php - 如何将 JSON 字符串转换为 PHP 变量？

Question

我试图使用 cURL 从 url 解析 json 数据，然后使用 json_decode 来访问对象，但失败了。我已经搜索过如何访问但失败了。

这是我访问的链接，希望能够解决问题。

http://www.dyn-web.com/tutorials/php-js/json/decode.php
https://stackoverflow.com/questions/12429029/php-get-values-from-json-encode
https://stackoverflow.com/questions/20193575/fetch-json-data-using-php
https://stackoverflow.com/questions/12429029/php-get-values-from-json-encode
https://stackoverflow.com/questions/4433951/decoding-json-after-sending-using-php-curl

这是 var_dump($obj) 的结果；

array(1) {
  ["login"]=>
  array(2) {
    ["error"]=>
    bool(false)
    ["user"]=>
    array(5) {
      ["br_code"]=>
      int(0)
      ["mem_id"]=>
      int(202)
      ["username"]=>
      string(8) "johndoe"
      ["email"]=>
      string(33) "<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="fd97929593999298bd9a909c9491d39e9290" rel="noreferrer noopener nofollow">[email protected]</a>"
      ["created_at"]=>
      string(19) "2017-08-07 15:35:39"
    }
  }
}

这是我的 PHP 代码

<?php
session_start();

$ch = curl_init('http://localhost/sample/login.php');

$username= $_SESSION["USERNAME"];
$password= $_SESSION["PASSWORD"];

$credentials = [
    'username' => $username,
    'password' => $password
];

curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $credentials);
curl_setopt($ch, CURLOPT_USERNAME, "{$username}:{$password}");
// execute!
$response = curl_exec($ch);
// close the connection
curl_close($ch);

$obj= json_decode($response, true);

// echo $obj; // Notice: Array to string conversion in

// echo $obj[0]; //  Undefined offset

// echo $obj->user; //Trying to get property of non-object in

var_dump($obj);



?>

Answer 1

看起来是一个数组的数组

例如:用户名

var_dump($obj['login']['user']['username']);

其他值依此类推

$my_username =  $obj['login']['user']['username']);