Question

我正在尝试使用PHP从MySQL数据库生成以下JSON。我如何在PHP中进行操作让我们说使用explode函数将数组放入JSON对象中。我不知道如何在对象中获取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对象中的数组：

[
{
  "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();
?>

使用php在json对象内部的数组和json对象内的对象

1 个答案: