dongzuan4491 2017-10-31 00:00
浏览 109
已采纳

如何使用移动检测api(useragentinfo.co)

i am getting too much information from api, i can't use this, I want only "device_model" and "device_type" from this result, and how i can do it? What i am getting this result:

{"device_model":"Emulator","os_version":"7","browser":"Chrome","browser_version":"60","os":"Windows","device_type":"Desktop","useragent":"Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36","device_brand":"Unknown","is_bot":false}

Mobile detect api

<?php
$useragent = $_SERVER['HTTP_USER_AGENT'];
// get api token at https://useragentinfo.co/
$token = "#API_TOKEN";
$url = "https://useragentinfo.co/api/v1/device/";

$data = array('useragent' => $useragent);

$headers = array();
$headers[] = "Content-type: application/json";
$headers[] = "Authorization: Token " . $token;

$curl = curl_init($url);
curl_setopt($curl, CURLOPT_HEADER, false);
curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
curl_setopt($curl, CURLOPT_HTTPHEADER, $headers);
curl_setopt($curl, CURLOPT_POST, true);
curl_setopt($curl, CURLOPT_POSTFIELDS, json_encode($data));

$json_response = curl_exec($curl);

$status = curl_getinfo($curl, CURLINFO_HTTP_CODE);

if ($status != 200 ) {
    die("Error: call to URL $url failed with status $status, response $json_response, curl_error " . curl_error($curl) . ", curl_errno " . curl_errno($curl));
}

curl_close($curl);

echo $json_response;

?>

</div>
  • 写回答

1条回答

  • duanchai0028 2017-10-31 00:07
    关注

    If you are getting a standard json response from the api, you can just json_decode() it and use it like this:

    <?php
    
    // Do your curl request and get json_response here...
    
    // Decode json response
    $result = json_decode($json_response);
    
    // All of the fields are then accessed like this:
    // $result->device_model (for e.g.)
    
    ?>
    
    <b>Device Model</b>: <?php echo $result->device_model ?>
    
    <br>
    
    <b>Device Type</b>: <?php echo $result->device_type ?>
    etc...
    

    If you prefer to work with array instead, you can do this:

    // Decode json response
    $result = json_decode($json_responsem true);
    
    // Then use it like this:
    // $result['device_model'] etc...
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 执行 virtuoso 命令后,界面没有,cadence 启动不起来
  • ¥50 comfyui下连接animatediff节点生成视频质量非常差的原因
  • ¥20 有关区间dp的问题求解
  • ¥15 多电路系统共用电源的串扰问题
  • ¥15 slam rangenet++配置
  • ¥15 有没有研究水声通信方面的帮我改俩matlab代码
  • ¥15 ubuntu子系统密码忘记
  • ¥15 保护模式-系统加载-段寄存器
  • ¥15 电脑桌面设定一个区域禁止鼠标操作
  • ¥15 求NPF226060磁芯的详细资料