如何从多个 MySQL 表中生成有效的 json 输出

Question

我在本地服务器中使用 phpMyAdmin 创建了一个 mySQL 数据库。在这个数据库中，我存储了我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，我创建了三个表：一个包含我朋友的 id 和名称，一个包含团队的 id 和名称，另一个包含 friends_id 和 teams_id（这是关系表）。下面的 MySQL 脚本更清楚地表明了这一点：

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,
  `name` varchar(30) NOT NULL,
  PRIMARY KEY (`id`)
)

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

我想用这些数据给出一个 json 输出，因此我运行以下 PHP 脚本：

<?php

$dbServername = 'localhost';
$dbUsername = 'root';
$dbPassword = '';
$dbName = 'Friends';

$conn = mysqli_connect($dbServername, $dbUsername, $dbPassword, $dbName);

header('Content-Type: application/json');

$sql = 'SELECT * FROM friends;';
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
$arr = [];
if ($resultCheck > 0) {
   while ($row = mysqli_fetch_assoc($result)) {
        $arr[] = $row; 
   }
} 
echo json_encode($arr, JSON_PRETTY_PRINT);

$sql = 'SELECT * FROM teams;';
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
$arr = [];
if ($resultCheck > 0) {
   while ($row = mysqli_fetch_assoc($result)) {
        $arr[] = $row; 
   }
} 
echo json_encode($arr, JSON_PRETTY_PRINT);

$sql = 'SELECT * FROM relations;';
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
$arr = [];
if ($resultCheck > 0) {
   while ($row = mysqli_fetch_assoc($result)) {
        $arr[] = $row; 
   }
} 
echo json_encode($arr, JSON_PRETTY_PRINT);
?>

然而，这并没有给出一个有效的 json 输出，而是三个不同的 json 数组。例如，像这样：

[..., 
{
        "id": "3",
        "name": "David Belton",
    },
...]

[..., 
{
        "id": "4",
        "name": "Boston Celtics",
    },
...]

[..., 
{
        "friends_id": "3",
        "teams_id": "4"
    },
...]

如何打印所有具有有效 json 输出的表？

Answer 1

因为这个问题与How to join arrays with MySQL from 3 tables of many-to-many relationship 有关，所以我只发布了一个 MySQL 答案。

查询

SELECT
  CONCAT(
      "["
    , GROUP_CONCAT(json_records.json)
    , "]"
  )  AS json
FROM (

  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 IN(SELECT id FROM friends) #select the friends you need
  GROUP BY
     friends.id
) 
 AS json_records

结果

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

演示

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

Answer 2

您不需要编写多个 sql 查询来检索它。您可以执行以下操作

<?php
$dbServername = 'localhost';
$dbUsername = 'root';
$dbPassword = '';
$dbName = 'Friends';

$conn = mysqli_connect($dbServername, $dbUsername, $dbPassword, $dbName);

header('Content-Type: application/json');

$sql = 'SELECT * FROM friends INNER JOIN relations ON friends.id=relations.friends_id
INNER JOIN teams ON relations.teams_id=teams.id';
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
$arr = [];
if ($resultCheck > 0) {
   while ($row = mysqli_fetch_assoc($result)) {
        $arr[] = $row; 
   }
} 
echo json_encode($arr, JSON_PRETTY_PRINT);
?>