php - 使用 php 的 json 对象内的数组和 json 对象内的对象

Question

我正在尝试使用 PHP 从 MySQL 数据库生成以下 JSON。我该如何在 PHP 中执行此操作，比如说使用爆炸函数将数组放入 JSON 对象中。我不知道如何在对象内获取 JSON 对象。我只需要分离 PHP 文件即可实现以下目的。

<?php
include 'database.php';
$pdo = Database::connect();
$sql = 'SELECT * FROM users';
$q = $pdo->prepare($sql);
$q->execute(array($sql));
$array = array();
 while ($row = $q->fetch(PDO::FETCH_ASSOC)){
     array_push($array, $row);
 }
$json = json_encode($array);
echo $json;
Database::disconnect();?>

JSON 对象内的数组:

[
{
  "firstName":"John", 
  "lastName":"Doe",
  "images": ['image1','image2','image3']
}, 
{
  "firstName":"Anna",   
  "lastName":"Smith",
  "images": ['image1','image2','image3']
},
{
 "firstName":"Peter", 
 "lastName":"Jones",
 "images": ['image1','image2','image3']
 }
]

对象内的 JSON 对象:

[
{
  "firstName":"John", 
  "lastName":"Doe",
  "cover": {
            "cover_id": "0858699703",
            "source": "www.myimages.co.zw/images/photo",
            "offset_y": "0"
          }
}, 
{
  "firstName":"Anna",   
  "lastName":"Smith"
  "cover": {
            "cover_id": "0858699703",
            "source": "www.myimages.co.zw/images/photo",
            "offset_y": "0"
          }
},
{
 "firstName":"Peter", 
 "lastName":"Jones"
 "cover": {
            "cover_id": "0858699703",
            "source": "www.myimages.co.zw/images/photo",
            "offset_y": "0"
          }
 }
]

Answer 1

 <?php
include 'database.php';
$pdo = Database::connect();
$sql = 'SELECT * FROM test';
$q = $pdo->prepare($sql);
$q->execute(array($sql));
$array = array();
 while ($row = $q->fetch(PDO::FETCH_ASSOC)){

     $row_array['name'] = $row['name'];
     $row_array['surname'] = $row['surname'];
     $row_array['images'] =  explode(" ", $row['images']);

     array_push($array, $row_array);
 }
$json = json_encode($array);
echo $json;
Database::disconnect();
?>