Question

我在CakePHP中有以下代码。我需要使用guzzle

在laravel中使用相同的代码

$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);

我已经使用guzzle将它转换为laravel但是没有用。按照我在laravel中转换的代码：

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

注意：使用GuzzleHttp \ Client;写在文件的顶部。

提前感谢您的帮助！

Answer 1

你能尝试这样吗，

$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);

您可以使用查询代替 form_params 将参数作为查询字符串发送。

用于ssl的laravel guzzle httpsocket参数验证peer＆amp;主办

1 个答案: