为什么 json_encode/decode 会以这种方式配置我的数组？ PHP答案

【问题标题】：Why does json_encode/decode configure my array this way? PHP为什么 json_encode/decode 会以这种方式配置我的数组？ PHP
【发布时间】：2012-07-28 18:40:51
【问题描述】：

我正在尝试发送一个 JSON 文件来进行一些测试。我有一个创建多维数组的简单测试文件。这是测试文件：

<?php
error_reporting(E_ALL);
ini_set('display_errors', '1');

$user['Mongo'] = null; 
$user['Facebook'] = 12345;
$user['Twitter'] = null;
$user['Foursquare'] = null;
$user['Google'] = null;
$user['Name'] = "Bill Gates";
$user['Sex'] = 'M';
$user['Age'] = 26;
$user['Birthday'] = "1985-08-13";
$user['Friends'][0]['Mongo'] = null;
$user['Friends'][0]['Facebook'] = 123456;
$user['Friends'][0]['Twitter'] = null;
$user['Friends'][0]['Foursquare'] = null;
$user['Friends'][0]['Google'] = null;
$user['Friends'][0]['Name'] = "John Smith";
$user['Friends'][0]['Relationship'] = "Open";
$user['Friends'][1]['Mongo'] = null;
$user['Friends'][1]['Facebook'] = 1234567;
$user['Friends'][1]['Twitter'] = null;
$user['Friends'][1]['Foursquare'] = null;
$user['Friends'][1]['Google'] = null;
$user['Friends'][1]['Name'] = "Martina McBride";
$user['Friends'][1]['Relationship'] = "Open";

$user_json = json_encode($user);

$call = curl_init('http://MY_IP_HERE/user_login.php');

curl_setopt($call, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($call, CURLOPT_POSTFIELDS, $user_json);
curl_setopt($call, CURLOPT_RETURNTRANSFER, true);
curl_setopt($call, CURLOPT_HTTPHEADER, array('Content-Type: application/json', 'Content-Length: '.strlen($user_json)));

$result = curl_exec($call);
curl_close($call);

echo $result;
?>

我正在尝试像这样检索文件：

<?php
include_once('interaction_class.php');

error_reporting(E_ALL);
ini_set('display_errors', '1');

$fp = fopen('php://input', 'r');
$rawData = stream_get_contents($fp);

$user = json_decode($rawData);

if ($user['Mongo'] == null)
{
    $user_id = $interaction->new_guest($user);
}
//...Other stuff...

我在 if 语句行中收到有关 stdCLass 类对象的错误。所以，我做了一个 var_dump()，结果如下：

object(stdClass)#1 (10) 
{ 
    ["Mongo"] => NULL 
    ["Facebook"] => int(12345) 
    ["Twitter"] => NULL 
    ["Foursquare"] => NULL 
    ["Google"] => NULL 
    ["Name"] => string(15) "Bill Gates" 
    ["Sex"] => string(1) "M" 
    ["Age"] => int(26) 
    ["Birthday"] => string(10) "1985-08-13" 
    ["Friends"] => array(2) 
    { 
        [0] => object(stdClass)#2 (7) 
        { 
            ["Mongo"] => NULL 
            ["Facebook"] => int(123456)
            ["Twitter"] => NULL 
            ["Foursquare"] => NULL 
            ["Google"] => NULL 
            ["Name"] => string(10) "John Smith" 
            ["Relationship"] => string(4) "Open" 
        } 
        [1] => object(stdClass)#3 (7) 
        { 
            ["Mongo"] => NULL 
            ["Facebook"] => int(1234567) 
            ["Twitter"] => NULL 
            ["Foursquare"] => NULL 
            ["Google"] => NULL 
            ["Name"] => string(15) "Martina McBride" 
            ["Relationship"] => string(4) "Open" 
        } 
    } 
}

我的问题是，为什么我像这样执行 json_decode() 后无法访问信息：

$thing['Key']

为什么要把它解码成一个对象而不是一个数组？

提前感谢您的帮助！

【问题讨论】：

标签： php post json

【解决方案1】：

PHP函数json_decode()默认返回对象：http://www.php.net/manual/en/function.json-decode.php

如果您想使用数组，请将布尔值“true”添加为函数中的可选第二个参数，如下所示：

$user = json_decode($rawData, true);

这会将数组返回到变量$user

如果你想继续加载对象，你可以使用$user['Mongo']而不是$user->Mongo

希望这会有所帮助，祝你好运！

【讨论】：