从URL获取JSON对象

Question

我有一个返回JSON对象的URL，如下所示：

{
    "expires_in":5180976,
    "access_token":"AQXzQgKTpTSjs-qiBh30aMgm3_Kb53oIf-VA733BpAogVE5jpz3jujU65WJ1XXSvVm1xr2LslGLLCWTNV5Kd_8J1YUx26axkt1E-vsOdvUAgMFH1VJwtclAXdaxRxk5UtmCWeISB6rx6NtvDt7yohnaarpBJjHWMsWYtpNn6nD87n0syud0"
} 

我想从URL获取JSON对象，然后获取access_token值。

那么如何通过PHP检索它呢？

Answer 1

您可以使用PHP的json_解码函数：

$url = "http://urlToYourJsonFile.com";
$json = file_get_contents($url);
$json_data = json_decode($json, true);
echo "My token: ". $json_data["access_token"];

Answer 2

$url = 'http://.../.../yoururl/...';
$obj = json_decode(file_get_contents($url), true);
echo $obj['access_token'];

Php还可以将属性与破折号一起使用：

garex@ustimenko ~/src/ekapusta/deploy $ psysh
Psy Shell v0.4.4 (PHP 5.5.3-1ubuntu2.6 — cli) by Justin Hileman
>>> $q = new stdClass;
=> <stdClass #000000005f2b81c80000000076756fef> {}
>>> $q->{'qwert-y'} = 123
=> 123
>>> var_dump($q);
class stdClass#174 (1) {
  public $qwert-y =>
  int(123)
}
=> null

Answer 3

$json = file_get_contents('url_here');
$obj = json_decode($json);
echo $obj->access_token;

要做到这一点，file_get_contents需要启用allow_url_fopen。这可以在运行时通过包括：

ini_set("allow_url_fopen", 1);

您还可以使用curl获取url。要使用curl，可以使用以下示例：

$ch = curl_init();
// IMPORTANT: the below line is a security risk, read https://paragonie.com/blog/2017/10/certainty-automated-cacert-pem-management-for-php-software
// in most cases, you should set it to true
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_URL, 'url_here');
$result = curl_exec($ch);
curl_close($ch);

$obj = json_decode($result);
echo $obj->access_token;