Question

我有以下代码。我需要使用laravel相同的代码使用狂饮但不工作狂饮

$url = "https://xyz?"; 
$query ='first_name='. $data->FirstName .'&gender=""'. '&home_phone='. $data->HomePhone.'&ip_address='. $data->IPAddress.'&last_name='. $data->LastName.'&user_defined_url='. $result; 

$HttpSocket = new HttpSocket(array('ssl_verify_peer' => false, 'ssl_verify_host' => false)); 
$post_response = $HttpSocket->get($url,$query); 
$response = explode('&',$post_response->body); 

我已经转换它laravel。下面我的代码，我转换laravel：

$client   = new Client(['verify' => false ]); 
$post_response = $client->get($url, $query); 
$response  = explode('&',$post_response->body); 

注意：使用GuzzleHttp \客户端;在文件顶部被写入。

在此先感谢您的帮助！

Answer 1

你可以尝试这样的，

$client = new Client(); 
$post_response = $client->request('GET', $url, [ 
    'verify' => false, 
    'form_params' => [ 
     'first_name' => $data->FirstName, 
     'gender' => "", 
     'home_phone' => $data->HomePhone, 
     'ip_address' => $data->IPAddress, 
     'last_name' => $data->LastName, 
     'user_defined_url' => $result 
    ] 
]); 
$response = explode('&',$post_response->body); 

而不是form_params可以使用查询发送的参数作为查询字符串。