php从url输出josn 从URL获取JSON对象

您需要阅读有关json_decode函数的信息/manual/zh/function.json-decode.php

干得好

$json = '{"expires_in":5180976,"access_token":"AQXzQgKTpTSjs-qiBh30aMgm3_Kb53oIf-VA733BpAogVE5jpz3jujU65WJ1XXSvVm1xr2LslGLLCWTNV5Kd_8J1YUx26axkt1E-vsOdvUAgMFH1VJwtclAXdaxRxk5UtmCWeISB6rx6NtvDt7yohnaarpBJjHWMsWYtpNn6nD87n0syud0"}';

//OR $json = file_get_contents('http://someurl.dev/...');

$obj = json_decode($json);

var_dump($obj-> access_token);

//OR

$arr = json_decode($json, true);

var_dump($arr['access_token']);