试图回显 json 数据的 php 代码

Question

我有这个输出，我只是想回显 token 或将 token 存储在变量中

object(stdClass)#1 (9) {
 ["loggedIn"]=> bool(true) 
["token"]=> string(32) "xe5b6bHPPo-_nAvB78NfTpopq9e44bYb" 
["secret"]=> string(20) "icCPoDvcf1Vvm2i7-6GM" 
["expirationTimestamp"]=> string(24) "2016-11-26T16:32:29.728Z" 
["type"]=> string(4) "USER" 
["data"]=> object(stdClass)#2 (17)}

这是我的代码:

<?php
$ch = curl_init();

curl_setopt($ch, CURLOPT_URL, "https://security.voluum.com/login");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "GET");


$headers = array();
$headers[] = "Authorization: Basic plFubTHhcl93MzA4QHJJkQ9vqmNvwTpMNYYUIz5kcm9s";
curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);

$result = curl_exec($ch);
if (curl_errno($ch)) {
    echo 'Error:' . curl_error($ch);
}

var_dump(json_decode($result));
var_dump(json_decode($result, true));

echo $result->token;
curl_close ($ch);

?>

上面的代码是否正确？如何从输出中提取单个数据？请帮助我

Answer 1

$res_decoded = json_decode($result);
echo $res_decoded->token;