Question

如何使用php读取JSON数据响应？响应t来自第三方进行用户身份验证。首先，我只想要displayName和preferredUsername数据。

Json回复：

 {
      "stat": "ok",
      "profile": {
        "providerName": "testing",
        "identifier": "http://testing.com/58263223",
        "displayName": "testing",
        "preferredUsername": "testing",
        "name": {
          "formatted": "testing"
        },
        "url": "http://testing.com/testing/",
        "photo": "https://securecdn.testing.com/uploads/users/5826/3223/avatar32.jpg?1373393837",
        "providerSpecifier": "testing"
      }
    }

Answer 1

您可以使用json_decode（http://php.net/manual/en/function.json-decode.php）功能对结果进行解码，然后检索值：

$json_data = '{
      "stat": "ok",
      "profile": {
        "providerName": "testing",
        "identifier": "http://testing.com/58263223",
        "displayName": "testing",
        "preferredUsername": "testing",
        "name": {
          "formatted": "testing"
        },
        "url": "http://testing.com/testing/",
        "photo": "https://securecdn.testing.com/uploads/users/5826/3223/avatar32.jpg?1373393837",
        "providerSpecifier": "testing"
      }
    }';

$json = json_decode($json_data);

echo $json->profile->displayName;
echo $json->profile->preferredUsername;

Answer 2

 <?php
$json='{
      "stat": "ok",
      "profile": {
        "providerName": "testing",
        "identifier": "http://testing.com/58263223",
        "displayName": "testing",
        "preferredUsername": "testing",
        "name": {
          "formatted": "testing"
        },
        "url": "http://testing.com/testing/",
        "photo": "https://securecdn.testing.com/uploads/users/5826/3223/avatar32.jpg?1373393837",
        "providerSpecifier": "testing"
      }
    }';


    $data=json_decode($json ,true);
     $preferredUsername=$data['profile']['preferredUsername'];
     $displayName=$data['profile']['displayName'];

    ?>

Answer 3

json_decode正是您所寻找的：

$json = '[
    {
        "displayName": "testing",
        "preferredUsername": "testing",
    }
]';

$jsonArray = json_decode($json);

foreach($jsonArray as $value){
    $displayName = $value->Display Name;
    $preferredUsername = $value->Preferred User;
}

使用php读取Json数据响应

3 个答案: