从json_decode提取数据:json_decode不工作

Question

我有以下代码：

<?php 

$consumerKey = '';
$consumerSecret = '';
$url = '';

$data = array(
'grant_type'  => 'password',
    'username'    => '',
    'password'    => ''
);


$curl = curl_init($url);
curl_setopt($curl, CURLOPT_HTTPHEADER, array('Accept: application/json','Accept-Language: en_US'));
curl_setopt($curl, CURLOPT_USERPWD, $consumerKey.':'.$consumerSecret);
curl_setopt($curl, CURLOPT_POSTFIELDS, http_build_query($data));
    $result = curl_exec($curl);
    $status = curl_getinfo($curl, CURLINFO_HTTP_CODE);
    $result = json_decode($result);
    curl_close($curl);
?>

它返回json结果之后的示例，但不返回json_decode的访问令牌。curl_setopt($curl，CURLOPT_RETURNTRANSFER，1)；不返回下面的令牌。谢谢你对我的照顾。

{"access_token":"ffdd8dfb-2013-32ee-bc3e-dc5689d6c8fb","refresh_token":"7bf1ddad-d696-3d83-a524-37dac002164a","scope":"default","token_type":"Bearer","expires_in":3600}

Answer 1

在你的回答- @Andreas和@YvesLeBorg的鼓舞下，我重做了代码，得到了我想要的东西。我把这个用来帮助别人。谢谢。

<?php 

$consumerKey = '';
$consumerSecret = '';
$url = '';
$curl = curl_init($url);
$data = array(
    'grant_type'  => 'password',
    'username'    => '',
    'password'    => ''
);
$curl = curl_init($url);
curl_setopt($curl, CURLOPT_HTTPHEADER, array('Accept: application/json','Accept-Language: en_US'));
curl_setopt($curl, CURLOPT_USERPWD, $consumerKey.':'.$consumerSecret);
curl_setopt($curl, CURLOPT_POSTFIELDS, http_build_query($data));
curl_setopt($curl,CURLOPT_RETURNTRANSFER,TRUE);
    $result = curl_exec($curl);
    $status = curl_getinfo($curl, CURLINFO_HTTP_CODE);
    $result = json_decode($result);
    $access_token = $result->access_token;
echo $access_token;
    curl_close($curl);