCREMENT,PRIMARY KEY (`id`)
)

CREATE TABLE `relations` (
  `friends_id` int(4) NOT NULL,`teams_id` int(4) NOT NULL,)

显然,我在这些表中插入了一些值,但是我没有在这里提供大量的源代码以节省一些空间.其中一小部分如下：

INSERT INTO `friends` (`id`,`name`)
VALUES
    (1,'David Belton'),(2,'Alex James');

INSERT INTO `teams` (`id`,'Cleveland Cavaliers'),'Boston Celtics');

INSERT INTO `relations` (`friends_id`,`teams_id`)
VALUES
    (1,1),2);

在运行从数据库中获取数据并打印它们的PHP脚本之后,我希望为每个朋友提供以下类型的有效json输出：

{
    "id": "1","name": "Alex James","team": ["Boston Celtics","Cleveland Cavaliers"] 
}

如何为每个拥有MysqL的人制作这个最喜欢的球队？

附：
我预先假设在使用PHP检索数据之前,最好在MysqL中完成.

“eazy”方法是使用CONCAT生成JSON.
并使用GROUP_CONCAT将多个团队记录合并为一个JSON数组.
此方法也适用于不支持创建JSON函数的旧MysqL版本.

询问

SELECT 
 CONCAT(
     "{",'"id"',":",'"',friends.id,",'"name"',friends.name,'"team"',"[",GROUP_CONCAT('"',teams.name,'"'),"]","}"
   ) AS json
FROM 
 friends 
INNER JOIN 
 relations 
ON 
 friends.id = relations.friends_id
INNER JOIN
 teams 
ON
 relations.teams_id = teams.id
WHERE 
 friends.id = 1

结果

|                                                            json |
|-----------------------------------------------------------------|
| {"id":"1","name":"David Belton","team":["Cleveland Cavaliers"]} |

演示

http://www.sqlfiddle.com/#!9/4cd244/19

编辑了更多的朋友

询问

SELECT
  CONCAT(
      "[",GROUP_CONCAT(json_records.json) # combine json records into a string,"]"
  )  AS json
FROM (

  SELECT 
     CONCAT(
       "{","}"
     ) AS json 
  FROM 
    friends 
  INNER JOIN 
    relations 
  ON 
    friends.id = relations.friends_id
  INNER JOIN
    teams 
  ON
    relations.teams_id = teams.id
  WHERE 
    friends.id IN(SELECT id FROM friends) #select the friends you need or just simply friends.id IN(1,2)
  GROUP BY
     friends.id
) 
 AS json_records

结果

|                                                                                                                                             json |
|--------------------------------------------------------------------------------------------------------------------------------------------------|
| [{"id":"1","team":["Cleveland Cavaliers"]},{"id":"2","name":"Alex James","team":["Boston Celtics","Cleveland Cavaliers"]}] |

演示

http://www.sqlfiddle.com/#!9/4cd244/61

原文链接：https://www.f2er.com/php/138922.html

php – 如何使用3个多对多关系表从MySQL连接数组

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