从URL获取JSON对象 [英] Get JSON object from URL

问题描述

我有一个返回如下JSON对象的URL:

I have a URL that returns a JSON object like this:

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

我想获取access_token值.那么如何通过PHP检索它呢?

I want to get the access_token value. So how can I retrieve it through PHP?

推荐答案

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

为此，file_get_contents要求启用allow_url_fopen.可以在运行时完成以下操作:

For this to work, file_get_contents requires that allow_url_fopen is enabled. This can be done at runtime by including:

ini_set("allow_url_fopen", 1);

您还可以使用 curl 来获取网址.要使用curl，您可以使用在此处找到的示例:

You can also use curl to get the url. To use curl, you can use the example found here:

$ch = curl_init();
curl_setopt($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;

这篇关于从URL获取JSON对象的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！